Big Idea Math

Big Ideas Math Integrated Math  Chapter 3 Maintaining Mathematical Proficiency

Page 58   Exercise 1  Problem 1

Question 1.

Plot the point on a coordinate plane that is on the y-axis and 5 units down from the origin.

Answer:

To find- plotting of a point that is on the y− axis and 5 units down from the origin.

Using property of coordinate plane we will plot required point.

∵  Required point is on the y− axis and 5 units down the origin.

Here point O is the origin and point A is our required point having coordinate b (0,−5).

Read and Learn More Big Ideas Math Integrated Math 1 Student Journal Solutions

The final answer is that the plotting of the point which is on the y− axis and 5 units down the origin is

Big Ideas Math Integrated Math 1 Student Journal Chapter 3 Solutions Page 58  Exercise 2  Problem 2

Question 2.

Evaluate the expression 16 – 4x at x = -4.

Answer:

Page 58  Exercise 3  Problem 3

Question 3.

Evaluate the expression 12x + 7 at x = -2.

Answer:

Given: 12 x + 7; x = −2

To find- value of the expression at given value.

We will put the given value of x in the given expression to get the required answer.

Put x = −2 in the given expression, we get

= 12(−2) + 7

=− 24 + 7

= − 17.

12 x + 7 = − 17.

The final answer is that the value of the expression 12x + 7 at x = −2 is−17.

Page 58  Exercise 4  Problem 4

Question 4.

Evaluate the expression -9- 3x at x = 5.

Answer:

Given: −9−3x ; x = 5

To find- value of the expression at given value.

We will put the given value of x in the given expression to get the required answer.

Put x = 5in the given expression, and we get

= −9−3(5)

=−9−15

= −24.

−9−3x = −24.

The final answer is that the value of the expression −9 − 3x at x = 5 is−24.

Chapter 3 Maintaining Mathematical Proficiency Solutions Big Ideas Math Integrated Math 1 Page 59   Exercise 5  Problem 5

Question 5.

Determine whether the given relation represents a function.

Answer:

Given: An input-output table

To find- Whether the given relation represents a function.

From the given input-output table, we can see that all the values of x is related with element 8 of y

⇒  Every element of x is related with exactly one element of y.

⇒ The given relation is a function.

The final answer is that the given relation

Is a function.

Page 59  Exercise 5  Problem 6

Question 6.

Given the input-output table below:

Determine whether the given relation represents a function.

Answer:

Given:  An input-output table

To find- Whether the given relation represents a function.

From the given input-output table, we can see that a single element of x i.e. 8 is related with more than one element of y i.e. 0,1,2,3 and 4.

⇒  Element of x is related with multiple element of y.

⇒ The given relation is not a function.

The final answer is that the given relation

Is not a function.

Big Ideas Math Integrated Math 1 Chapter 3 Step-By-Step Solutions Page 59   Exercise 5  Problem 7

Question 7.

Given the input-output table below:

Determine whether the given relation represents a function.

Answer:

Given:  An input-output table

To find- whether the given relation represents a function.

From the given input-output table, we can see that there is an element of x i.e.3 which is related to two elements of y i.e.10and 11.

⇒ There is an element in x which is related to multiple elements in y.

⇒ The given relation is not a function.

The final answer is that the

Is not a function.

Page 59  Exercise 5  Problem 8

Question 8.

Analyze the given graph to determine whether the relation it represents is a function:

Determine whether the given relation represents a function.

Answer:

Given: A graph representing a relation

To find- whether the given relation represents a function.

From the given graph, we can easily see that there is an element of x i.e.5 which is related to two elements of y i.e.4 and 5.

⇒ There is an element in xwhich is related to multiple elements in y.

⇒ The given relation is not a function.

The final answer is that the given relation

Is not a function.

Page 59  Exercise 5  Problem 9

Question 9.

Given the set of ordered pairs (-2,5), (-1,8), (0,6), (1,6), and (2,7), determine whether the relation represents a function.

Answer:

Given: (−2,5),(−1,8),(0,6),(1,6),(2,7)

We have to determine whether each given relation represents a function.

Since x− coordinates represent input and y−Coordinates represent the output of the relation.

Arrange the given relation

From above we can see that each input has exactly one output.

Hence, the given relation represents a function.

The given relation (−2,5),(−1,8),(0,6),(1,6),(2,7) represents a function. Since each input has unique output.

Page 59  Exercise 5  Problem 10

Question 10.

Given the set of ordered pairs (-2,0), (-1,0), (-1,1), (0,1), (1,2), and (2,2), determine whether the relation represents a function.

Answer:

Given: (−2,0),(−1,0),(−1,1),(0,1),(1,2),(2,2)

We have to determine whether each given relation represents a function.

Since x-coordinates represent input and y-coordinates represent the output of the relation.

(−2,0),(−1,0),(−1,1),(0,1),(1,2),(2,2)  [Given relation]

In we can see that in the above-given relation input −1 has two outputs which are 0 and 1.

But a relation to be a function, each input has exactly one output.

which does not satisfy here in the given relation.

Hence, the given relation does not represent a function.

The given relation (−2,0),(−1,0),(−1,1),(0,1),(1,2),(2,2) does not represent a function because in the given relation input −1 have two output 0 and 1.

Solutions For Big Ideas Math Integrated Math 1 Chapter 3 Maintaining Mathematical Proficiency Exercises Page 60  Exercise 6   Problem 11

Question 11.

Determine whether each of the following relations represents a function. Explain your reasoning.

1. Relation 1: (2,6), (2,7), (1,8), (3,6)
2. Relation 2: (1,3), (2,5), (4,7)

Answer:

Function:

It is a rule in which for an input we get an output.

An important condition for the function is that for an input, x there should not be more than one different output.

Example of a function:  y = x³

Example of not a function: (2,6)(2,7),(1,8),(3,6)

A relation is a function where each input has exactly one output. Example of a function:(1,3),(2,5),(4,7)

Page 60   Exercise 6   Problem 12

Question 12.

Determine whether each of the following relations represents a function. Explain your reasoning.

1. Given the equation, y = 2x³, determine if it represents a function.
2. Given the set of ordered pairs (2,6), (2,7), (1,8), and (3,6), determine if it represents a function.

Answer:

Function:

It is a rule in which for an input we get an output.

An important condition for the function is that for an input, x there should not be more than one different output,y.

Example of a function: y = 2x³

Example of not a function: (2,6)(2,7),(1,8),(3,6)

A relation is a function where each input has exactly one output. Example of not a function: (2,6)(2,7),(1,8),(3,6)

Page 63  Exercise 7  Problem 13

Question 13.

Given the graph below, determine whether the relation it represents is a function. Use the vertical line test to justify your answer.

Answer:

Given: The graph

We have to find that the given graph represents a function or not.

Here, we use the vertical line test to check the given graph is a function or non-function.

From the graph, we can see that the input  x = 2 has two different outputs  y = 2 and y = 3

Also, by using the vertical line test draw a vertical line from x = 2 in the given graph

Here, the line pass through two points i.e y = 2 and y = 3.

Which does not satisfy the condition of function.

Hence, the given graph is not a function.

The given graph does not represent a function since it does not satisfy the condition of a function because when a vertical line is drawn it will pass through more than one point i.e. two points which can be shown in the below graph.

Big Ideas Math Integrated Math 1 Student Journal Chapter 3 Guide Page 63  Exercise 8   Problem 14

Question 14.

Given the graph below, determine whether the relation it represents is a function. Use the vertical line test to justify your answer.

Answer:

Given: The graph

We have to find that the given graph represents a function or not.

Here, we use the vertical line test to check the given graph is a function or non-function

Applying vertical line test in the given graph

We can see in the above graph that when we draw a vertical line from any point it will not pass through more than one point.

Hence, the given graph represents a function.

When we apply the vertical line test we get that it will not pass through more than one point. Hence, the given graph represents a function

Page 63  Exercise 9   Problem 15

Question 15.

Given the graph below, determine the domain and range of the function represented by the graph.

Answer:

Given: The graph

To find – The domain and range of the function represented by the graph.

First, from the graph determine the coordinates points.

x− coordinates represent the domain.

y−coordinates represent the range.

Find the coordinates from the graph

The coordinate points are (0,4),(1,4),(2,4),(3,3),(4,3),(5,3)

Find the domain and range

Since, x− coordinates represent the domain.

Domain: D→{0,1,2,3,4,5}

And, y− coordinates represent the range.

Range: R → {4,3}

Hence, the domain D  → {0,1,2,3,4,5} and Range R → {4,3} from the given graph.

Chapter 3 Maintaining Mathematical Proficiency Explained Big Ideas Math Integrated Math 1 Page 63  Exercise 10  Problem 16

Question 16.

Given the graph below, determine the domain and range of the function represented by the graph.

Answer:

Given: The graph

To find – The domain and range of the function represented by the graph.

First, from the graph determine the coordinates points.

x−coordinates represent the domain.

y−coordinates represent the range.

The given is

From the graph, we can say that it is a continuous graph.

So, since x− coordinates represent the domain.

Domain:  D→[−3,3]

And, y−coordinates represent the range.

Range: R→[0,3]

Hence, the domain and range of the given graph are Domain D→[−3,3],  and  Range R→[0,3]

Page 63  Exercise 11  Problem 17

Question 17.

Given the function, y = 12x, where y represents the number of pages printed and x represents the time taken to print in minutes, identify the independent and dependent variables.

Answer:

Given: y = 12x

To find –  Independent and dependent variables from the given function.

Here, y is the number of pages and x is the time taken to print in minutes.

y = 12x  [Given function]

Since y is the number of pages and x is the time taken to print in minutes.

Here, the independent variable is ‘x ‘ time taken to print in minutes.

And, the dependent variable is ′y′ the number of pages printed.

The independent variable is ′x′and the dependent variable is ′y′in the given function y = 12x.

Page 63  Exercise 12  Problem 18

Question 18.

Given the function, y = 12x with a domain of {1,2,3,4}, find the range of the function.

Answer:

Given: The function y = 12x

The domain of the given function is 1,2,3 and 4.

We have to find the range of the given function.

By putting different values of x = 1,2,3,4 find the value of y.

Here, x represents domain and y represents the range.

Put the value of x = 1,2,3,4 in the given function y=12x

From the above table, we get

Range = 12,24,36,48

Hence, the range of the given function y = 12x  whose domain is 1,2,3 and 4is 12,24,36,48 .

Page 64   Exercise 13  Problem 19

Question 19.

Given the length l = 2x and the breadth b = x of a rectangle, where x = 1, 2, 3, 4, 5:

1. Complete the table for the perimeter P of the rectangle using the formula P = 2 (l+b).
2. Plot the points on a graph where the x-axis represents x and the y-axis represents P.
3. Determine whether the pattern is linear or non-linear.

Answer:

Given:

Length of the rectangle, l=2x

The breadth of the rectangle, b=x

x = 1,2,3,4,5

We have to complete the given table

And, draw the graph and find the pattern whether it is linear or non-linear.

Using the formula of the perimeter of the rectangle complete the given table.

Then put the points on the graph and find the pattern.

Perimeter of rectangle, P = 2(l + b)

Put l = 2x and b = x in the above formula

P = 2(2x + x)

P = 2(3x)

P = 6x

Now, put x = 1,2,3,4,5 in above equation to find the value of P

For x = 1

P=6(1)

⇒  P = 6

For x=2

P = 6(2)

P = 12

For x = 3

P = 6(3)

P = 18

For x = 4

P = 6(4)

P = 24

For x = 5

P = 6(5)

P = 30

Make the table

Now, mark the points on the graph x-axis represents x.

And, the y−axis represents P.

Hence, the graph is in a linear pattern.

Hence, the table is

And it makes a linear pattern shown in the below graph.

Worked Examples For Big Ideas Math Integrated Math 1 Chapter 3 Maintaining Mathematical Proficiency Page 64   Exercise 13  Problem 20

Question 20.

Given the length l = 2x and the breadth b = x of a rectangle, where x = 1,2,3,4,5:

1. Complete the table for the area A of the rectangle using the formula A = l x b.
2. Plot the points on a graph where the x-axis represents x and the y-axis represents A.
3. Determine whether the pattern is linear or non-linear.

Answer:

Given:

Length of the rectangle, l = 2x

The breadth of the rectangle, b = x

x = 1,2,3,4,5

We have to complete the given table

And, draw the graph and find the pattern whether it is linear or non-linear.

Using the formula of the area of the rectangle complete the given table.

Then put the points on the graph and find the pattern.

Area of the rectangle, A = l × b

Put l = 2x and b = x in the above formula

A = 2x × x

A= 2 x²

Now, put x = 1,2,3,4,5 in the above equation

For x = 1

A = 2(1)²

A = 2 × 1×1

A = 2

For x = 2

A = 2(2)²

A = 2 × 2 × 2

A = 8

For x = 3

A = 2(3)²

A = 2 × 3 × 3

A = 18

For x = 4

A = 2(4)²

A = 2 × 4 ×4

A = 32

For x = 5

A=2(5)²

A = 2 × 5 × 5

A = 50

The table is

Now, mark the points on the graph

The x-axis represents the value of x.

And, y-axis represents the value of A.

Since the above graph makes a parabola and a parabola is a non-linear graph.

Hence, the complete table is

The graph is

Since the above graph makes a parabola and parabola is a non-linear graph.

Page 64  Exercise 13  Problem 21

Question 21.

1. Given the radius r = 1,2,3,4,5 of a circle, where C is the circumference of the circle:
2. Complete the table for the circumference C of the circle using the formula C = 2πr.
3. Plot the points on a graph where the x-axis represents r and the y-axis represents C.
4. Determine whether the pattern is linear or non-linear.

Answer:

Given:

The radius of circle, r = 1,2,3,4,5C is the circumferences of the circle.

We have to complete the given table

And, draw the graph and find the pattern whether it is linear or non-linear.

Using the formula of the circumferences of the circle to complete the given table.

Then put the points on the graph and find the pattern.

Circumferences of a circle, C = 2πr

Put r = 1,2,3,4,5 in the above formula

For r = 1

C = 2×\(\frac{22}{7}\)×1

C = 6.28

For r = 2

C = 2×\(\frac{22}{7}\)×2

C = 12.57

For r = 3

C = 2×\(\frac{22}{7}\)×3

C = 18.85

For  r= 4

C =  2×\(\frac{22}{7}\)×4

C = 25.14

For r = 5

C = 2×\(\frac{22}{7}\)×5

C = 31.42

Make the table

Now, plot the points on the graph

The x-axis represents the radius of the circle, r

The y−axis represents the circumferences of the circle, C

Hence, the graph makes a linear pattern.

Hence, the table is

And, the graph makes a linear pattern which can be shown below.

Big Ideas Math Integrated Math 1 Chapter 3 Detailed Answers Page 64   Exercise 13  Problem 22

Question 22.

Given the radius r and the area A of a circle, complete the table and determine whether the function is linear or non-linear by drawing the graph.

1. Use the formula A = πr² to find the area for each given radius r = 1,2,3,4,5.

2. Complete the table:

3. Plot the points on a graph where the x-axis represents the radius r and the y-axis represents the area A.

4. Determine whether the graph represents a linear or non-linear function.

Answer:

The given table is

Where r is the radius and A is the area of the circle.

Now, find the area with respect to different radius and do draw the graph to check whether the function is linear or not.

We know the formula for area of circle

A = πr²

So, for the given radius, the area can be calculated with the help of above formula i.e.

So, the graph is

This shows that the graph of the data is not in a straight line i.e. non-linear which cannot be written in the form of y = mx + b.

Hence, it is concluded that the graph of the given data including radius with respect to area of a circle is not a linear function

Page 65  Exercise 14  Problem 23

Question 23.

Given the graphs from Big Ideas Math Integrated Math 1 Student Journal 1st Edition Chapter 3 Maintaining Mathematical Proficiency Page 64 Exercise 13 Problems (19-22), determine whether each graph represents a function. Use the vertical line test to justify your answer.

1. Problem 19: The perimeter of a rectangle for different lengths x and breadth 2x.
2. Problem 20: The area of a rectangle for different lengths x and breadth 2x.
3. Problem 21: The circumference of a circle for different radii r.
4. Problem 22: The area of a circle for different radii r.

Answer:

Here, from the given data, we have to draw the graph of dependent variable with respect to independent variable.

If no vertical line drawn would intersect the curve more than once, then it will be a function.

For Page 64   Exercise 13  Problems (19 – 22)  lets plot the graph.

For  Page 64   Exercise 13  Problem 19    the perimeter of the rectangle for different length x and breadth 2x.

So, the graph is

Since no vertical line intersects the line more than once. So, the graph does represent a function.

For Page 64   Exercise 13  Problem 20 , the area can be calculated as A = L × B.

Since no vertical line intersect the curve more than once. So, the graph does represent a function.

For  Page 64   Exercise 13  Problem 21, circumference(C) can be calculated for radius r is C = 2πr

Since no vertical line intersects the curve more than once. So, the graph does represent a function.

For Page 64   Exercise 13  Problem 22, area(A) of the circle can be calculated as A = πr²

Since, no vertical line can intersect the line more than once.

So, the graph does represent a function.

Hence, for the Page 64   Exercise 13  Problems(19 – 22) there is no vertical line that intersect the line more than once. So, the graph does represent a function

Page 65  Exercise 15  Problem 24

Question 24.

To determine whether a function is linear or not, consider the example of the perimeter of a rectangle with length x and breadth 2x. The perimeter P can be calculated using the formula P = 2(L + B).

1. Given the formula for the perimeter, p = 2(L + B), where L = x and B = 2x, rewrite the formula in terms of x.
2. Complete the table below for the perimeter when x = 1,2,3,4,5.
3. Plot the graph using the values from the table.
4. Determine if the function is linear or non-linear based on the graph.

Answer:

To explain whether a function is linear or not, let’s take an example to explain it.

Consider an example of perimeter of a rectangle of length x and breadth 2x such that perimeter can be calculated as P = 2(L + B).

The graph of the data for perimeter of rectangle

The graph Is

Since, the graph of the function is straight line and can be written in the form of y = mx + b and no vertical line drawn is intersecting the curve more than once.

So, it is a linear function.

And if any vertical line drawn would intersect the curve more than once, then it will be a nonlinear function.

Hence, the function is linear or not depends on whether the graph of the data makes a straight line or not and also to check any vertical line drawn would intersect the curve more than once or not

Page 65  Exercise 16  Problem 25

Question 25.

Consider the following real-life problems and determine whether they represent linear or nonlinear functions. Explain your reasoning and sketch the graphs to illustrate each example.

Answer:

To define real life problem of linear and nonlinear functions. We will consider examples to understand it.

The function represents linear :

Example:- A 20-gallon bathtub is draining at a rate of 2.5 gallons per minute.

The number g of gallons remaining is a function of the number m of minutes.

Since the water in the bathtub (y) is decreasing by a constant amount (x).

So, the graph of the function is

Since the graph is a straight line and so is a linear function.

The function represents Nonlinearly:

Example:- Diamond cutter is making a pendant and that pendant is going to be circular.

A customer decide to buy that pendant and ask how large that pendant will be.

In this situation, the area of the pendant can be calculated with the help of the formula

A = πr² and so for different radius, we get the graph as

Since the graph is not a straight line and so is a nonlinear function.

Hence, the real-life examples of linear and nonlinear functions are

1.  A 20-gallon bathtub is draining at a rate of 2.5 gallons per minute. The number g of gallons remaining is a function of the number m of minutes

2.  Diamond cutter is making a pendant and that pendant is going to be circular.

3. A customer decide to buy that pendant and ask how large that pendant will be.

Page 68  Exercise 17  Problem 26

Question 26.

The graph of a given function is shown below.

Graph:

(Include the graph from Big Ideas Math Integrated Math 1 Student Journal 1st Edition Chapter 3 Maintaining Mathematical Proficiency Page 68 Exercise 17 Problem 26 Graph 1)

1. Check whether the function is linear or not with the help of the given graph.
2. Explain your reasoning based on the characteristics of the graph.

Answer:

The graph of the given function is

Check whether the function is linear or not with the help of given graph.

Since, from the graph, the rate of change of x with respect to y is constant and so make a straight line.

For a function to be linear, the line of graph must be straight where there is no vertical line that intersect the curve more than once and it can be seen that there exist no vertical line.

That intersect the curve more than once and so the given graph is of linear function.

Hence, the given graph of the function is a graph of linear function

Page 68  Exercise 18  Problem 27

Question 27.

Given the data in the table below.

1. Check the rate of change of values between x and y.
2. Plot the graph of the function using the given data.
3. Determine whether the graph represents a linear function.

Answer:

The given data in the form of a table is

Now, check the rate of change of values between x and y.

From the given table, the rate of change between x and y values is constant i.e. the graph of the function is

This implies that the graph of the function is linear.

Hence, the table of the data is a linear function from the graph

Big Ideas Math Integrated Math 1 Student Journal Chapter 3 Maintaining Proficiency Answers Page 68  Exercise 19  Problem 28

Question 28.

Given the data in the table below:

1. Check the rate of change of values between x and y.
2. Plot the graph of the function using the given data.
3. Determine whether the graph represents a linear function.

Answer:

The given data in the form of table is

Now, check the rate of change of values between x and y.

From the given table, the rate of change between x and y values is not constant i.e. the graph of the function is

This implies that the graph of the function is nonlinear.

Hence, the table of the data is a nonlinear function from the graph

Page 68  Exercise 20  Problem 29

Question 29.

Given equation y = 3-2x:

1. Check whether the equation can be written in the form y = mx + b, where m is the slope and b is the y-intercept.
2. Determine the slope m and the y-intercept b.
3. Confirm whether the equation represents a linear function.
4. Plot the graph of the equation and verify its linearity.

Answer:

The given equation is

y = 3−2x

Now, check the equation can be written in the form of y = mx + b, where m is the slope.

Since the equation can be written as

y = −2x + 3

Where m is -2 and b is 3. Therefore, the given equation is linear function.

Also, from the graph

This implies that the graph of the function is linear.

Hence, the equation y = 3−2x is a linear function.

Page 68  Exercise 21  Problem 30

Question 30.

Given the equation \(y=-\frac{3}{4} x^3\):

1. Check whether the equation can be written in the form y = mx + b, where m is the slope and b is the y-intercept.
2. Determine if the equation represents a linear or nonlinear function.
3. Explain why the equation is considered nonlinear.
4. Plot the graph of the equation and verify its nonlinearity.

Answer:

The given equation is

y = \(\frac{-3}{4}\) x³

Now, check the equation can be written in the form of y = mx + b, where m is the slope.

Since the given equation can not be written in the form of y = mx + b.

Therefore, the given equation is nonlinear.

Also, from the graph

This implies that the equation is nonlinear function.

Hence, the given equation is a nonlinear equation.

Page 68  Exercise 22  Problem 31

Question 31.

Given the graph of the function:

(Big Ideas Math Integrated Math 1 Student Journal 1st Edition Chapter 3 Maintaining Mathematical Proficiency Page 68 Exercise 22 Problem 31 Graph)

1. Observe the given graph and determine whether it has a continuous domain or a discrete domain.
2. Explain the reasoning behind your conclusion.
3. State the domain of the given graph.

Answer:

The graph of the given function is

Now, observe the graph and find whether it has a continuous domain or a discrete domain.

Hence, the domain of the given graph is the continuous domain.

Page 68  Exercise 23  Problem 32

Question 32.

Given the graph of the function:

(Big Ideas Math Integrated Math 1 Student Journal 1st Edition Chapter 3 Maintaining Mathematical Proficiency Page 68 Exercise 23 Problem 32 Graph)

1. Observe the given graph and determine whether it has a continuous domain or a discrete domain.
2. Identify the domain of the function from the graph.
3. Explain why the domain is classified as continuous or discrete.

Answer:

The graph of the given function is

Now, observe the graph and find whether it has a continuous domain or a discrete domain.

Since the domain is a set of input values for which function exist and input values are the x-values.

Therefore, from the graph, the domain(D) is D = {1,2,3}

Also, the domain of the function is a discrete domain as only integer values are being used in the graph.

Hence, the domain of the given graph is the discrete domain.

Big Ideas Math Integrated Math  Chapter 2 Maintaining Mathematical Proficiency

Big Ideas Math Integrated Math 1 Student Journal Chapter 2 Solutions Page 27  Exercise 1 Problem  1

Question 1.

Graph the absolute value of the given number:

|-2|

Answer:

We have to graph the number ∣−2∣

Given a number, ∣−2∣

The absolute value is 2.

Therefore, ∣−2∣ = 2

The graph of ∣−2∣ is

Page 27  Exercise 2  Problem 2

Question 2.

Graph the result of the following expression:

-3 + |-3|

Answer:

We have to graph the number −3 + ∣−3∣

Given, −3 + ∣−3∣

The absolute value of −3 is 3

So, −3+∣−3∣ = −3+3

= 0

The graph of −3+∣−3∣ is

Read and Learn More Big Ideas Math Integrated Math 1 Student Journal Solutions

Page 27  Exercise 3  Problem 3

Question 3.

Determine whether -4 is greater than or less than -7.

Answer:

We have to decide  −4 is greater or lesser than − 7

Is greater than−7.

So − 4 > − 7

The desired solution for this is − 4 > − 7

Chapter 2 Maintaining Mathematical Proficiency Solutions Big Ideas Math Integrated Math 1 Page 27  Exercise 4  Problem 4

Question 4.

Determine whether -8 is greater than or less than -5.

Answer:

Now we have to decide −8 is greater or lesser than − 5.

−8 is lesser than −5

−8 lies left side of −5, so −8 is less than −5

So − 8 < − 5

The desired solution of this is − 8 < − 5

Page 27  Exercise 5  Problem 5

Question 5.

Determine whether |-5| is greater than, less than, or equal to 5.

Answer:

Page 27  Exercise 6  Problem 6

Question 6.

Determine whether -7 is greater than, less than, or equal to |-6|.

Answer:

Given

-7

Now we have decided −7 is greater or lesser or equal to  ∣−6∣.

As we know any number in modulus treated as positive number.

So here ∣−6∣ is treated as + 6.

So As we know

−7 < ∣−6∣

The desired solution of this is −7 < ∣−6∣.

Big Ideas Math Integrated Math 1 Chapter 2 Step-By-Step Solutions Page 27 Exercise 7  Problem 7

Question 7.

Using a number line, represent the numbers a, b, -a, and -b.

Answer:

Given

The numbers a, b, -a, and -b.

First, we have to make a number line representing a,b,−a,−b.

After that move on basis that has been described in Tips section.

Now we have to represent numbers a,b,−a,−b.

As we know number present right-hand side of any number are greater then number presented left-hand side are smaller.

So here is a greater than b and −b is greater than −a.

−b is greater than−a.

Page 27  Exercise 8  Problem 8

Question 8.

Using a number line, represent the numbers a, b, -a, and -b. Additionally, represent the absolute values |-a| and |-b|.

Answer:

Given

a, b, -a, and -b.

First, we have to make a number line.

After that represent a,b,∣−a∣,∣−b∣

Now we have to represent a,b,−a,−b on number line.

So here − b < − a.

As we know any number in modulus treated as a positive number.

So here by the number line representation ∣−b∣ is greater than ∣−a∣

The desired solution is ∣−b∣ is greater than ∣−a∣.

Solutions For Big Ideas Math Integrated Math 1 Chapter 2 Maintaining Mathematical Proficiency Exercises Page 28  Exercise 9  Problem 9

Question 9.

Determine the type of inequality and present it graphically.

Answer:

We first find the type of inequality.

In this question greater than equal to type inequality available.

In the above diagram point is cover and goes to infinity so it is greater than equal to type inequality. or we can say x≥1

The desired solution is x ≥ 1.

Page 28  Exercise 9  Problem 10

Question 10.

Determine the type of inequality and write the equation of the inequality based on the given graph.

Answer:

First, we find the type of inequality.

Then write the equation of inequality.

It is greater than type inequality in word  x > 1

 The Desired solution is x > 1 

Big Ideas Math Integrated Math 1 Student Journal Chapter 2 guide Page 28  Exercise 9  Problem 11

Question 11.

Determine the type of inequality and write the equation of the inequality based on the given graph.

Answer:

For the given graph.

The inequality equation is x ≤ 1

In words are x is less than or equal to 1

The inequality equation is x≤1, In words are “x is less than or equal to 1”

Page 28  Exercise 9  Problem 12

Question 12.

Determine the type of inequality and write the equation of the inequality based on the given graph.

Answer:

For the given graph.

The inequality equation is x<1

In words are x is less than 1

The inequality equation is x<1 , In words are “x is less than 1”

Page 29  Exercise 10  Problem 13

Question 13.

Describe how mathematical inequalities are used in everyday situations. Provide examples of situations where inequalities are applied and explain how they represent real-life constraints.

Answer:

Inequalities are used all the time in the world around us—we just have to know where to look.

Figuring out how to interpret the language of inequalities is an important step toward learning how to solve them in everyday contexts.

Think about the following situations.

Speed limits on the highway minimum payments on credit card bills the number of text messages you can send each month.

From your cell phone the amount of time it will take to get from home to school.

All of these can be represented as mathematical inequalities.

These situations on a day-to-day basis are confronted with mathematical inequalities but we may not notice them because they are so familiar.

Chapter 2 Maintaining Mathematical Proficiency Explained Big Ideas Math Integrated Math  Page 31  Exercise 11  Problem 14

Question 14.

Translate the following sentence into a mathematical inequality:

“Seven is less than or equal to the difference of a number q and 6.”

Answer:

The sentence as an inequality is.

Seven is less than or equal to the difference of a number q and 6.

Hence the equation as inequality is

⇒ 7 ≤ q−  6

Hence the equation as inequality is  7≤ q − 6

Page 31  Exercise 12  Problem 15

Question 15.

Translate the following sentence into a mathematical inequality:

“The sum of a number u and 14 is more than 6.”

Answer:

The sentence is an inequality.

The sum of a numbers u and 14 is more than 6.

Hence The equation inequality is

u + 14 > 6

The equation as inequality is  u + 14 > 6

Page 31  Exercise  13  Problem 16

Question 16.

Graph the linear inequality x ≥ 3.

Answer:

Given

x ≥ 3

To graph, a linear inequality in one variable (say, x or y ), first get variables alone on one side.

Then consider the related equation obtained by changing the inequality sign to an equality sign.

The graph of this equation is a line.

Here the equation is x ≥ 3

Then the graph is

where dot represent the  x = 3

The graph of the equation is 

Page 32  Exercise 14  Problem 17

Question 17.

Graph the linear inequality x > -1.

Answer:

Given

x > -1

To graph, a linear inequality in one variable (say, x or y ), first get variables alone on one side.

Then consider the related equation obtained by changing the inequality sign to an equality sign. The graph of this equation is a line.

Here the equation is x > −1

Then the graph is

where circle represent x ≠ − 1

The graph of the equation is

Page 32  Exercise 15  Problem 18

Question 18.

Graph the linear inequality x < 1 and describe the characteristics of the graph.

Answer:

Given

x < 1

x<1 is made dark as shown in above graph

We write values below or above 1

we write whether 1 is included in graph

We write about circle on1

We write values included in graph

x<1 is made darker in the above graph.

We can see 0,−1,−2 below 1

1 is not included in graph,it is open.

Circle on 1 is not filled

We can see 0,−1,−2 that is lesser values included in graph

The graph for x<1 is as shown in graph

Where we see 1 is not included.

Values less than 1 are included in the graph

Page 32 Exercise 16  Problem 19

Question 19.

Observe the given graph and determine the inequality it represents. Describe your observations and the steps to determine the inequality.

Answer:

We write observations looking at graph such as whether 1 is included or not.

We write about circle on 1

We then write about values included in the graph  are above or below 1

We arrive at decision whether we use less than or greater than or equal to sign

Then we write inequality as per our observation

we can see 1 is not included in the figure so it is open. Circle on 1 is not filled.

We see values above 1 like 2,3,….. etc.

This means inequality in the graph is x>1

1 is not included so it is open in the graph.

The graph shows values above 1.

Hence the inequality that represents the graph is x>1

Page 32  Exercise 17  Problem 20

Question 20.

Observe the given graph and determine the inequality it represents. Describe your observations and the steps to determine the inequality.

Answer:

We write observations looking at graph whether 0 is included or not.

We write about circles on 0.

We then write about values included in the graph  are above or below 0.

We arrive at decision of whether we use less than or greater than or equal to sign.

Then we write inequality as per our observation

We can see 0 is not included in the figure so it is open circle on 0 is not filled.

We see values below 0 like −1,−2,……etc.

This means inequality in the graph is x<0

0 is not included in the graph so it is open.

The graph shows values below 0.

Hence the inequality that represents the graph is x<0

Worked Examples For Big Ideas Math Integrated Math 1 Chapter 2 Maintaining Mathematical Proficiency Page 32  Exercise 18  Problem 21

Question 21.

Observe the given graph and determine the inequality it represents. Describe your observations and the steps to determine the inequality.

Answer:

We write observations looking at graph whether −3 is included or not.

We write about circle on −3.

We then write about values included in graph e are above or below −3.

We arrive at decision whether we use less than or greater than or equal to sign.

Then we write inequality as per our observation

−3 is included in the graph circle on −3 is filled so we use equal to sign in inequality of graph.

We see values above −3 like −1,−2,……. etc.

This means inequality in the graph is x≥−3

−3 is included in the graph.

The graph shows values above −3.

Hence the inequality that represents the graph is x ≥ −3

Page 32  Exercise 19  Problem 22

Question 22.

Observe the given graph and determine the inequality it represents. Describe your observations and the steps to determine the inequality.

Answer:

We write observations looking at graph whether 3 is included or not.

We write about circle on 3

We then write about values included in the graph  are above or below 3

We arrive at decision of whether we use less than or greater than or equal to sign.

Then we write inequality as per our observation

3 is included in the graph circle on 3 is filled so we use equal to sign in the inequality of graph.

We see values below 3 like 1,2,…. etc.

This means inequality in the graph is x≤3

3 is included in the graph.

The graph shows values below 3.

Hence the inequality that represents the graph is x ≤ 3.

Page 33  Exercise 20  Problem 23

Question 23.

Solve the following inequalities using addition and subtraction. Show all steps clearly.

1. Solve inequality x – 3 < 5 using addition.
2. Solve inequality x + 5 > 10 using subtraction.

Answer:

We write an example as solving  x−3 < 5 for addition.

We add 3 to each side of above inequality.

Thus adding 3 solves inequality.

Now we write an example as solving x+5>10 we subtract 5 from each side of inequality.

Thus subtracting 5 solves inequality.

Hence we use addition and subtraction to solve inequality

We solve inequality by using addition.

For example  x − 3 < 5.

We add 3 to each side of above inequality

x − 3 + 3 < 5 + 3

x < 8

We solve inequality using subtraction.

For example x + 5 > 10

Subtracting 5 from both sides of above inequality.

x + 5 − 5 > 10 − 5

x > 5

We add 5to both sides of  x − 3 < 5 Solving we get  x < 8 We subtract 5 from both sides of  x + 5 > 10 Solving we get  x > 5

Big Ideas Math Integrated Math 1 Chapter 2 Detailed Answers Page 34  Exercise 21  Problem 24

Question 24.

Given the passing efficiency formula: \(P=\frac{8.4 Y+100 C+330 T-200 N}{A}\)

Determine a set of values for A, C, Y, T, and N that make the inequality P < 0 true. Verify the solution by substituting the values into the formula and showing that the inequality holds.

Answer:

Given :   P<0

To find –  A passing record that makes the inequality true.

First of all we will assume the values of different variables.

Then we will check for these values so that the given inequality is satisfies.

Passing efficiency formula is  P=\(\frac{8.4 Y+100 C+330 T-200 N}{A}\)

Let’s consider

A = 240

C = 130

Y = 1200

T = 14 and

N = 150

⇒   P\( = \frac{8.4(1200)+100(130)+330(14)-200(150)}{240}\)

⇒  P = \(\frac{10080+13000+4620-30000}{240}\)

⇒  P =  \(\frac{27700-30000}{240}\)

⇒   P = \(\frac{−2300}{240}\)

⇒  P =  − 9. 58

⇒  P < 0

Let’s fill the table now.

The final answer is that the passing record that makes the given inequality true is A = 240, C = 130, Y = 1200, T = 14, and  N = 150

Page 34  Exercise 21  Problem 25

Question 25.

Given the inequality:

P + 100 ≥ 250

and passing efficiency formula: \(P=\frac{8.4 Y+100 C+330 T-200 N}{A}\)

Determine a set of values for A, C, Y, T, and N that make the inequality true. Verify the solution by substituting the values into the formula and checking the inequality.

Answer:

Given:  P + 100 ≥ 250

To find – A passing record that makes the inequality true.

First of all we will assume the values of different variables.

Then we will check for this values so that the given inequality satisfies.

Passing efficiency formula is  P = \(\frac{8.4 Y+100 C+330 T-200 N}{A}\)

Let’s consider

A = 300

C = 210

Y = 2500

T = 25 and

N = 8.

⇒  P = \(\frac{8.4(2500)+100(210)+330(25)-200(8)}{300}\)

⇒  P = \(\frac{21000+21000+8250-1600}{300}\)

⇒  P = \(\frac{48650}{300}\)

⇒  P =  162.16

⇒ P + 100 = 262.16

⇒  P + 100 > or < 250.

Let’s fill the table now.

The final answer is that the passing record that makes the given inequality true is  A = 300, C = 210, Y = 2500, T =25, and  N = 8.

Page 34  Exercise 21  Problem 26

Question 26.

Given the inequality:

P – 250 > -80

and passing efficiency formula: \(P=\frac{8.4 Y+100 C+330 T-200 N}{A}\)

Determine a set of values for A, C, Y, T, and N that make the inequality true. Verify the solution by substituting the values into the formula and checking the inequality.

Answer:

Given: P−250>−80

To find –   A passing record that makes the inequality true.

First of all we will assume the values of different variables.

Then we will check for these values so that the given inequality satisfies.

Passing efficiency formula is   P = \(\frac{8.4 Y+100 C+330 T-200 N}{A}\)

Let’s consider

A = 250

C = 180

Y = 3000

T = 25 and

N = 6.

⇒  P \(=\frac{8.4(3000)+100(180)+330(25)-200(6)}{250}\)

⇒  P = \(\frac{25200+18000+8250-1200}{250}\)

⇒  P =  \(\frac{50250}{250}\)

⇒  P =  201

⇒  P −  250 = 201 −  250

⇒  P  − 250 = − 49

⇒  P − 25 0 > − 80

Let’s fill the table now.

The final answer is that the passing record that makes the given inequality true is A = 250, C = 180, Y = 3000, T = 25, and N =6.

Page 34  Exercise 22  Problem 27

Question 27.

Explain how to use addition and subtraction to solve inequalities. Provide examples to illustrate each method.

Answer:

Let’s first understand the use of addition to solve an inequality by taking an example.

If x − 2 ≤ 5 is an inequality.

Adding 2  to both sides, we get

x − 2 + 2 ≤ 5 + 2

⇒ x ≤ 7

which is the solution of the given inequality.

We will now understand the use of subtraction to solve an inequality by taking an example.

If x + 7 > 5 is an inequality.

Subtracting 7 from both sides, we get

x + 7 − 7 > 5 − 7

⇒  x> − 2

which is the solution of the given inequality.

The final answer is that we can use addition or subtraction such that only the variable term remains on either side of the inequality and then it will be easy to solve an inequality.

Page 36  Exercise 23  Problem 28

Question 28.

Given the inequality -3 > -3 + h:

1. Solve the inequality by adding 3 to both sides and simplify.
2. Represent the solution on a number line.

Answer:

Given: Inequality is − 3 > − 3 + h.

Add 3 on both side of the inequality and simplify it.

Now represent the inequality on a number line.

We have

− 3 > − 3 + h

⇒  − 3 + 3 > − 3 + h + 3 (Adding3 both side)

⇒  0 > h

⇒  h < 0

Representing the inequality on a number line.

The solution of the inequality −3 > − 3 + h is h<0 and its representation on number line is given below.

Page 37   Exercise 24  Problem 29

Question 29.

Given the inequality s-(-1)≥2:

1. Simplify the inequality.
2. Subtract 1 from both sides of the inequality and simplify further.
3. Represent the solution on a number line.

Answer:

Given:  Inequality is−(−1)≥2.

Simplify it and then subtract 1.

On both side of an inequality.

Now represent the inequality on a number line.

We have

s − (−1) ≥ 2

⇒  s + 1 ≥ 2

⇒  s + 1− 1 ≥ 2 − 1 (Subtracting 1 on both side)

⇒  s ≥1

Representing the inequality on a number line.

The solution of the inequality −(−1) ≥ 2 is s ≥ 1 and its representation on the number line is given below.

Page 37  Exercise 25  Problem 30

Question 30.

Given the inequality 6 – 9 + u < -2:

1. Simplify the inequality.
2. Add 3 to both sides of the inequality and simplify further.
3. Represent the solution on a number line.

Answer:

Given: Inequality is6−9+u<−2.

Simplify it and then add 3 both side of an inequality.Now represent the inequality on number line.

We have

6 − 9 + u < − 2

⇒ −3 + u < − 2

⇒ −3 + u + 3 <− 2 + 3 (Adding 3 both side)

⇒ u < 1

Representing the inequality on a number line.

The solution of the inequality  6−9 + u <− 2 is u < 1 and its representation on number line is given below.

Page 38  Exercise 26  Problem 31

Question 31.

Complete the table and determine whether the inequality 6<3x is true or false for each given value of x. Then, solve the inequality by dividing both sides by 3 and choose the correct solution from the given graph.

Answer:

Complete the table and decide whether the inequality  6 < 3x is true or false for each case.

Divide by 3 on both sides of the inequality 6 < 3x and find the solution.

Now choose the correct solution from the given graph.

We have

6 < 3x

⇒  \(\frac{6}{3}\) < \(\frac{3x}{3}\)

⇒  2 < x

⇒   x >  2

The correct solution from the given graph is

The solution of the inequality is x >2 and its representation on number line is given below.

Page 38  Exercise 27  Problem 32

Question 32.

Complete the table and analyze it to find the solution for each of the given inequalities. Then, write the solution for each inequality and represent it on a number line.

Answer:

Given: 

Inequalities are:

⇒ 1.2  x < 4

⇒  2.3 ≥ 3x

⇒  3.2x < 8

⇒  4.6 ≥ 3x

Complete the table and analyze it to find the solution of given inequality.

We have 2 x < 4

Therefore the solution is x < 2

The rule is divide the coefficient of x both sides of the equality.

Therefore the solution is x < 2

We have 3 ≥ 3x

We have 2 x < 8

We have

All are represented in tables.

Page 39  Exercise 28   Problem 33

Question 33.

1. Consider the effects of dividing both sides of an inequality by positive and negative numbers.
2. Given the inequality 3<4, divide both 2 and determine whether the inequality sign changes or remains the same.
3. Given the inequality 3<4, divide both sides by -2 and determine whether the inequality sign changes or remains the same.
4. Explain the rules for dividing inequalities by positive and negative numbers.

Answer:

Lets consider the inequality.

3 < 4

⇒   \(\frac{3}{2}\)<\(\frac{4}{2}\) (Dividing both side by 2)

⇒  1.5<2 (Sign of inequality remains same)

Again consider the inequality.

3 < 4

⇒  \(\frac{3}{−2}\)<\(\frac{4}{−2}\)

(Dividing both side by−2)

⇒  −1.5 > −2 (Sign of inequality changes)

If we divide by a positive number on both sides of inequality the sign of inequality remains the same.

If we divide by a negative number on both side of inequality the sign of inequality changes.

Page 39  Exercise 30   Problem 35

Question 35.

Solve the inequality 7x < -21 using the rule for dividing both sides of the inequality by the coefficient of the variable. Then represent the solution on a number line.

Answer:

Given

7x < -21

We have to solve each inequality 7x < − 21

From  Page 38  Exercise 27  Problem 32,  Tables  we get- Part 1  a rule that is 

To find the solution of the inequality we have to divide both sides of the inequality by the coefficient of the variable.

From  Page 38  Exercise 27  Problem 32, Tables we get – Part 1

The solution of the inequality is x < 2

The solution of the inequality is 1 ≥ x

The solution of the inequality is  x < 4

The solution of the inequality is  2 ≥ x

The rule is to divide both sides of the inequality by the coefficient of x.

Given, 7x < − 21 divided by 7 on both sides of an inequality.

x<− \(\frac{21}{7}\)

x < − 3

The solution is x < − 3

Page 39  Exercise 30   Problem 36

Question 36.

Solve the inequality 12 ≤ 4x using the rule for dividing both sides by the coefficient of the variable. Then represent the solution on a number line.

Answer:

We have to solve each inequality 12 ≤ 4x

From   Page 38  Exercise 27  Problem 32,  Tables we get Part 1 a rule to find the solution of the inequality.

The rule is to get the solution of the inequality we have to divide both sides of the equality by the coefficient of the variable.

If the coefficient of x is negative then the inequality sign will be inverted.

Here the coefficient of x is 4.

From  Page 38  Exercise 27  Problem 32, Tables we get – Part  1

The solution of the inequality is  x < 2

The solution of the inequality is 1 ≥ x

The solution of the inequality is  x < 4

From  Page 38  Exercise 27  Problem 32,   Tables we get  – Part 2

The solution of the inequality is x ≤ 2

The solution of the inequality is x >− 2

The solution of the inequality is −1 ≤ x

The solution of the inequality is x > − 4

The solution of the inequality is x ≥ − 2

We conclude that:

The rules of solution a inequality are

Divide the both sides by the coefficient of x

If the coefficient of x is a negative number then invert the inequality sign

Given,12 ≤ 4x

4x ≥ 12

Divide by 4 on both sides of an inequality.

x  ≥ \(\frac{12}{4}\)

x ≥ 3

The solution is x ≥ 3

Big Ideas Math Integrated Math 1 Student Journal Chapter 2 Maintaining Proficiency Answers Page 39  Exercise 30  Problem 37

Question 37.

Solve the inequality 10 < -5x using the division rule of inequality. Show all steps clearly.

Answer:

We are given an inequality 10 <−5x.

We need to use the division rule of inequality to solve it.

We are going to apply the division rule of inequality on given inequality.

And then solve it until we get the variable on one side and the number on the other.

10<−5x

⇔  \(\frac{10}{−5}\)>\(\frac{−5x}{−5}\)

(∵if x<y and z<0 then \(\frac{x}{z}\) >\(\frac{y}{z}\))

⇔ − 2 > x

By using the division rule of inequality for 10<−5x, we get −2 > x.

Page 39  Exercise 30   Problem 38

Question 38.

Solve the inequality -3 x ≤ 0 using the division rule of inequality. Show all steps clearly and represent the solution on a number line.

Answer:

We are given an inequality −3 x ≤ 0

We need to use the division rule of inequality to solve it.

We are going to apply the division rule of inequality on given inequality.

And then solve it until we get the variable on one side and the number on the other.

From   we get  Page 38  Exercise 27  Problem 32  (Part 1  and  Part 2)

The solution of the inequality is  x < 2

The solution of the inequality is  x ≤ 1

The solution of the inequality  is  x < 4

The solution of the inequality is  x ≥ − 2

The solution of the inequality is  x > − 2

The solution of the inequality is x ≥ − 1

The solution of the inequality is  x > − 4

The solution of the inequality is  x ≥ − 2

The rules are:

Divide both sides of the inequality by the coefficient of x.

If the coefficient of x is a negative number then invert the inequality sign.

Find the interval of inequality using division rule

−3x < 0

⇔  \(\frac{−3x}{−3}\) > \(\frac{0}{−3}\)

(∵if x<y and z<0 then \(\frac{x}{z}\) >\(\frac{y}{z}\))

⇔  x  >  0

By using the division rule of inequality for −3x ≤ 0  we get x  ≥ 0.

Page 41  Exercise 31  Problem 39

Question 39.

Solve the inequality -67 ≤ 37f using the multiplication property of inequality. Then, represent the solution on a number line.

Answer:

We are given an inequality \(\frac{−6}{7}\) ≤ \(\frac{3}{7}\) f

We have to solve it and graph the solution.

We will do this by using the multiplication property of inequality.

To obtain the required solution, we will solve till we get the variable on one side and numbers on another side.

Finally, we will represent this solution on a number line.

On the number line, fill in the circle for (≤or≥) and leave the circle unfilled for (<or>).

\(\frac{−6}{7}\)≤\(\frac{3}{7}\) f

⇔  \(\frac{−6}{7}\) ×\(\frac{7}{3}\)

(∵If x≤y, and z>0,then xz≤yz)

⇔  \(\frac{3}{7}\)f×\(\frac{7}{3}\)

⇔  2 ≤ f

For inequality \(\frac{−6}{7}\)≤\(\frac{3}{7}\) f

Interval notation of solution:  (−2,∞)

The representation of solution on the number line:

The required solution to the inequality\(\frac{−6}{7}\)≤\(\frac{3}{7}\) f is −2 ≤ f . Graph of solution

Page 41  Exercise 32  Problem 40

Question 40.

Solve the inequality -4m ≥ -16 using the division property of inequality. Then, represent the solution on a number line.

Answer:

We are given an inequality −4m≥−16

We have to solve it and graph the solution.

We will do this by using the division property of inequality.

To obtain the required solution, we will solve till we get the variable on one side and number on another side.

Finally, we will represent the solution on a number line.

On the number line, fill in the circle for (≤or≥) and leave the circle unfilled for (<or>)

−4m ≥−16

⇔  \(\frac{−4m}{4}\) ≤ \(\frac{−16}{−4}\)

(∵If x≥y and z<0, then  \(\frac{x}{z}\) ≤ \(\frac{y}{z}\))

⇔  m≤4

For inequality −4m≥−16

Interval notation of solution: (−∞,4)

The representation of solution on the number line:

The required solution to inequality −4m≥−16 is m≤4. Graph of solution

Page 42  Exercise 33  Problem 41

Question 41.

Solve the inequality \(1 \leq-\frac{1}{4} y\) using the multiplication property of inequality. Then, represent the solution on a number line.

Answer:

We are given an inequality 1≤−\(\frac{−1}{4}\) y

We have to solve it and graph the solution.

We will do this by using the multiplication property of inequality.

To obtain the required solution, we will solve till we get the variable on one side and number on another side.

Finally, we will represent the solution on a number line.

On the number line, fill in the circle for (≤or≥)and leave the circle unfilled for (<or>).

1≤−\(\frac{−1}{4}\) y

⇔ 1×−4≥−\(\frac{−1}{4}\) y × −4

(∵ If x≤y and z<0, then xz≥yz)

⇔ −4 ≥ y

For inequality   1≤−\(\frac{−1}{4}\) y

Interval notation of solution: (−∞,−4)

The representation of solution on the number line:

The required solution to inequality 1≤−\(\frac{−1}{4}\) y  is −4 ≥ y. Graph of solution

Page 42  Exercise 34  Problem 42

Question 42.

Solve the inequality \(-\frac{2}{3}<-4 x\) using the division property and the multiplication property of inequality. Then, represent the solution on a number line.

Answer:

We are given an inequality \(\frac{−2}{3}\)<− 4x

We have to solve it and graph the solution.

We will do this by using the division property and the multiplication property of inequality.

To obtain the required solution, we will solve until we get the variable on one side and the number on another side.

Finally, we will represent this solution on a number line.On the number line, fill in the circle for (≤or≥) and leave the circle unfilled for (<or>).

−\(\frac{−2}{3}\) <−4x

⇔ −\(\frac{−2}{3}\)<−4x × 3<−4x × 3

(∵ If x<y and z>0,then xz<yz)

⇔ −2<−12x

⇔ −\(\frac{−2}{12}\) > \(\frac{−12}{−12}\) x

(∵ If x < y and z<0, then \(\frac{x}{z}\)>\(\frac{y}{z}\))

⇔ \(\frac{1}{6}\)> x

For inequality \(\frac{−2}{3}\) <−4x

Interval notation of solution: (−∞,\(\frac{1}{6}\))

The representation of solution on the number line:

The required solution to inequality  <−4x is \(\frac{−2}{3}\) \(\frac{1}{6}\) >x. Graph of solution:

Big Ideas Math Integrated Math  Chapter 1 Maintaining Mathematical Proficiency

Big Ideas Math Integrated Math 1 Student Journal Chapter 1 Solutions Page 1  Exercise 1 Problem 1

Question:

Simplify the following expression step-by-step:

$-1+(-3)$

−1+(−3)

Solution

Given

−1+(−3)

In this question {−1 + (−3)} first, we have to multiply the sign appears before the number.

If any question don’t have any sign before the number then by default it is a positive sign (+).

After that we do the addition or subtraction.

Read and Learn More Big Ideas Math Integrated Math 1 Student Journal Solutions

Multiply the sign presented before the no.

−1 + (−3)

We know that

(+) × (−)=(−)

So here after the multiplication

− 1 − 3

Here we do that addition or subtraction operation.

− 1 − 3 = −4

The desired solution is −1 + (−3) = −4.

Chapter 1 Maintaining Mathematical Proficiency Solutions Big Ideas Math Integrated Math 1 Page 1  Exercise 2  Problem 2

Question: Simplify the following expression step-by-step:

$-4-7$

−4−7

Answer:

Given expression is −4−7

If any question doesn’t have any sign before the number then by default it is positive sign.

After that, we do addition or subtraction.

We have to do subtraction operation

− 4 − 7 = −11

The desired solution is −4 − 7 = 11

Chapter 1 Maintaining Mathematical Proficiency Solutions Big Ideas Math Integrated Math 1 Page 1  Exercise 3  Problem 3

Question: Find two pairs of integer numbers whose sum is

$-6$

−6.

Question:

The record monthly high temperature for March is 56°F, and the record monthly low temperature for March is -4°F.

Using the formula for the range of a parameter, calculate the range of temperatures for the month of March.

Big Ideas Math Integrated Math 1 Chapter 1 Step-By-Step Solutions Page 1  Exercise 4  Problem 4

Given

The record monthly high temperature for March is 56°F.

The record monthly low temperature for March is −4°F.

Now use the formula for range of a parameter.

Range of a parameter

= Highest value of the parameter − The lowest value of parameter

=  56°F − (−4°)F

=  56°F + 4°F

=  60°F

The range of temperatures for the month of March is 60°F.

Big Ideas Math Integrated Math 1 Chapter 1 Step-By-Step Solutions Page 1 Exercise 5  Problem 5

Determine the algebraic sign of the product of the given numbers and find the final value.

Given:

$-8\times (-5)$

−8×(−5)

Determine the algebraic sign of the given numbers.

Here 8 has a negative algebraic sign and 5 also has a negative algebraic sign.

Determine the algebraic sign of the final answer.

In this case it is a positive sign.

Simply multiply the given numbers and write the final answer with appropriate sign.

We have

⇒ − 8 ×  (−5)

= −8 × (−5)

= 40

The value of −8 × (−5) is 40.

Solutions For Big Ideas Math Integrated Math 1 Chapter 1 Maintaining Mathematical Proficiency Exercises Page 1  Exercise 6  Problem 6

Question:

Write

$\frac{14}{2}$

  as

$\frac{1}{2}\mathrm{}$

×14 and determine the algebraic sign of the given numbers.

Write  \(\frac{14}{2}\)  as  \(\frac{1}{2}\) × 14

Determine the algebraic sign of the given numbers.

Here \(\frac{1}{2}\) has a positive algebraic sign and 14 also has a positive algebraic sign.

Determine the algebraic sign of the final answer.

In this case it is a positive sign.

Simply multiply the given numbers and write the final answer with appropriate sign.

We have

⇒  \(\frac{14}{2}\)

=  \(\frac{1}{2}\) × 14

= \(\frac{1}{2}\) × (2 × 7)

Cancel out of 2

= 7

The value of \(\frac{14}{2}\) is 7

Big Ideas Math Integrated Math 1 Student Journal Chapter 1 Guide Page 1  Exercise 7  Problem 7

Question: Simplify the following expression and determine the algebraic sign of the final answer:

$\frac{–30}{–3}$

-3-30

Given ​

$\frac{–30}{–3}$

Write −30/(−3) = \(\frac{−30}{−3}\)

Now \(\frac{−30}{−3}\)

= \(\frac{1}{−3}\) × (−30)

Determine the algebraic sign of the given numbers.

Here 1 has a negative algebraic sign and 30 also has a negative algebraic sign.

Determine the algebraic sign of the final answer.

In this case it is a positive sign.

Simply multiply the given numbers and write the final answer with appropriate sign.

We have

−30/(−3)

= \(\frac{−30}{−3}\)

= \(\frac{1}{−3}\) × (−30)

= \(\frac{1}{3}\) × 3 × 10

= 10

The value of −30/(−3) is 10.

Big Ideas Math Integrated Math 1 Student Journal Chapter 1 guide Page 1  Exercise 8  Problem 8

Question:

Find two pairs of integers whose product is -20 by factorizing the number

$-20\mathrm{}$

Answer:

Given number

$-20$

To find the integers whose product is -20 can be found by factorizing number −20

Factors of number -20 are given as

−20  = −(2 × 2 × 5)

Therefore two pairs of integers are

−(4 × 5)​ = −20

−(2 × 10)  = −20

Two pair of integers are : −(4×5) and −(2×5)

Page 1  Exercise 9  Problem 9

Question:

A football team loses 3 yards in 3 consecutive games.

Calculate the total yardage gained by negating the magnitude of the total yardage loss.

Answer:

Given that a football team loses 3 yards in 3 consecutive game.

We have to find total yardage gained by negating the magnitude of total yardage loss.

Here a football team losses 3 yards in 3 consecutive game.

Total yardage loss therefore total yardage gained = − (Total yardage loss)

= −(9)

The total yardage gained is − 9.

Page 2  Exercise 10  Problem 10

Question:

Explain how simple linear equations can be used to solve real-life problems. Provide specific examples.

Uses of simple equations to solve real-life problems

There are various real-life situations where unknown quantity can be expressed as linear equation.

Some are given as below-

To compare ages of two persons and find their ages.

To find the runs scored by a batsman.

Number of trees planted by an organization.

Cost of travelling in a taxi from place A to place B.

Cost of buying mangoes from a market.

To find profit/loss of a shop.

The final answer is that some of the real-life problem that can be solved using simple equations are:

To compare ages of two persons and find their ages

To find the runs scored by a batsman.

Number of trees planted by an organization.

Cost of travelling in a taxi from place A to place B.

Cost of buying mangoes from a market.

To find profit/loss of a shop.

Chapter 1 Maintaining Mathematical Proficiency explained Big Ideas Math Integrated Math 1 Page 2   Exercise 11   Problem 11

Question:

Measure all the four angles of a given quadrilateral using a protractor and check the precision of your measurements.

For a given quadrilateral have 4 sides and 4 angles denoted as ∠A, ∠B, ∠C, ∠D

We have to measure all the four angles using protractor and check how precise our measurement.

By using a protractor measuring the angle of the quadrilateral given below:

m∠A = 100 degrees

m ∠ B = 91 degrees

m∠C = 92 degrees

m∠D = 67 degrees

m∠A + m∠B + m∠C + m∠D = 100 + 91 + 92 + 67

=  360 degrees

Measurement of angles of given quadrilateral:

​m∠A = 100 degree

m∠B = 91 degree

m∠C = 92 degree

m∠D = 67 degree

m∠A + m∠B + m∠C + m∠D = 360degrees

Free Big Ideas Math Integrated Math 1 Chapter 1 Solutions Maintaining Mathematical Proficiency Page 2  Exercise 11 Problem 12

Question:

Measure all four angles of a given quadrilateral using a protractor and verify the precision of your measurements.

Answer:

For a given quadrilateral have 4 sides and 4 angles denoted as ∠A, ∠B, ∠C, ∠D

We have to measure all the four angle using protractor and check how precise our measurement.

By using protractor measure the angle of the quadrilateral given below:

m∠A = 67 degree

m∠B = 147 degree

m∠C = 56 degree

m∠D = 90 degree

m∠A + m∠B + m∠C + m∠D = 67 + 147 + 56 + 90 = 360 degree

Measurement of angles of given quadrilateral:

​m∠A = 67 degree

m∠B = 147 degree

m∠C = 56 degree

m∠D = 90 degree

m∠A + m∠B + m∠C + m∠D = 67 + 147 + 56 + 90 = 360 degree

Page 2  Exercise 11  Problem 13

Question:

Measure all four angles of a given quadrilateral using a protractor and verify the precision of your measurements.

For a given quadrilateral have 4 sides and 4 angles denoted as ∠A, ∠B, ∠C, ∠D

We have to measure all the four angles using protractor and check how precise our measurement.

By using a protractor measuring the angle of quadrilateral given below:

​m∠A  =  94 degrees

m∠B  =  79 degrees

m∠C  =  72 degrees

m∠ D  = 115 degrees
​
m∠A + m∠B + m∠C + m∠D =  94  + 79 + 72 + 115 = 360 degrees

Measurement of angles of given quadrilateral:

​m∠A  =  94 degrees

m∠B  =  79 degrees

m∠C  =  72 degrees

m∠ D  = 115 degrees
​
m∠A + m∠B + m∠C + m∠D =  94  + 79 + 72 + 115 = 360 degrees

Worked Examples For Big Ideas Math Integrated Math 1 Chapter 1 Maintaining Mathematical Proficiency Page 2  Exercise 12  Problem 14

For a given quadrilateral have 4 sides and 4 angles denoted as ∠A,∠B,∠C,∠D

We have to measure all four angles using a protractor and check how precise our measurement.

By using a protractor measure of the angles of the quadrilaterals are given below:

Making the three different quadrilaterals.

1)

2)

3)

Measuring the angles of the three different quadrilaterals using protractor and verifying the values

Hence, The measurements are accurate so, by using a protractor we can measure the angles of the quadrilateral.

And the sum of all four of any quadrilateral is 360°

Big Ideas Math Integrated Math 1 Chapter 1 Detailed Answer Page 3   Exercise 13  Problem 15

A quadrilateral is given with its measurement of angle as:

​m∠A = 85degrees

m∠B = 80 degrees

m∠C = 100 degrees

m∠D = x degrees
​
We can find the m∠D using conjecture of  Page 2 Exercise 11 Problem 11

Sum of the angle measures of a quadrilateral are equals to 360 degrees.

Hence writing the simple equation :

85 + 80 + 100 +  x  = 360

⇒ x = 95

Hence m∠D = 95 degrees

Measure angle D of given quadrilateral is 95 degrees.

Page 3  Exercise 13  Problem 16

A quadrilateral is given with its measurement of angle as:

​m∠A = 72degrees

m∠B = 60 degrees

m∠C = 78degrees

m∠D = x degrees
​
We can find the m∠D

Using conjecture of Page 2 Exercise 11 Problem 11

Sum of the angle measures of a quadrilateral are equals to 360 degrees.

Hence writing the simple equation :

72 + 60 + 78 + x = 360

⇒  x = 150

m∠D = 150 degrees

Measure angle D of given quadrilateral is 150 degrees.

Big Ideas Math Integrated Math 1 Student Journal Chapter 1 Maintaining Proficiency Answers Page 3  Exercise 13  Problem 17

A quadrilateral is given with its measurement of angle as:

​m∠A = 90degrees

m∠B = 30 degrees

m∠C = 90degrees

m∠D = x degrees
​
We can find the m∠D = using conjecture of Page 2 Exercise 11 Problem 11

Sum of the angle measures of a quadrilateral are equals to 360 degrees.

Hence writing the simple equation :

90 + 30 + 90 + x = 360

⇒  x = 150

m∠D  = 150 degrees

Measure angle D of given quadrilateral is 150 degrees.

Page 6  Exercise 14  Problem 18

We have a simple equation as −15 + w = 6

We can find solution of equation by using core concept of the addition property of equality.

We have

−15 + w = 6…….(1)

Applying addition property of equality.

−15 + w + 15 = 6 + 15

⇒  w = 21

Hence solution of equation is w =  21

To verify the solution, put w = 21 in equation (1)

We get, −15 + 21 = 6

6 = 6

Therefore the solution w = 21 satisfying the equality of given equation.

The solution of given equation −15 + w = 6 is w = 21

Big Ideas Math Integrated Math 1 Student Journal Chapter 1 Maintaining Proficiency answers Page 6  Exercise 15  Problem 19

We have a simple equation as:  z−5 = 8

We can find solution of equation by using core concept

We have

z−5 = 8……………(1)

Applying addition property of equality.

⇒  z −5 + 5 = 8 + 5

⇒  z = 13

Hence solution of equation is  z =  13

To verify the solution, put  z = 13 in equation (1)

We get, ​13 − 5 = 8

⇒  8 = 8
​
Therefore the solution z=13

Satisfying the equality of given equation.

The solution of given equation z−5 = 8 is z = 13

Page 6  Exercise 16  Problem 20

We have simple equation as: −2 = y − 9

We can find solution of equation by using core concept of addition property of equality.

We have

−2 = y − 9……………. (1)

Applying addition property of equality.

⇒ −2 + 9 = y − 9 + 9

⇒  7 = y

Hence solution of equation is y = 7

To verify the solution put y = 7 in equation (1)

we get, −2 = 7−9

−2 = −2

Therefore the solution y  = 7 satisfying the equality of given equation.

The solution of given equation −2 = y−9  is  y = 7

Page 6  Exercise 17  Problem 21

We will first form the linear equation with variable p.

We will solve the linear equation to get the value of p.p will be the desired result.

Price of headphone = $ p

Price of headphone after applying coupon= $ (p − 17.95)

Price paid after applying coupon for headphone = $ 71.80

Price of headphone after applying coupon = Price paid for headphone

(p- 17.95) = 71.80

p = 71.80 + 17.95

p = $ 89.75

Original price of headphone = $ 89.75

Original price of headphone = $ 89.75

Page 6  Exercise 18  Problem 22

We will let the number of browines made for the party be x.

We will first form the linear equation with variable x.

We will solve the linear equation to get the value of x.

x will be the desired result.

Let, x brownies made for the party.

Number of brownies left = 16

Brownies left = 2/5 of total brownies

Then

\(\frac{2}{5}\) of   x = 16

\(\frac{2}{5}\) of   x = 16

x = 16

2x = 16 × 5

2x = 80

x = \(\frac{80}{2}\)

x= 40

I made 40 brownies for the party.

40 brownies were made for the party.

Big Ideas Math Integrated Math 1 Student Journal Chapter 1 Maintaining Proficiency answers Page 7  Exercise 19  Problem 23

We will form the equation 30°+9x°+(30+x)°=180° and solve it to get the value of x.

Sum of angles of triangle = 180°

30° + 9x° + (30+x)° = 180°

60° + 10x° = 180°

10x° = 180°− 60°

10x° = 120°

x° = \(\frac{120°}{10}\)

x = 12°

First angle = 30°

Second angle = ​9x°

= 108°

Third angle = (30+x)°

= 42°

We can check the answer by substituting value of x

30° + (9 × 12)° + (30 + 12)° = 180°

Angles of triangles are 30°,108°,42°

Page 7  Exercise 19  Problem 24

We will substitute the sum of angles = 180°

50°+ (x + 20)°+ (x + 10)°= 180°

By solving this, we will get the value of x.

Sum of angles of triangle = 180°

50° + (x + 20)° + (x + 10)° = 180°

80° + 2x = 180°

2x = 180° − 80°

2x = 100°

x = 50°

First angle  =  50°

Second angle = 70°

Third angle = 60°

We can check the value of x by substituting

50° + (50 + 20)° + (50 + 10)° = 180°

Angles of triangle are 50°,70°,60°

Page 7 Exercise 19 Problem 25

We will take the sum the value of all angles of diagram.

Sum of all angles will be equal to 360°

We get 50° + (2x + 30)° + (2x + 20)° + x° = 360°

On solving we will get the value of x and angles.

Sum of all angles  = 360°

50° + (2x + 30)°+(2x + 20)° + x° = 360°

100° + 5x° = 360°

5x° = 360° − 100°

5x° = 260°

x° = \(\frac{260°}{5}\)

x° = 52°

First angle = 50°

Second angle = 134°

Third angle = 124°

Fourth angle = 52°

We can check by substituting the value of x

50° + (2 × 52 + 30)° + (2 × 52 + 20)° + 52° = 360°

x° = 52°

Angles are 50°,52°,124°,134°

Page 7  Exercise 19  Problem 26

We will substitute the sum of all values of angles that is 360°

We will get the linear equation as (x+35)°  + (x + 42)° + (x − 17)° + x° = 360°.

We will solve the linear equation to get the value of x and angles.

On solving we will get the value of x and angles.

Sum of all angles = 360°

(x + 35)° + (x + 42)° + (x−17)° + x° = 360°

60° + 4x° = 360°

4x° = 360° − 60°

4x°= 300°

x° = \(\frac{300°}{4}\)

x° = 75°

First angle = 75°

Second angle = 110°

Third angle = 117°

Fourth angle  = 58°

We can check by substituting the value of x

(75 + 35)° + (75 + 42)° + (75 − 17)° + 75° = 360°

Angles are 75°,110°,117°,58°

Page 8  Exercise 20 Problem  27

We will draw an irregular polygon of 3 sides of different lengths and angles.

We draw a triangle

Page 8  Exercise 20  Problem  28

We will measure the angles of an irregular polygon drawn in a part.

Angles are 50°,70°,60°

Page 8  Exercise 20  Problem  29

We will find the different value of angles by letting variable x.

Let, the value of x be 50∘

Then , other two angles will be (x + 20)° ,(x + 60)°

Angles in variable are x°,(x + 20)°,(x + 10)°

Page 8   Exercise 20  Problem  30

We take any polygon drawn by our friend and measure the sides.

Polygon by my friend is

Angles are 75°,88°,110°,87°

Page 8  Exercise 20  Problem  31

We will measure the angle by a protractor and then by making equation and find out equation are helpful in finding angles.

Let, Angles are (x + 35)°,(x + 48),°(x + 70)°,(x + 47)°

(x+35)° + (x + 48)°+ (x + 70)°+ (x + 47)° = 360°

x = 40°

Angles are 75°,88°,110°,87°

Yes, Answer seems reasonable.

Both method that is protractor and equations gave the same answer.

Angles are 75°,88°,110°,87°

Page 8  Exercise 21   Problem  32

We have to tell a real life problem which can be solved by the use of the multistep solution.

So say you work part-time with your three friends to mow a lawn, you get total 40 dollars.

All of you incurred 10 dollars in total to get to the work site.

Now how much money each of you will get can simply be calculated by using the multi-step equation.

Let the money each of you get be x

Now the above situation can be written in the equation form as below

3x + 10 = 40

Here 10 is the cost you incurred to reach the work site.

Subtract 10 from both sides, and we get

⇒ 3x + 10 − 10 = 40 − 10

⇒ 3x = 30

In the next step we divide both sides by 3 we get

⇒  \(\frac{3x}{3}\)

= \(\frac{30}{3}\)

⇒  x = 10

Hence each of you gets $10

This is just one of the many ways we can use the multistep equations in our real life to solve day-to-day problem

The use of multistep equation in real life are multi-faceted one such example of their utility has been given above.

Page 8  Exercise 22  Problem  33

We have to prove that the formula for the sum of interior angles is

S = 180 (n−2)

The steps of the proof of the formula will be to first prove that the sum of all angles of the triangles the polygon is made of is equal to 180×n.

Then we will prove that sum of the angles other than the interior angle is 360∘

Then upon subtracting the 360 from the total angle 180×n we will get the sum of the interior angles as

S = 180(n−2)

ABCDE is an n sides polygon, Taking any point inside the polygon and joining we get n triangles.

The sum of these triangles will be 180 × n

Thus we can write that sum of angles at O  +  sum of interior angles  = 180 × n (1)

The sum of angles at O will be 360∘because total angle of a circle is always 360∘and the angles at O make a circle.

The equation 1can then be written as

⇒ 360+ sum of interior angles  = 180 × (n)

⇒  Sum of interior angles  = 180n − 360

⇒  Sum of interior angles = 180(n−2)

Hence the theorem is proved.

The formula for the calculation of interior angle of the polygon: S = 180 (n − 2)  is correct and we have proved it above.

Page 10  Exercise 23  Problem 34

We have to solve the equation.

3x + 4 = 19 and find the value of x

We will solve it as below

First, subtract 4 on both sides, we get

⇒  3x + 4 − 4 = 19 − 4

We get as

⇒  3x = 15

Then we divide both sides by 3, we get

⇒  \(\frac{3x}{3}\) = \(\frac{15}{3}\)

⇒  x = 5

Thus the value of the variable x is

x = 5

The solution of the equation: 3x + 4 = 19  is  x = 5

Page 10  Exercise 24  Problem 35

We have to solve the equation :

17 = z − (−9) and find the value of the variable z.

First, we will solve the brackets given on the left-hand side of the equation.

Then we will subtract 9 from both sides of the equation so that the right side is free of constants.

We have to solve the equation :

17 = z − (−9) and find the value of the variable z

First, we will solve the brackets given on the left-hand side of the equation.

Then we will subtract 9 from both sides of the equation so that the right side is free of constants.

The solution of the equation: 17 = z−(−9) is  z = 8

Page 10   Exercise 25   Problem 36

We have to solve the equation:

15 = 2 + 4 − d

We have to find the value of the variable d

First, we will solve the right-hand side addition in the given equation.

We will then subtract 6 on both sides of the equation then multiply −1 to get the variable in the positive value.

15 = 2 + 4 − d

15 = 6 − d

Solving further we get

15 − 6 = 6 − d − 6

9 = −d

9 × −1 = − d × − 1

d = −9

Thus the value of the variable d is

The solution of the equation: 15 = 2 + 4 − d is d = −9

Page 10  Exercise 26  Problem 37

We have to solve the equation:

\(\frac{q+(−5)}{3}\) = 8 and find the value of the variable q

First multiply both the sides of the equation with the number 3 to remove the denominator present on the left side of the equation.

Now after opening the bracket we will add 5 to both the sides of the equation.

\(\frac{q+(−5)}{3}\) = 8

\(\frac{q+(−5)}{3}\) × 3 = 8 × 3

q+ (−5) = 24

q−5 + 5 = 24 + 5

q = 29

Thus the value of the variable q is  ⇒ q = 29

The solution of the equation\(\frac{q+(−5)}{3}\) = 8  is q  =  29

Page 10  Exercise 27  Problem 38

We have to solve the equation :

5z−2z−4=−7 and find the value of the variable,z.

First, we will add the terms of the variable z

5z − 2z − 4 = −7

3z − 4 = − 7

Add 4 on both the sides of the equation.

3x − 4 + 4 = − 7 + 4

Divide both the sides of the equation by 3

3x = −3

\(\frac{3x}{3}\) = \(\frac{−3}{3}\)

x = −1

Thus the value of the variable x is

x = −1

The solution of the equation: 5z − 2z − 4 = −7 is z = −1

Page 11  Exercise 28    Problem 39

We have to solve the equation:

3(z + 7) = 21 and find the value of the variable z

First, solve the bracket on the left-hand side of the equation:

Subtract 21 from both sides of the equation:

Since the right-hand side will be 0, after dividing both sides of the equation with 3

3(z + 7) = 21

⇒  3(z + 7) = 21

⇒  3z + 21 = 21

Solving further we get

3z + 21  − 21 = 21 − 21

3z = 0

\(\frac{3z}{3}\)= \(\frac{0}{3}\)

z = 0

Thus the value of the variable z is ⇒ z = 0

The solution of the equation : 3(z+7) = 21  is z = 0

Page 11  Exercise 29  Problem 40

To find the value of z for given equation

⇒ −4(z − 12) = 42

First applying the distributive property to the given equations, then simplifying using the mathematical operations, and then the value of  z   can be obtained.

Applying distributive property on left-hand side of the given equation.

−4(z−12) = 42

−4 × z−{(−4) × 12} = 42

As ​(−) × (−) = (+)

(−) × (+) = (−)
​
Therefore equation obtained is

− 4z − (−48) = 42

− 4z + 48 = 42

Subtracting 48 on both sides of the equation, we get

− 4z = 42 − 48

− 4z = −6

Multiplying −1 on both sides of the equation, we get

⇒ 4z=6

Dividing both sides of the equation by 4 and simplifying

\(\frac{4z}{4}\) = \(\frac{6}{4}\)

z  =  \(\frac{3}{2}\) or  z = 1.5

The value of z is 1.5 for equation − 4(z−12) = 42

Page 11  Exercise 30  Problem 41

For the given equation is  33 = 12r−3(9−r), we have to calculate the value of r.

Value of r can be obtained by using distributive property in the given equation and mathematical properties like

​(−) × (−) = (+)

(−) × (+) = (−)
​
Given equation is

33 = 12r − 3 (9 − r)

Using distributive property on right-hand side of the equation and simplifying using mathematical operations

33 = 12r + {(−3) × 9 + (−3) × (−r)}

33 = 12r + (−27 + 3r)

On simplification, we get

​33 = 12r − 27 + 3r

​Adding the terms containing unknown r, we get

33 = 15r − 27

Adding 27 on both side of the above-obtained equation and simplifying

33 + 27 = 15r − 27 + 27

60 = 15r

Dividing both sides of the above equation by 15, then simplifying

\(\frac{60}{15}\)=\(\frac{15r}{15}\)

4 = r  or  r = 4

The desired value of ris 4for the given equation, 33 = 12r − 3(9 − r).

Page 11  Exercise 31  Problem 42

Value of g is to be calculated for given equation 7 + 3 (2g − 6) = −29

The first distributive property is applied in the equation and then the mathematical operations are used to obtain the desired value of unknown g.

The given equation is

7 + 3(2g−6) = −29

Using distributive property on the left-hand side of the equation and simplifying

7 + {3 × 2g − 6 × 3} = −29

7 + 6g − 18 = −29

6g − 11 = −29

Adding 11 on both sides of the equation and then simplifying

6g − 11 + 11 = − 29 + 11

6g = −18

Dividing both sides of the equation by 6 and we get

\(\frac{6g}{6}\)= −\(\frac{−18}{6}\)

g = −3

Hence, the value of g is −3 for the equation 7 + 3(2g−6) =−29

Page 11  Exercise 32  Problem 43

Odd integer is given with an expression 2n+1where n is any integer, then three consecutive integers are

2n + 1,2n + 1 + 2, 2n + 1 + 4

Therefore 2n + 1,2n + 3,2n + 5

It is given that sum of these consecutive odd integers is 63.

On solving the value for n, consecutive odd integers are calculated.

Consecutive odd integers for an odd integer having the expression 2n + 1 are

2n + 1,2n + 3,2n + 5

As sum of consecutive odd integers is 63, therefore

(2n + 1) + (2n + 3) + (2n + 5) = 63

Simplifying above equation 6n + 9 = 63

Subtracting 9 from both sides of the equation

6n + 9 − 9 = 63 − 9

On simplifying we get 6n = 54

Dividing both sides of the equation by 6 and we get

\(\frac{6n}{6}\)= \(\){54}{6}[\latex]

n = 9

Hence, the odd integers are

First odd integer = 2n + 1

= 2 × 9 + 1

= 19

2n + 1 = 19

Second odd integer = 2n+3

= 2 × 9 + 3

= 21

2n+3 = 21

Third odd integer = 2n + 5

= 2 × 9 + 5

= 23

2n + 5 = 23

Hence, three consecutive odd integers having sum 63are 19,21and 23 having odd integer with expression 2n + 1.

Page 11 Exercise 33 Problem 44

Given:  Cousin is 8 years older than your brother.

Three years ago, cousin was twice as old as brother.

To find- Present age of brother and cousin.

First of all we will define the present ages of cousin and brother.

Then we will define their ages three years ago.

Using given condition to find the value of x and then ages of cousin and brother.

Let the present age of brother be x years.

⇒   Present age of cousin will be (x+8)years.

Three years ago

Age of brother = (x−3) years and

Age of cousin = (x + 8 − 3)years

=  (x + 5)years.

According to given condition, we get

x + 5 = 2(x − 3)

⇒ x + 5 = 2x − 6

⇒  5 + 6 = 2x − x

⇒  11 = x

⇒  Present age of brother = 11years and

Present age of cousin = 11 + 8

=  19 years.

The final answer is that the present age of brother and his cousin are 11 and 19 years respectively.