Big Ideas Math Integrated Math 1 Student Journal Solutions Chapter 4 Maintaining Mathematical Proficiency Exercise

Big Ideas Math Integrated Math  Chapter 4 Maintaining Mathematical Proficiency

 

Page 90  Exercise 1  Problem 1

Question 1.

Given the graph below, identify the coordinates of point A.

1. Determine the x-coordinate of point A by finding the perpendicular distance from the y-axis.
2. Determine the y-coordinate of point A by finding the perpendicular distance from the x-axis.
3. Write the ordered pair that corresponds to point A.

Answer:

we will find the x-coordinate of a point & the y-coordinate of a point by getting perpendicular distances from the axis.

Point A

The x-coordinate of A: 2

The y-coordinate of A: 6

The ordered pair corresponds to point A is (2,6)

The ordered pair corresponds to point A is (2,6)

 

Page 90  Exercise 2  Problem 2

Question 2.

Given the graph below, identify the coordinates of point E.

1. Determine the x-coordinate of point E by finding the perpendicular distance from the y-axis.
2. Determine the y-coordinate of point E by finding the perpendicular distance from the x-axis.
3. Write the ordered pair that corresponds to point E.

Answer:

We will find the x-coordinate of a point & the y-coordinate of a point by getting perpendicular distances from the axis.

point E

The x-coordinate of E: 0

The y-coordinate of E: − 4

The ordered pair corresponds to point E is (0,−4)

The ordered pair corresponds to point E is (0,−4)

 

Page 90  Exercise 3  Problem 3

Question 3.

Given the graph below, identify which points are located in Quadrant III of the Cartesian coordinate system.

1. List all the points from the graph.
2. Determine which points are located in Quadrant III.
3. Write the coordinates of the point(s) that are located in Quadrant III.

Answer:

Cartesian coordinates system

We will check which points in the given graph is in Quadrant III

Points in III Quadrant

F is only in III Quadrant

F is only in III Quadrant

 

Page 90  Exercise 4  Problem 4

Question 4.

Given the graph below, identify which points are located in Quadrant IV of the Cartesian coordinate system.

1. List all the points from the graph.
2. Determine which points are located in Quadrant IV.
3. Write the coordinates of the point(s) that are located in Quadrant IV.

Answer:

Cartesian coordinates system

We will check which points in the given graph is in Quadrant IV

Points in IV Quadrant

D is only in IV Quadrant

D is only in IV Quadrant

 

Page 90  Exercise 5  Problem 5

Question 5.

Given the graph below, identify which points are located on the negative x-axis of the Cartesian coordinate system.

1. List all the points from the graph.
2. Determine which points are located on the negative x-axis.
3. Write the coordinates of the point(s) that are located on the negative x-axis.

Answer:

Cartesian coordinates system

We will check which points are located on the negative x-axis

Points are located on the negative x-axis

G is the only point located on the negative x-axis

G is the only point located on the negative x-axis

 

Page 90  Exercise 6  Problem 6

Question 6.

Solve the equation x – y = -12 for the variable y.

Answer:

Given

x − y = −12

By using the isolation of variable y of the given equation, we will solve the equation for y.

x−y = −12

Subtracting x to both sides

x − y − x = −12−x

−y = − 12− x

Dividing by -1 on both sides

y = 12 + x

Solution of x − y = −12 for  y = 12 + x

 

Page 90  Exercise 7  Problem 7

Question 7.

Solve the equation 8x + 4y = 16 for the variable y.

Answer:

Given

8x + 4y = 16

By using the isolation of variable y of the given equation, we will solve the equation for y.

8x + 4y = 16

Dividing by 4 on both sides

\(\frac{8x+4y}{4}=\frac{16}{4}\)

2x + y = 4

Subtracting 2x  from both sides

2x + y −2x = 4 − 2x

y = 4 − 2x

Solution of  8x + 4y = 16 for   y = 4 − 2x

 

Page 90  Exercise 8  Problem 8

Question 8.

Solve the equation 3x – 5y + 15 = 0 for the variable y.

Answer:

Given

3x−5y + 15 = 0

By using the isolation of variable y of the given equation, we will solve the equation for y.

3x − 5y + 15 = 0

Adding 5y on both sides

3x − 5y + 15 + 5y = 0 + 5y

3x + 15 = 5y

Dividing by 5 on both sides

\(\frac{3x+15}{5}\)= y

Interchanging both sides

y = \(\frac{3x+15}{5}\)

Solution of 3x−5y+15=0 for y = \(\frac{3x+15}{5}\)

 

Page 90  Exercise 9  Problem 9

Question 9.

Solve the equation 0 = 3y – 6x + 12 for the variable y.

Answer:

Given

0 = 3y−6x+12

By using the isolation of variable y of the given equation, we will solve the equation for y.

0 = 3y − 6x + 12

Subtracting 3y from both sides

0 − 3y = 3y − 6x + 12 − 3y

−3y = −6x + 12

Dividing by −3 both sides

\(\frac{−3y}{−3}\)=\(\frac{−6x+12}{−3}\)

y = 2x − 4

Solution of 0 = 3y − 6x + 12 for y = 2x − 4

 

Page 91 Exercise 10  Problem 10

Question 10.

Given two points (0,-1) and (2,3) on a graph, find the slope of the line passing through these points. Using the slope and one of the points, determine the y-intercept and write the equation of the line. Finally, plot the line and identify its slope and y-intercept.

Answer:

Given

(0,-1) and (2,3)

From the given graph, we can find the slope of the line where 2 points are known using slope & given point we will find both intercept of line

The slope of the line from points (0,−1) & (2,3)

m = \(\frac{−1−3}{0−2}\)

m = 2

y = 2x + c

Putting point (2,3) in the equation

3 = 2 × 2 + c

c = −1

Equation of line y = 2x − 1

Plot

y-Interceptc = −1 , Slope m = 2, Liney = 2x−1

 

Page 91 Exercise 10  Problem 11

Question 11.

Given two points (0,2) and (4,-2) on a graph, find the slope of the line passing through these points. Using the slope and one of the points, determine the y-intercept and write the equation of the line. Finally, plot the line and identify its slope and y-intercept.

Answer:

Given

(0,2) and (4,-2)

From the given graph, we can find the slope of the line where 2 points(0,2)&(4,−2) are known using slope & given point we will find both intercepts of the line

The slope of the line from points (0,2) & (4,−2)

m = \(\frac{−2−2}{4−0}\)

m = −1

y = −x + c

Putting point(0,2) in the equation

2 = 0 + c

c = 2

Equation of line y = −x + 2

Plot

y-Intercept c = 2 , Slope m = −1, Line y = −x + 2

 

Page 91 Exercise 10  Problem 12

Question 12.

Given two points (−3,3) and (3,−1) on a graph, find the slope of the line passing through these points. Using the slope and one of the points, determine the y-intercept and write the equation of the line. Finally, plot the line and identify its slope and y-intercept.

Answer:

Given

(−3,3) and (3,−1)

From the given graph, we can find the slope of the line where 2 points (−3,3) & (3,−1)are known using slope & given point we will find both intercept of line

The slope of the line from points (−3,3)&(3,−1)

m = \(\frac{−1−3}{3+3}\)

m = \(\frac{−4}{6}\)

m = \(\frac{−2}{3}\)

y= \(\frac{−2}{3}\) x + c

From(3,−1)

−1 = \(\frac{−2}{3}\) (3) + c

Cancellation to 3

−1= −2+c

c = 2−1

c = 1

y = \(\frac{−2}{3}\)x+1

Plot

y-Intercept c=1, Slope m = \(\frac{−2}{3}\), Line y = \(\frac{−2}{3}\) x + 1

 

Page 91  Exercise 10  Problem 13

Question 13.

Given two points (4,0) and (2,−1) on a graph, find the slope of the line passing through these points. Using the slope and one of the points, determine the y-intercept and write the equation of the line. Finally, plot the line and identify its slope and y-intercept.

Answer:

Given

(4,0) and (2,−1)

From the given graph, we can find the slope of the line where 2 points(4,0)&(2,−1) are known using slope and given point we will find both intercepts of line

The slope of the line from points (4,0) & (2,−1)

Slope m = \(\frac{−1−0}{2−4}\)

m = \(\frac{1}{2}\)

y=\(\frac{1}{2}\)x + c

From point (4,0)

0 = \(\frac{1}{2}\)(4) + c

c = −2

y = \(\frac{x}{2}\) − 2

Plot

y-Intercept c = −2, Slope m = \(\frac{1}{2}\), Line y = \(\frac{x}{2}\) − 2

 

Page 92  Exercise 11  Problem 14

Question 14.

Given two points (0,20) and (2000,80) on a graph, find the slope of the line passing through these points. Using the slope and one of the points, determine the y-intercept and write the equation of the line.

Answer:

Given

(0,20) and (2000,80)

From the given graph, we can find the slope of the line where 2 points (0,20) & (2000,80)are known using slope & given point we will find both intercepts of line

The slope of the line from points (0,20) & (2000,80)

m = \(\frac{3}{100}\)

y-Intercept

c = 20

Using these

y = \(\frac{3}{100}\) x+20

Equation of a given line y = \(\frac{3}{100}\) x+20

 

Page 94  Exercise 12  Problem 15

Question 15.

Find the equation of a line given its slope m = 2 and y-intercept c = −3

Answer:

Given

m = 2 and y-intercept c = −3

Using slope & y-intercept we will find the equation of the line

y = mx + c

Putting m & c values

y = 2x − 3

Equation of the line y = 2x − 3

 

Page 94 Exercise 12 problem 16

Question 16.

Find the equation of a line given its slope m = −3 and y-intercept c = 7.

Answer:

Given

m = −3 and y-intercept c = 7.

Using slope & y-intercept we will find the equation of the line

y = mx + c

Putting m & c values

y =−3x + 7

Equation of the line  y = −3x + 7

 

Page 94  Exercise 13  Problem 17

Question 17.

Find the equation of a line given its slope \(m = /frac{1}{3}\) and y-intecept c = 2.

Answer:

Given

\(m = /frac{1}{3}\) and y-intecept c = 2.

Using slope & y-intercept we will find the equation of the line

y = mx + c

Putting m & c values

y = \(\frac{1}{3}\) x + 2

Equation of the line y = \(\frac{1}{3}\) x + 2

 

Page 94  Exercise 14  Problem 18

Question 18.

Given two points (0,0) and (2,4) on a graph, find the slope of the line passing through these points. Using the slope and one of the points, determine the y-intercept and write the equation of the line.

Answer:

We will get the slope from the given points & put the slope value in the line equation.

By using the known slope equation we use one point to get the value of y-intercept by satisfying that point

Given points (0,0) & (2,4)

Slope  m = \(\frac{4−0}{2−0}\)

m = 2

y = 2x + c

Using (0,0)

c = 0

y = 2x

Equation of the line in slope-intercept form y = 2x

 

Page 94 Exercise 15 Problem 19

Question 19.

Given two points (0,5) and (3,−4) on a graph, find the slope of the line passing through these points. Using the slope and one of the points, determine the y-intercept and write the equation of the line.

Answer:

We will get the slope from the given points & put the slope value in the line equation.

By using the known slope equation we use one point to get the value of y-intercept by satisfying that point

Given points (0,5) & (3,-4)

Slope  m = \(\frac{−4−5}{3−0}\)

m = −3

y = −3x + c

Using point (0,5)

5 = 0 + c

c = 5

y = −3x + 5

Equation of the line in the slope-intercept form y = −3x + 5

 

Page 94  Exercise 16  Problem 20

Question 20.

Given two points (−2,−3) and (0,−2) on a graph, find the slope of the line passing through these points. Using the slope and one of the points, determine the y-intercept and write the equation of the line.

Answer:

We will get the slope from the given points & put the slope value in the line equation.

By using the known slope equation we use one point to get the value of y-intercept by satisfying that point

Given points(-2,-3) & (0,-2)

Slope  m = \(\frac{−2+3}{0+2}\)

m = \(\frac{1}{2}\)

y = \(\frac{1}{2}\)x + c

Using point (0,-2)

c = −2

y = \(\frac{1}{2}\) x−2

Equation of the line in slope-intercept form y = \(\frac{1}{2}\) x−2

 

Page 94 Exercise 17 Problem 21

Question 21.

Given two points (0,3) and (4,0) on a graph, find the slope of the line passing through these points. Using the slope and one of the points, determine the y-intercept and write the equation of the line.

Answer:

We will get the slope from the given points & put the slope value in the line equation.

By using the known slope equation we use one point to get the value of y-intercept by satisfying that point

Given points (0,3) & (4,0)

Slope m = \(\frac{−3}{4}\)

y = \(\frac{−3}{4}\)x+c

Using (0,3)

c = 3

y = \(\frac{−3}{4}\) x + 3

Equation of the line in slope-intercept form y = \(\frac{−3}{4}\) x + 3

 

Page 95  Exercise 18  Problem 22

Question 22.

Given two points (2,1) and (0,−7) on a graph, find the slope of the line passing through these points. Using the slope and one of the points, determine the y-intercept and write the equation of the line.

Answer:

We will get the slope from the given points & put the slope value in the line equation.

By using the known slope equation we use one point to get the value of the y-intercept by satisfying that point

Given points (2,1) & (0,-7)

Slope m = \(\frac{−7−1}{0−2}\)

m = \(\frac{−8}{−2}\)

m = 4

y = 4x+c

Using point (0,-7)

c = −7

y = 4x − 7

Equation of the line in slope-intercept form y = 4x − 7

 

Page 95  Exercise 19 Problem 23

Question 23.

Given two points (0,2) and (4,3) on a graph, find the slope of the line passing through these points. Using the slope and one of the points, determine the y-intercept and write the equation of the line.

Answer:

We will get the slope from the given points & put the slope value in the line equation.

By using the known slope equation we use one point to get the value of y-intercept by satisfying that point

Given points (0,2) & (4,3)

Slope m = \(\frac{3−2}{4}\)

m = \(\frac{1}{4}\)

y = \(\frac{1}{4}\)x + c

Using point (0,2)

c = 2

y= \(\frac{1}{4}\) x+2

Equation of the line in slope-intercept form y = \(\frac{1}{4}\) x + 2

 

Page 95  Exercise 20  Problem 24

Question 24.

Given two points (8,0) and (0,8) on a graph, find the slope of the line passing through these points. Using the slope and one of the points, determine the y-intercept and write the equation of the line.

Answer:

We will get the slope from the given points & put the slope value in the line equation.

By using the known slope equation we use one point to get the value of the y-intercept by satisfying that point

Given points (8,0) & (0,8)

Slope m = \(\frac{−8}{8}\)

m = −1

y = −x + c

Using (0,8)

c = 8

y = −x + 8

Equation of the line in slope-intercept form y = −x + 8

 

Page 95  Exercise 21  Problem 25

Question 25.

Given two points (0,3) and (2,−5) on a graph, find the slope of the line passing through these points. Using the slope and one of the points, determine the y-intercept and write the equation of the line.

Answer:

We will get the slope from the given points & put the slope value in the line equation.

By using the known slope equation we use one point to get the value of y-intercept by satisfying that point

Given points (0,3) & (2,-5)

Slope m = \(\frac{−5−3}{2−0}\)

m = −4

y = −4x + c

Using point (0,3)

c = 3

y = −4x + 3

Equation of the line in the slope-intercept form y = −4x + 3

 

Page 95  Exercise 22  Problem 26

Question 26.

Given the linear function f(x)= c + dx and the following conditions:

1. f(0) = 5
2. f(9) = -4

Determine the values of the constants c and d. Write the final equation of the linear function.

Answer:

We will put given conditions in the linear function by which we will get values of constants (c,d)

f(x) = c + dx

Given f(0) = 5

f(0) = c

c = 5

f (9) = c + 9d

−4 = 5 + 9d

d = \(\frac{−9}{9}\)

d = −1

f(x) = 5 − x

A linear function f with the given values, f(x) = 5 − x

 

Page 95  Exercise 23  Problem 27

Question 27.

Given the linear function f(x)= c + dx and the following conditions:

1. f(0) = 10
2. f(7) = -4

Determine the values of the constants c and d. Write the final equation of the linear function.

Answer:

We will put given conditions in the linear function by which we will get values of constants (c,d)

Given f(0) = 10

f(0) = c

c = 10

f(7) = c + 7d

−4 = 10 + 7d

d = \(\frac{−14}{7}\)

d = −2

f(x) = 10−2x

A linear function f with the given values f(x) = 10 − 2x

 

Page 95  Exercise 24  Problem 28

Question 28.

Given the linear function f(x)= c + dx and the following conditions:

1. f(0) = 2
2. f(-2) = -2

Determine the values of the constants c and d. Write the final equation of the linear function.

Answer:

We will put given conditions in the linear function by which we will get values of constants (c,d)

f(x) = c + dx

Given f(0) = 2

f(0) = c

c = 2

f(−2) = c−2d

−2 = 2 −2d

d = \(\frac{−4}{−2}\)

d = 2

f(x) = 2 + 2d

A linear function f with the given values f(x) = 2 + 2d

 

Page 95  Exercise 25  Problem 29

Question 29.

An electrician charges a fixed initial fee and an hourly rate for their services. The total cost function f(x) is given by f(x)=c+dx, where x is the number of hours worked.

Given the following conditions:

1. The initial fee (when x=0) is 50.
2. After 4 hours of work, the total cost is 190.

Determine the values of the constants c and d. Write the final linear model that represents the total cost as a function of the number of hours worked.

Answer:

f(x) = c + dx in this f(x) is electrician charges function of hours of work x

We will put given conditions in the linear function f(x) = c + dx by which we will get values of constants (c,d)

f(x) = c + dx  , At x = 0 initially

f(0) = c

50 = c

After 4 hours of work

f(4) = c + 4d

190 = 50 + 4d

4d = 140

d = 35

f(x) = 50 + 35x

A linear model that represents the total cost as a function of the number of hours worked. f(x) = 50 + 35x

 

Page 95  Exercise 25  Problem 30

Question 30.

An electrician charges a fixed initial fee and an hourly rate for their services. The total cost function f(x) is given by f(x)= 50 + 35x, where x is the number of hours worked.

1. Identify the slope (or gradient) of the linear function.
2. Determine the electrician’s charge per hour.

Answer:

In the function that we find d is the charges per hour which is the gradient(slope) of the linear function

Linear model

In this slope or gradient is 35

The electrician charge per hour = 35

The electrician charge per hour = 35

Page 96  Exercise 26  Problem 31

Question 31.

Given the slope m and a point (x₁,y₁) through which a line passes, use the point-slope form of the equation of a line to find the equation of the line.

1. The slope m=4
2. The point (x₁,y₁) = (2,3)

Write the equation of the line using the point-slope form y −y₁ =m(x−x₁).

Answer:

Given slope & point, we will put them in the equation of the line when one point & slope is given

y − y₁  =  m(x−x₁)  by putting slope & one point values in this equation we get the equation of line

Equation of line when slope & one point is given y−y₁ =  m(x−x₁)

 

Page 96  Exercise  27  Problem 32

Question 32.

Given the slope m and a point (x₁,y₁) through which a line passes, use the point-slope form of the equation of a line to find the equation of the line.

1. The slope m = \(\frac{1}{2}\)
2. The point (x₁,y₁) = (3,2)

Answer:

The slope is given & from the given graph we will get the points & use them to find the equation by putting them in the slope-point form of the equation

Slope m = \(\frac{1}{2}\)

Given point (3,2)

Using the slop-point form of the equation

y−y₁ = m(x−x₁)

y−2 = \(\frac{1}{2}\) (x−3)

Simplifying  2y − 4 = x−3

2y  =  x + 1

Equation of the line 2y = x + 1

 

Page 96  Exercise 28   Problem 33

Question 33.

Given the slope m=−2 and a point (−4,6) through which the line passes, use the point-slope form of the equation of a line to find the equation of the line. Then simplify the equation to the standard form.

Answer:

The slope is given & from the given graph we will get the point & use them to find equation by putting them in the slope-point form of equation

Slope m =−2

Given point (-4,6) using the slop-point form of the equation

y−y₁ = m(x−x₁)

y−6 = −2(x + 4)

Simplifying

y−6 = −2x−8

y + 2x + 2 = 0

Equation of the line y + 2x + 2 = 0

 

Page 96  Exercise 29  Problem 34

Question 34.

Given the slope m and a point (x₁,y₁) through which the line passes, use the point-slope form of the equation of a line to find the equation of the line.

1. Find the equation of the line given the following conditions:
   • Slope: m = 3
   • Point: (2,−1)
2. Find the equation of the line given the following conditions:
   • Slope: m = \(-\frac{1}{2}\)
   • Point: (4,5)

Answer:

The slope is given & from the given graph we will get the point & use them to find equation by putting them in the slope-point form of equatio

Slope is m

Given point (x₁,y₁)

Equation of the line when one point & slope is given

y − y₁ = m(x−x₁)

Equation of the line y−y₁ = m(x − x₁)

 

Page 97  Exercise 30  Problem 35

Question 35.

Given the slope m=25 representing the savings rate in dollars per month and a point (4,175) representing the savings after 4 months, use the point-slope form of the equation to find the equation of the line that models the savings over time.

1. Write a point-slope from the equation y – y₁ = m(x-x₁) using the given slope and points
2. Simplify the equation to the slope-intercept form.

Answer:

The slope is given & from the given graph we will get the point & use them to find equation by putting them in the slope-point form of equation

Given slope gradient of saving is $25 / month

m = 25

Given point (4,175)

y−y₁= m(x−x₁)

y−175 = 25(x−4)

Simplifying

y − 175 = 25x−100

y = 25x + 75

where y is saving & x is number of months

The equation that represents the balance A after x months. y = 25x + 75

 

Page 97  Exercise 30  Problem 36

Question 36.

Given the slope m=4 and the y-intercept 175, find the equation of the line in slope-intercept form and plot it on a graph to verify if it matches the given Plot.

Answer:

The slope is given & from the given graph we will get the point & use them to find the equation by putting them in the slope-point form of the equation.

We will plot equation on the graph & match with given plot

y = 4x + 175

Plotted graph is same as given

 

Page 97  Exercise 31  Problem 37

Question 37. Given the slope

m = 3 and a y-intercept = 175, use the slope-intercept form of the equation to find the equation of the line that models the given scenario.

1. Identify the slope and y-intercept:
   • Given: m = 4
   • y-intercept = 175
2. Write the slope-intercept form of the equation y = mx + b using the given slope and y-intercept.
3. Plot the equation on a graph and verify if it matches the given plot.

Answer:

The slope is given & from the given graph we will get the point & use them to find equation by putting them in the slope-point form of equation.

y−y₁ = m(x−x₁) by putting slope & one point values in this equation we get the equation of line

Equation of line when slope & one point is given  y−y₁ = m(x−x₁)

 

Page 97   Exercise  31  Problem 38

Question 38.

Given the slope m = 5 and a point (0,0) which is the origin, use the point-slope form of the equation to find the equation of the line.

1. Identify the slope and point:
   • Given: m = 5
   • (x₁,y₁) = (0,0)
2. Write the point-slope form of the equation y-y₁ = m(x-x₁) using the given slope and point.
3. Simplify the equation to the slope-intercept form.

Answer:

The slope is given & from the given graph we will get the point & use them to find equation by putting them in the slope-point form of equation

Let’s assume m = 5 & given point is the origin

Equation of the line when one point & slope is given

y−y₁ = m(x−x₁)

y−0 =  5(x−0)

y = 5x

An equation of a line when you are given the  Slope m = 5 and a Point(0,0) on the line  y = mx

 

Page 99  Exercise  32  Problem 39

Question 39.

Given the slope m = -1 and a point (-2,1), use the point-slope from of the equation to find the equation of the line.

1. Identify the slope and pont:
   • Given: m = -1
   • (x₁,y₁) = (-2,1)
2. Write the point-slope form of the equation y−y₁ = m(x−x₁) using the given slope and point.
3. Simplify the equation to the standard form.

Answer:

The slope is given & from the given graph we will get the point & use them to find equation by putting them in the slope-point form of equation

Given slope m = −1

Point(-2,1) equation of the line when one point & slope is given

y−y₁ =  m(x−x₁)

y−1 = −(x+2)

y = −x−1

Equation of a line  y = −x−1

 

Page 99   Exercise 32  Problem 40

Question 40.

Given the slope m = \(\frac{4}{3}\) and a point (-2,1), use the point-slope form of the equation to find the equation of the line.

1. Identify the slope and point:
   • Given: m = \(\frac{4}{3}\)
   • (x₁,y₁) = (-2,1)
2. Write the point-slope form of the equation y−y₁ = m(x−x₁) using the given slope and point.
3. Simplify the equation to the standard form.

Answer:

The slope is given & from the given graph we will get the point & use them to find equation by putting them in the slope-point form of equation

Given slope

m = \(\frac{4}{3}\)

Point(-2,1) equation of the line when one point & slope is given

y−y₁ = m(x−x₁)

y−1 = \(\frac{4}{3}\) (x+2)

3y−3 = 4x + 8

3y = 4x + 11

Equation of a line 3y = 4x + 11

 

Page 99  Exercise 33  Problem 41

Question 41.

Given the slope m = \(\frac{1}{2}\) and a point (-2,1), use the point-slope from of the equation to find the equation of the line.

1. Identify the slope and pont:
   • Given: m = \(\frac{1}{2}\)
   • (x₁,y₁) = (-2,1)
2. Write the point-slope form of the equation y−y₁ = m(x−x₁) using the given slope and point.
3. Simplify the equation to the standard form.

Answer:

The slope is given & from the given graph we will get the point & use them to find equation by putting them in the slope-point form of equation

Given slope

m = − \(\frac{−1}{2}\)

Point(-2,1) equation of the line when one point & slope is given

y−y₁ =  m(x−x₁)

y−1 = −\(\frac{−1}{2}\) (x+2)

2y−2 = −x−2

2y = −x

Equation of line 2y = −x

 

Page 99   Exercise 34   Problem 42

Question 42.

Given the points (1, 2) and (4, -1), find the equation of the line in slope-intercept form.

1. Calculate the slope m using the given points \(m=\frac{y_2-y_1}{x_2-x_1}\) where (x₁,y₁) = (1,2) and (x₂,y₂) = (4,-1)
2. Write the equation of the line using the point-slope from y-y₁ = m(x-x₁) and simplify it to slope-intercept from y = mx + c.

Answer:

We will get the slope from the given points & put the slope value in the line equation.

By using the known slope equation we use one point to get the value of y-intercept by satisfying that point

Given points (1,2) & (4,-1)

Slope m = \(\frac{2+1}{1−4}\)

m = −1

y = −x+c using point (1,2)

2 = −1+c

c = 3

y = −x + 3

Equation of the line in slope-intercept form y = −x + 3

 

Page 99  Exercise 35  Problem 43

Question 43.

Given the points (−4,1) and (4,−1), find the equation of the line in slope-intercept form.

1. Calculate the slope m using the given points: \(m=\frac{y_2-y_1}{x_2-x_1}\) where (x₁,y₁) = (-4,1) and (x₂,y₂) = (4,-1)
2. Write the equation of the line using the point-slope from y-y₁ = m(x-x₁) and simplify it to slope-intercept from y = mx + c.

Answer:

We will get the slope from the given points & put the slope value in the line equation.

By using the known slope equation we use one point to get the value of the y-intercept by satisfying that point

Given points (−4,1) & (4,−1)

m = \(\frac{1+1}{−4−4}\)

m = \(\frac{−1}{4}\)

y = \(\frac{−1}{4}\) x+c

Using point (4,-1)

−1 = \(\frac{−1}{4}\) × 4 + c

c = 0

y = \(\frac{−1}{4}\) x

Equation of the line in slope-intercept form y = \(\frac{−1}{4}\) x

 

Page 99  Exercise 36  Problem 44

Question 44.

Given points (-1,-9) and (1,-3), find the equation of the line in slope-intercept form.

1. Calculate the slope m using the given points: \(m=\frac{y_2-y_1}{x_2-x_1}\) where (x₁,y₁) = (-1,-9) and (x₂,y₂) = (1,-3)
2. Write the equation of the line using the slope-intercept form y = mx + c and determine the value of c by substituting one of the given points.

Answer:

We will get the slope from the given points & put the slope value in the line equation.

By using the known slope equation we use one point to get the value of y-intercept by satisfying that point

Given points (-1,-9) & (1,-3)

Slope m = \(\frac{−3+9}{1+1}\)

m = 3

y = 3x + c

Using point (1,-3) −3 = 3 × 1 + c

c = −6

y = 3x − 6

Equation of the line in slope-intercept form y = 3x − 6

 

Page 99  Exercise 37  Problem 45

Question 45.

Given the points (1,2) and (2,0), find the equation of the line in slope-intercept form.

1. Calculate the slope m using the given points: \(m=\frac{y_2-y_1}{x_2-x_1}\) where (x₁,y₁) = (1,2) and (x₂,y₂) = (2,0)
2. Write the equation of the line using the point-slope from y-y₁ = m(x-x₁) and simplify it to slope-intercept from y = mx + c.

Answer:

We will get the slope from the given points & put the slope value in the line equation.

By using the known slope equation we use one point to get the value of the y-intercept by satisfying that point

Given points (1,2) & (2,0)

Slope m = \(\frac{2−0}{1−2}\)

m = −2

y = −2x + c

Using point (2,0)

0 = −2 × 2 + c

c = 4

y = −2x + 4

Equation of the line in the slope-intercept form y = −2x + 4

 

Page 100  Exercise 38  Problem 46

Question 46.

Given the linear function f(x)= c+dx and the following conditions:

1. f(1) = 7
2. f(-2) = 1

Find the values of the constants c and d and write the linear function.

Answer:

We will put given conditions in the linear function by which we will get values of constants (c,d)

We will solve 2 equation to get 2 unknowns (c,d)

f(x) = c + dx

Using given conditions

f(1) = c + d

7 = c + d

f(−2) = c−2d

1 = c−2d

Solving both equations d = 2 & c = 5

f(x) = 5 + 2x

A linear function f with the given values f(x) = 5+2x

 

Page 100  Exercise 39  Problem 47

Question 47.

Given the linear function f(x)= c+dx and the following conditions:

1. f(−1) = 2
2. f(3) = 3

Find the values of the constants c and d and write the linear function.

Answer:

We will put given conditions in the linear function by which we will get values of constants (c,d)

We will solve 2 equation to get 2 unknowns (c,d)

f(x) = c + dx

Using given conditions

f(−1) = c−d

2 = c−d

f(3) = c + 3d

3 = c + 3d

Solving both equations

d = \(\frac{1}{4}\) &c = \(\frac{9}{4}\)

f(x)= \(\frac{9}{4}\)+\(\frac{1}{4}\)x

A linear function f with the given values f(x)= \(\frac{9}{4}\)+\(\frac{1}{4}\)x

 

Page 100  Exercise 40  Problem 48

Question 48.

Given the linear function f(x)= c+dx and the following conditions:

1. f(0) = -2
2. f(4) = -1

Find the values of the constants c and d and write the linear function.

Answer:

We will put given conditions in the linear function by which we will get values of constants (c,d)

We will solve 2 equation to get 2 unknowns (c,d)

f(x) = c + dx

Using given conditions

f(0) = c

−2 = c

f(4) = c + 4d

−1 = −2 + 4d

d = \(\frac{1}{4}\)

f(x) = −2  + \(\frac{1}{4}\)x

A linear function f with the given values f(x) = −2+\(\frac{1}{4}\)x

 

Page 100  Exercise 41  Problem 49

Question 49.

Given the linear function f(x)= c+dx and the following conditions:

1. f(3) = 5
2. f(2) = 6

Find the values of the constants c and d and write the linear function.

Answer:

We will put given conditions in the linear function by which we will get values of constants (c,d).

We will solve 2 equation to get 2 unknowns (c,d).

f(x) = c + dx

Using given conditions

f(3) = c + 3d​

5 = c + 3d

f(2)= c + 2d

6 = c + 2d

Solving both equations

d = −1 & c =  8

f(x) = 8−x

A linear function f with the given values f(x) = 8−x

 

Page 100  Exercise 42  Problem 50

Question 50.

Given a set of points (−3,−98) and (−1,18), determine if the data in the table can be modeled by a linear equation. Follow these steps:

1. Find the slope (m) of the line passing through the given points.
2. Use the slope and one of the points to find the y-intercept (c).
3. Write the equation of the line in slope-intercept form.
4. Check if the equation satisfies another point from the data table to confirm if the data can be modeled by a linear equation.

Answer:

We will get the slope from the given points & put the slope value in the line equation.

By using the known slope equation we use one point to get the value of y-intercept by satisfying that point.

After getting we will check whether given data satisfy the equation, if yes then the data in the table can be modeled by a linear equation

Equation from points (-3,-98) & (-1,18)

Slope

m = \(\frac{18+98}{−1+3}\)

m = 58

y = 58x + c

From point (-1,18)

18 =−58 + c

y = 58x + 76

y = 58x + 76

For 0

y = 58 × 0 + 76

y = 76 not on line data in the table can not be modeled by a linear equation

Data in the table can not be modeled by a linear equation

 

Page 100  Exercise 43  Problem 51

Question 51.

Given the conditions, we need to create a linear function D(h) representing the miles driven after h hours. We are given that the gradient of the distance (d) is 60 miles per hour, and at h=3, the distance D=265 miles.

1. Determine the linear function D(h) that represents the miles driven after h hours.
2. Calculate the odometer reading after 7 hours of continuous driving.

Answer:

We will put given conditions in the linear function by which we will get values of constants (c,d)

After using values of h we can get D corresponding h using the function

D(h) = c + dh

Where given gradient of distance d = 60

At h = 3 D=265

Using these D(h) = c + 60h

265 = c + 60 × 3

c = 85

Function D(h) = 85 + 60h

D(h) = 85 + 60h

At h = 7

D(h) = 85 + 60 × 7

D(7) = 505

Linear function D that represents the miles driven after h hours   D(h) = 85 + 60h , The odometer read after 7 hours of continuous driving  D(7) = 505