Big Ideas Math Integrated Math 1 Student Journal Solutions Chapter 2 Maintaining Mathematical Proficiency Exercise

Big Ideas Math Integrated Math  Chapter 2 Maintaining Mathematical Proficiency

 

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

 

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

 

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:

Given

|-5|

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

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

So here number 5 is equal to 5  Or  ∣−5∣ = 5

The desired solution of this is ∣−5∣ = 5.

 

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∣.

 

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∣.

 

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 

 

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.

 

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

 

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

 

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

 

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: