lesson 1 problem solving practice lines answer key

Solving Systems by Substitution

13.1: Math Talk: Is It a Match? (10 minutes)

CCSS Standards

Building On

• HSA-REI.D.10

Routines and Materials

Instructional Routines

• MLR8: Discussion Supports

This Math Talk encourages students to look for connections between the features of graphs and of linear equations that each represent a system. Given two graphs on an unlabeled coordinate plane, students must rely on what they know about horizontal and vertical lines, intercepts, and slope to determine if the graphs could represent each pair of equations. The equations presented and the reasoning elicited here will be helpful later in the lesson, when students solve systems of equations by substitution.

All four systems include an equation for either a horizontal or a vertical line. Some students may remember that the equation for such lines can be written as \(x = a\) or  \(y=b\) , where \(a\) and \(b\)  are constants. (In each of the first three systems, one equation is already in this form. In the last system, a simple rearrangement to one equation would put it in this form.) Activating this knowledge would enable students to quickly tell whether a system matches the given graphs.

Those who don't recall it can still reason about the system structurally. For instance, given a system with \(x=\text-5\) as one of the equations, they may reason that any point that has a negative \(x\) -value will be to the left of the vertical axis. The solution (if there is one) to this system would have to have -5 for the  \(x\) -value. The intersection of the given graphs is a point to the right of the vertical axis (and therefore having a positive \(x\) -value), so the graphs cannot represent that system.

To match graphs and equations, students need to look for and make use of structure (MP7) in both representations. In explaining their strategies, students need to be precise in their word choice and use of language (MP6). 

Because the warm-up is intended to promote reasoning, discourage the use of graphing technology to graph the systems.

Display one system at a time. Give students quiet think time for each problem and ask them to give a signal when they have an answer and a strategy. Keep all problems displayed throughout the talk. Follow with a whole-class discussion.

Student Facing

Here are graphs of two equations in a system.

Description: <p>Graph of 2 intersecting lines, origin O, in first quadrant. No labels or scale. Line 1 starts on vertical axis and trends downward and right. Line 2 is exactly vertical and intersects around the middle of Line 1.</p>

Determine if each of these systems could be represented by the graphs. Be prepared to explain how you know. 

\(\begin{cases} x + 2y = 8 \\x = \text-5 \end{cases}\)

\(\begin{cases} y = \text-7x + 13 \\y = \text-1 \end{cases}\)

\(\begin{cases} 3x = 8\\3x + y = 15 \end{cases}\)

\(\begin{cases} y = 2x - 7\\4 + y = 12 \end{cases}\)

Student Response

For access, consult one of our IM Certified Partners .

Activity Synthesis

Ask students to share their strategies for each problem. Record and display their responses for all to see. To involve more students in the conversation, consider asking:

• “Who can restate \(\underline{\hspace{.5in}}\) ’s reasoning in a different way?”
• “Did anyone have the same strategy but would explain it differently?”
• “Did anyone solve the problem in a different way?”
• “Does anyone want to add on to \(\underline{\hspace{.5in}}\) ’s strategy?”
• “Do you agree or disagree? Why?”

If no students mentioned solving the systems and then checking to see if the solution could match the graphs, ask if anyone approached it that way. For instance, ask: “How could we find the solution to the second system without graphing?” Give students a moment to discuss their ideas with a partner and then proceed to the next activity.

13.2: Four Systems (15 minutes)

• HSA-REI.C.6
• Think Pair Share

In this activity, students see the same four pairs of equations as those in the warm-up. This time, their job is to find a way to solve the systems. Some students may choose to solve by graphing, but the systems lend themselves to be solved efficiently and precisely by substitution.

As students work, pay attention to the methods students use to solve the systems. Identify those who solve by substitution—by replacing a variable or an expression in one equation with an equal value or equivalent expression from the other equation. Ask these students to share later.

Arrange students in groups of 2. Give students 6–8 minutes of quiet time to solve as many systems as they can and then a couple of minutes to share their responses and strategies with their partner.

Here are four systems of equations you saw earlier. Solve each system. Then, check your solutions by substituting them into the original equations to see if the equations are true.

A\(\begin{cases} x + 2y = 8 \\x = \text-5 \end{cases}\)

B\(\begin{cases} y = \text-7x + 13 \\y = \text-1 \end{cases}\)

C\(\begin{cases} 3x = 8\\3x + y = 15 \end{cases}\)

D\(\begin{cases} y = 2x - 7\\4 + y = 12 \end{cases}\)

Anticipated Misconceptions

Some students may not remember to find the value of the second variable after finding the first. They may need a reminder that the solution to a system of linear equations is a pair of values.

If some students struggle with the last system because the variable that is already isolated is equal to an expression rather than a number, ask what they would do if the first equation were \(y= \text{a number}\)  instead of \(y=2x-7\) .

If students don't know how to approach the last system, ask them to analyze both equations and see if the value of one of the variables could be found easily.

Select previously identified students to share their responses and strategies. Display their work for all to see. Highlight the strategies that involve substitution and name them as such.

Make sure students see that the last two equations can be solved by substituting in different ways. Here are two ways for solving the third system,  \(\begin{cases} 3x = 8\\3x + y = 15 \end{cases} \) , by substitution:

Finding the value of \(x\) and substituting it  into \(3x+8=15\) :

\(\begin {align} 3x&=8\\x&=\frac83\\ \\3x+y &=15\\ 3(\frac83) + y &=15\\8+y &=15\\y&=7 \end{align}\)

Substituting the value of \(3x\) into \(3x+8=15\) :

\(\begin {align} 3x+y &=15\\ 8 + y &=15\\y&=7 \end{align}\)

Here are two ways of solving the last system,  \(\begin{cases} y = 2x - 7\\4 + y = 12 \end{cases}\) , by substitution:

Substituting \(2x - 7\) for \(y\) in the equation  \(4 + y = 12\) :

\(\begin {align} 4+y&=12\\4 + (2x-7) &=12\\4 + 2x - 7 &=12\\ 2x -7 + 4 &=12\\ 2x-3&=12\\2x &=15\\x &=7.5\\ \\y&=2x - 7\\y&=2(7.5) - 7\\ y&=15-7\\y&=8 \end{align}\)

Rearranging or solving \(4+ y=12\) to get \(y = 8\) , and then substituting 8 for \(y\) in the equation  \(y=2x - 7\) :

\(\begin {align} y&=2x - 7\\8&=2x - 7\\ 15&=2x \\ 7.5 &=x \end{align}\)

In each of these two systems, students are likely to notice that one way of substituting is much quicker than the other. Emphasize that when one of the variables is already isolated or can be easily isolated, substituting the value of that variable (or the expression that is equal to that variable) into the other equation in the system can be an efficient way to solve the system.

13.3: What about Now? (10 minutes)

• MLR7: Compare and Connect

The activity allows students to practice solving systems of linear equations by substitution and reinforces the idea that there are multiple ways to perform substitution. Students are directed to find the solutions without graphing.

Monitor for the different ways that students use substitutions to solve the systems. Invite students with different approaches to share later.

Keep students in groups of 2. Give students a few minutes to work quietly and then time to discuss their work with a partner. If time is limited, ask each partner to choose two different systems to solve. Follow with a whole-class discussion.

Solve each system without graphing.

\(\begin{cases} 5x – 2y = 26 \\ y + 4 = x \end{cases}\)

\(\begin{cases} 2m – 2p = \text-6\\ p = 2m + 10 \end{cases}\)

\(\begin{cases} 2d = 8f \\ 18 - 4f = 2d \end{cases}\)

\(\begin{cases} w + \frac17z = 4 \\ z = 3w –2 \end{cases}\)

Are you ready for more?

Solve this system with four equations. \(\begin{cases}3 x + 2y - z + 5w= 20 \\ y = 2z-3w\\ z=w+1 \\ 2w=8 \end{cases}\)

When solving the second system, students are likely to substitute the expression \(2m+10\) for \(p\) in the first equation,  \(2m-2p=\text-6\) . Done correctly, it should be written as  \(2m-2(2m+10)=\text-6\) . Some students may neglect to write parentheses and write \(2m-4m+10=\text-6\) . Remind students that if \(p\) is equal to \(2m+10\) , then \(2p\)  is 2 times \(2m+10\) or \(2(2m+10)\) . (Alternatively, use an example with a sum of two numbers for  \(p\) : Suppose \(p=10\) , which means \(2p=2(10)\) or 20. If we express \(p\) as a sum of 3 and 7, or \(p=3+7\) , then \(2p=2(3+7)\) , not \(2\boldcdot 3 + 7\) . The latter has a value of 13, not 20.)

Some students who correctly write \(2m-2(2m+10)=\text-6\) may fail to distribute the subtraction and write the left side as  \(2m-4m+20\) . Remind them that subtracting by \(2(2m+10)\) can be thought of as adding \(\text-2(2m+10)\) and ask how they would expand this expression.

Select previously identified students to share their responses and reasoning. Display their work for all to see.

Highlight the different ways to perform substitutions to solve the same system. For example:

• In the second system, \(\begin{cases} 2m – 2p = \text-6\\ p = 2m + 10 \end{cases}\) , we could substitute \(2m+10\) for  \(p\) in the first equation, or we could substitute  \(p-10\) for  \(2m\) in the first equation.
• In the third system,  \(\begin{cases} 2d = 8f \\ 18 - 4f = 2d \end{cases}\) , we could substitute \(8f\) for  \(2d\) in the second equation, or we could substitute \(\frac14 d\) for  \(f\) in the second equation.

Lesson Synthesis

To emphasize that the method we choose for solving a systems may depend on the system, and that some systems are more conducive to be solved by substitution than others, present the following systems to students: 

\(\begin {cases} 3m + n = 71\\2m-n =30 \end {cases}\)

\(\begin {cases} 4x + y = 1\\y = \text-2x+9 \end {cases}\)

\(\displaystyle \begin{cases} 5x+4y=15 \\ 5x+11y=22 \end{cases}\)

Ask students to choose a system and make a case (in writing, if possible) for why they would or would not choose to solve that system by substitution. Consider asking students to use sentence starters such as these:

• I would choose to solve system \(\underline{\hspace{0.5in}}\) by substitution because . . .
• I would not choose to solve system \(\underline{\hspace{0.5in}}\) by substitution because . . .

With a little bit of rearrangement, all systems could be solved by substitution without cumbersome computation, but system 2 would be most conducive to solving by substitution.

Consider collecting students' responses or asking them to share their written arguments with a partner. 

13.4: Cool-down - A System to Solve (5 minutes)

Student lesson summary.

The solution to a system can usually be found by graphing, but graphing may not always be the most precise or the most efficient way to solve a system. 

Here is a system of equations:

\(\begin {cases} 3p + q = 71\\2p - q = 30 \end {cases}\)

The graphs of the equations show an intersection at approximately 20 for \(p\) and approximately 10 for \(q\) .

Without technology, however, it is not easy to tell what the exact values are.

Instead of solving by graphing, we can solve the system algebraically. Here is one way. 

If we subtract \(3p\) from each side of the first equation,  \(3p + q = 71\) , we get an equivalent equation:  \(q= 71 - 3p\) . Rewriting the original equation this way allows us to isolate the variable \(q\) . 

Because \(q\) is equal to  \(71-3p\) , we can substitute the expression  \(71-3p\) in the place of  \(q\) in the second equation. Doing this gives us an equation with only one variable, \(p\) , and makes it possible to find  \(p\) .

\(\begin {align} 2p - q &= 30 &\quad& \text {original equation} \\ 2p - (71 - 3p) &=30 &\quad& \text {substitute }71-3p \text{ for }q\\ 2p - 71 + 3p &=30 &\quad& \text {apply distributive property}\\ 5p - 71 &= 30 &\quad& \text {combine like terms}\\ 5p &= 101 &\quad& \text {add 71 to both sides}\\ p &= \dfrac{101}{5} &\quad& \text {divide both sides by 5} \\ p&=20.2 \end {align}\)

Now that we know the value of \(p\) , we can find the value of \(q\) by substituting 20.2 for \(p\) in either of the original equations and solving the equation.

\(\begin {align} 3(20.2) + q &=71\\60.6 + q &= 71\\ q &= 71 - 60.6\\ q &=10.4 \end{align}\) ​​​​​​

\(\begin {align} 2(20.2) - q &= 30\\ 40.4 - q &=30\\ \text-q &= 30 - 40.4\\ \text-q &= \text-10.4 \\ q &= \dfrac {\text-10.4}{\text-1} \\ q &=10.4 \end {align}​​​​​​\)

The solution to the system is the pair \(p=20.2\) and \(q=10.4\) , or the point \((20.2, 10.4)\) on the graph. 

This method of solving a system of equations is called solving by substitution , because we substituted an expression for \(q\) into the second equation.

A: ( 5 , 1 ) ( 5 , 1 )  B: ( −2 , 4 ) ( −2 , 4 )  C: ( −5 , −1 ) ( −5 , −1 )  D: ( 3 , −2 ) ( 3 , −2 )  E: ( 0 , −5 ) ( 0 , −5 )  F: ( 4 , 0 ) ( 4 , 0 )

A: ( 4 , 2 ) ( 4 , 2 )  B: ( −2 , 3 ) ( −2 , 3 )  C: ( −4 , −4 ) ( −4 , −4 )  D: ( 3 , −5 ) ( 3 , −5 )  E: ( −3 , 0 ) ( −3 , 0 )  F: ( 0 , 2 ) ( 0 , 2 )

Answers will vary.

ⓐ yes, yes  ⓑ yes, yes

ⓐ no, no  ⓑ yes, yes

x - intercept: ( 2 , 0 ) ( 2 , 0 ) ; y - intercept: ( 0 , −2 ) ( 0 , −2 )

x - intercept: ( 3 , 0 ) ( 3 , 0 ) , y - intercept: ( 0 , 2 ) ( 0 , 2 )

x - intercept: ( 4 , 0 ) ( 4 , 0 ) , y - intercept: ( 0 , 12 ) ( 0 , 12 )

x - intercept: ( 8 , 0 ) ( 8 , 0 ) , y - intercept: ( 0 , 2 ) ( 0 , 2 )

x - intercept: ( 4 , 0 ) ( 4 , 0 ) , y - intercept: ( 0 , −3 ) ( 0 , −3 )

x - intercept: ( 4 , 0 ) ( 4 , 0 ) , y - intercept: ( 0 , −2 ) ( 0 , −2 )

− 2 3 − 2 3

− 4 3 − 4 3

− 3 5 − 3 5

− 1 36 − 1 36

− 1 48 − 1 48

slope m = 2 3 m = 2 3 and y -intercept ( 0 , −1 ) ( 0 , −1 )

slope m = 1 2 m = 1 2 and y -intercept ( 0 , 3 ) ( 0 , 3 )

2 5 ; ( 0 , −1 ) 2 5 ; ( 0 , −1 )

− 4 3 ; ( 0 , 1 ) − 4 3 ; ( 0 , 1 )

− 1 4 ; ( 0 , 2 ) − 1 4 ; ( 0 , 2 )

− 3 2 ; ( 0 , 6 ) − 3 2 ; ( 0 , 6 )

ⓐ intercepts  ⓑ horizontal line  ⓒ slope–intercept  ⓓ vertical line

ⓐ vertical line  ⓑ slope–intercept  ⓒ horizontal line  ⓓ intercepts

• ⓐ 50 inches
• ⓑ 66 inches
• ⓒ The slope, 2, means that the height, h , increases by 2 inches when the shoe size, s , increases by 1. The h -intercept means that when the shoe size is 0, the height is 50 inches.
• ⓐ 40 degrees
• ⓑ 65 degrees
• ⓒ The slope, 1 4 1 4 , means that the temperature Fahrenheit ( F ) increases 1 degree when the number of chirps, n , increases by 4. The T -intercept means that when the number of chirps is 0, the temperature is 40 ° 40 ° .
• ⓒ The slope, 0.5, means that the weekly cost, C , increases by $0.50 when the number of miles driven, n, increases by 1. The C -intercept means that when the number of miles driven is 0, the weekly cost is $60
• ⓒ The slope, 1.8, means that the weekly cost, C, increases by $1.80 when the number of invitations, n , increases by 1.80. The C -intercept means that when the number of invitations is 0, the weekly cost is $35.;

not parallel; same line

perpendicular

not perpendicular

y = 2 5 x + 4 y = 2 5 x + 4

y = − x − 3 y = − x − 3

y = 3 5 x + 1 y = 3 5 x + 1

y = 4 3 x − 5 y = 4 3 x − 5

y = 5 6 x − 2 y = 5 6 x − 2

y = 2 3 x − 4 y = 2 3 x − 4

y = − 2 5 x − 1 y = − 2 5 x − 1

y = − 3 4 x − 4 y = − 3 4 x − 4

y = 8 y = 8

y = 4 y = 4

y = 5 2 x − 13 2 y = 5 2 x − 13 2

y = − 2 5 x + 22 5 y = − 2 5 x + 22 5

y = 1 3 x − 10 3 y = 1 3 x − 10 3

y = − 2 5 x − 23 5 y = − 2 5 x − 23 5

x = 5 x = 5

x = −4 x = −4

y = 3 x − 10 y = 3 x − 10

y = 1 2 x + 1 y = 1 2 x + 1

y = − 1 3 x + 10 3 y = − 1 3 x + 10 3

y = −2 x + 16 y = −2 x + 16

y = −5 y = −5

y = −1 y = −1

x = −5 x = −5

ⓐ yes  ⓑ yes  ⓒ yes  ⓓ yes  ⓔ no

ⓐ yes  ⓑ yes  ⓒ no  ⓓ no  ⓔ yes

y ≥ −2 x + 3 y ≥ −2 x + 3

y < 1 2 x − 4 y < 1 2 x − 4

x − 4 y ≤ 8 x − 4 y ≤ 8

3 x − y ≤ 6 3 x − y ≤ 6

Section 4.1 Exercises

A: ( −4 , 1 ) ( −4 , 1 )  B: ( −3 , −4 ) ( −3 , −4 )  C: ( 1 , −3 ) ( 1 , −3 )  D: ( 4 , 3 ) ( 4 , 3 )

A: ( 0 , −2 ) ( 0 , −2 )  B: ( −2 , 0 ) ( −2 , 0 )  C: ( 0 , 5 ) ( 0 , 5 )  D: ( 5 , 0 ) ( 5 , 0 )

ⓑ Age and weight are only positive.

Section 4.2 Exercises

ⓐ yes; no  ⓑ no; no  ⓒ yes; yes  ⓓ yes; yes

ⓐ yes; yes  ⓑ yes; yes  ⓒ yes; yes  ⓓ no; no

$722, $850, $978

Section 4.3 Exercises

( 3 , 0 ) , ( 0 , 3 ) ( 3 , 0 ) , ( 0 , 3 )

( 5 , 0 ) , ( 0 , −5 ) ( 5 , 0 ) , ( 0 , −5 )

( −2 , 0 ) , ( 0 , −2 ) ( −2 , 0 ) , ( 0 , −2 )

( −1 , 0 ) , ( 0 , 1 ) ( −1 , 0 ) , ( 0 , 1 )

( 6 , 0 ) , ( 0 , 3 ) ( 6 , 0 ) , ( 0 , 3 )

( 0 , 0 ) ( 0 , 0 )

( 4 , 0 ) , ( 0 , 4 ) ( 4 , 0 ) , ( 0 , 4 )

( −3 , 0 ) , ( 0 , 3 ) ( −3 , 0 ) , ( 0 , 3 )

( 8 , 0 ) , ( 0 , 4 ) ( 8 , 0 ) , ( 0 , 4 )

( 2 , 0 ) , ( 0 , 6 ) ( 2 , 0 ) , ( 0 , 6 )

( 12 , 0 ) , ( 0 , −4 ) ( 12 , 0 ) , ( 0 , −4 )

( 2 , 0 ) , ( 0 , −8 ) ( 2 , 0 ) , ( 0 , −8 )

( 5 , 0 ) , ( 0 , 2 ) ( 5 , 0 ) , ( 0 , 2 )

( 4 , 0 ) , ( 0 , −6 ) ( 4 , 0 ) , ( 0 , −6 )

( 3 , 0 ) , ( 0 , 1 ) ( 3 , 0 ) , ( 0 , 1 )

( −10 , 0 ) , ( 0 , 2 ) ( −10 , 0 ) , ( 0 , 2 )

ⓐ ( 0 , 1000 ) , ( 15 , 0 ) ( 0 , 1000 ) , ( 15 , 0 ) ⓑ At ( 0 , 1000 ) ( 0 , 1000 ) , he has been gone 0 hours and has 1000 miles left. At ( 15 , 0 ) ( 15 , 0 ) , he has been gone 15 hours and has 0 miles left to go.

Section 4.4 Exercises

−3 2 = − 3 2 −3 2 = − 3 2

− 1 3 − 1 3

− 3 4 − 3 4

− 5 2 − 5 2

− 8 7 − 8 7

ⓐ 1 3 1 3   ⓑ 4 12 pitch or 4-in-12 pitch

3 50 3 50 ; rise = 3, run = 50

ⓐ 288 inches (24 feet)  ⓑ Models will vary.

When the slope is a positive number the line goes up from left to right. When the slope is a negative number the line goes down from left to right.

A vertical line has 0 run and since division by 0 is undefined the slope is undefined.

Section 4.5 Exercises

slope m = 4 m = 4 and y -intercept ( 0 , −2 ) ( 0 , −2 )

slope m = −3 m = −3 and y -intercept ( 0 , 1 ) ( 0 , 1 )

slope m = − 2 5 m = − 2 5 and y -intercept ( 0 , 3 ) ( 0 , 3 )

−9 ; ( 0 , 7 ) −9 ; ( 0 , 7 )

4 ; ( 0 , −10 ) 4 ; ( 0 , −10 )

−4 ; ( 0 , 8 ) −4 ; ( 0 , 8 )

− 8 3 ; ( 0 , 4 ) − 8 3 ; ( 0 , 4 )

7 3 ; ( 0 , −3 ) 7 3 ; ( 0 , −3 )

horizontal line

vertical line

slope–intercept

• ⓒ The slope, 2.54, means that Randy’s payment, P , increases by $2.54 when the number of units of water he used, w, increases by 1. The P –intercept means that if the number units of water Randy used was 0, the payment would be $28.
• ⓒ The slope, 0.32, means that the cost, C , increases by $0.32 when the number of miles driven, m, increases by 1. The C -intercept means that if Janelle drives 0 miles one day, the cost would be $15.
• ⓒ The slope, 0.09, means that Patel’s salary, S , increases by $0.09 for every $1 increase in his sales. The S -intercept means that when his sales are $0, his salary is $750.
• ⓒ The slope, 42, means that the cost, C , increases by $42 for when the number of guests increases by 1. The C -intercept means that when the number of guests is 0, the cost would be $750.

not parallel

• ⓐ For every increase of one degree Fahrenheit, the number of chirps increases by four.
• ⓑ There would be −160 −160 chirps when the Fahrenheit temperature is 0 ° 0 ° . (Notice that this does not make sense; this model cannot be used for all possible temperatures.)

Section 4.6 Exercises

y = 4 x + 1 y = 4 x + 1

y = 8 x − 6 y = 8 x − 6

y = − x + 7 y = − x + 7

y = −3 x − 1 y = −3 x − 1

y = 1 5 x − 5 y = 1 5 x − 5

y = − 2 3 x − 3 y = − 2 3 x − 3

y = 2 y = 2

y = −4 x y = −4 x

y = −2 x + 4 y = −2 x + 4

y = 3 4 x + 2 y = 3 4 x + 2

y = − 3 2 x − 1 y = − 3 2 x − 1

y = 6 y = 6

y = 3 8 x − 1 y = 3 8 x − 1

y = 5 6 x + 2 y = 5 6 x + 2

y = − 3 5 x + 1 y = − 3 5 x + 1

y = − 1 3 x − 11 y = − 1 3 x − 11

y = −7 y = −7

y = − 5 2 x − 22 y = − 5 2 x − 22

y = −4 x − 11 y = −4 x − 11

y = −8 y = −8

y = −4 x + 13 y = −4 x + 13

y = x + 5 y = x + 5

y = − 1 3 x − 14 3 y = − 1 3 x − 14 3

y = 7 x + 22 y = 7 x + 22

y = − 6 7 x + 4 7 y = − 6 7 x + 4 7

y = 1 5 x − 2 y = 1 5 x − 2

x = 4 x = 4

x = −2 x = −2

y = −3 y = −3

y = 4 x y = 4 x

y = 1 2 x + 3 2 y = 1 2 x + 3 2

y = 5 y = 5

y = 3 x − 1 y = 3 x − 1

y = −3 x + 3 y = −3 x + 3

y = 2 x − 6 y = 2 x − 6

y = − 2 3 x + 5 y = − 2 3 x + 5

x = −3 x = −3

y = −4 y = −4

y = x y = x

y = − 3 4 x − 1 4 y = − 3 4 x − 1 4

y = 5 4 x y = 5 4 x

y = 1 y = 1

y = x + 2 y = x + 2

y = 3 4 x y = 3 4 x

y = 1.2 x + 5.2 y = 1.2 x + 5.2

Section 4.7 Exercises

ⓐ yes  ⓑ no  ⓒ no  ⓓ yes  ⓔ no

ⓐ yes  ⓑ no  ⓒ no  ⓓ yes  ⓔ yes

ⓐ no  ⓑ no  ⓒ no  ⓓ yes  ⓔ yes

y < 2 x − 4 y < 2 x − 4

y ≤ − 1 3 x − 2 y ≤ − 1 3 x − 2

x + y ≥ 3 x + y ≥ 3

x + 2 y ≥ −2 x + 2 y ≥ −2

2 x − y < 4 2 x − y < 4

4 x − 3 y > 12 4 x − 3 y > 12

• ⓑ Answers will vary.

Review Exercises

ⓐ ( 2 , 0 ) ( 2 , 0 )   ⓑ ( 0 , −5 ) ( 0 , −5 )   ⓒ ( −4.0 ) ( −4.0 )   ⓓ ( 0 , 3 ) ( 0 , 3 )

ⓐ yes; yes  ⓑ yes; no

( 6 , 0 ) , ( 0 , 4 ) ( 6 , 0 ) , ( 0 , 4 )

− 1 2 − 1 2

slope m = − 2 3 m = − 2 3 and y -intercept ( 0 , 4 ) ( 0 , 4 )

5 3 ; ( 0 , −6 ) 5 3 ; ( 0 , −6 )

4 5 ; ( 0 , − 8 5 ) 4 5 ; ( 0 , − 8 5 )

plotting points

ⓐ −$250  ⓑ $450  ⓒ The slope, 35, means that Marjorie’s weekly profit, P , increases by $35 for each additional student lesson she teaches. The P –intercept means that when the number of lessons is 0, Marjorie loses $250.  ⓓ

y = −5 x − 3 y = −5 x − 3

y = −2 x y = −2 x

y = −3 x + 5 y = −3 x + 5

y = 3 5 x y = 3 5 x

y = −2 x − 5 y = −2 x − 5

y = 1 2 x − 5 2 y = 1 2 x − 5 2

y = − 2 5 x + 8 y = − 2 5 x + 8

y = 3 y = 3

y = − 3 2 x − 6 y = − 3 2 x − 6

ⓐ yes  ⓑ no  ⓒ yes  ⓓ yes  ⓔ no

y > 2 3 x − 3 y > 2 3 x − 3

x − 2 y ≥ 6 x − 2 y ≥ 6

Practice Test

ⓐ yes  ⓑ yes  ⓒ no

( 3 , 0 ) , ( 0 , −4 ) ( 3 , 0 ) , ( 0 , −4 )

y = − 3 4 x − 2 y = − 3 4 x − 2

y = 1 2 x − 4 y = 1 2 x − 4

y = − 4 5 x − 5 y = − 4 5 x − 5

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• Authors: Lynn Marecek, MaryAnne Anthony-Smith
• Publisher/website: OpenStax
• Book title: Elementary Algebra
• Publication date: Feb 22, 2017
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Prentice Hall Algebra 1, Grade: 0 Publisher: Pearson Prentice Hall; Student edition

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Go Math Grade 8 Answer Key Chapter 14 Scatter Plots

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Lesson 1: Scatter Plots and Association

• Scatter Plots and Association – Page No. 436
• Scatter Plots and Association – Page No. 437

Scatter Plots and Association – Page No. 438

Lesson 2: Trend Lines and Predictions

• Trend Lines and Predictions – Page No. 442
• Trend Lines and Predictions – Page No. 443
• Trend Lines and Predictions – Page No. 444
• Model Quiz – Page No. 445

Mixed Review

• Mixed Review – Page No. 446

Guided Practice – Scatter Plots and Association – Page No. 436

Explanation: As Bob gets older, his height increases along with the straight line on the graph. So, the association is positive and linear.

Scatter Plots and Data Analysis Answer Key Question 2. Describe the association between Bob’s age and his height. Explain the association. Type below: _____________

Answer: The association is positive and linear. Bob’s height increases as he gets older. We would see that Bob’s height eventually stops increasing if the data continues.

Answer: There is an outlier at (35,18)

Explanation: There is a cluster in the “20 – 25” shots attempted range and a smaller cluster in the “5 – 14” shots attempted range. There is an outlier at (35,18)

ESSENTIAL QUESTION CHECK-IN

Question 4. Explain how you can make a scatter plot from a set of bivariate data. Type below: _____________

Answer: Bivariate data – data that has two variables per observation, An x variable and y variable. Scatterplot – The graph displaying categorical data, with an x and y-axis. Response Variable – the variable that is explained by the other. Explanatory Variable – the variable which explains the other.

14.1 Independent Practice – Scatter Plots and Association – Page No. 437

Sports Use the scatter plot for 5–8.

Question 5. Describe the association between the year and the distance jumped for the years 1960 to 1988. Type below: _____________

Answer: The data shows a positive linear association. If the year increases, the winning distance increases.

Question 6. Describe the association between the year and the distance jumped for the years after 1988. Type below: _____________

Answer: Between 1996 and 2004, there was a slight increase in distance over time. The data from 1988 to 2012 will show a negative association.

Question 7. For the entire scatter plot, is the association between the year and the distance jumped linear or nonlinear? _____________

Answer: The data show a rise between 1960 and 1988. The data also show a fall between 1988 and 2012. Therefore, overall, there is no linear pattern.

Scatter Plots and Data Unit Test Answer Key Question 8. Identify the outlier and interpret its meaning. Type below: _____________

Answer: The outlier is at (1968, 8.9). It represents a long jump of 8.9 meters in 1968 that exceeds the other jumps made in the surrounding years.

Question 9. Communicate Mathematical Ideas Compare a scatter plot that shows no association to one that shows a negative association. Type below: _____________

Answer: Randomly scattered data points with no apparent pattern define a scatter plot with no association. Data points that fall from left to right and has data set values that increase as the other decreases define a scatter plot with a negative association.

For 10–11, describe a set of real-world bivariate data that the given scatter plot could represent. Define the variable represented on each axis.

Answer: The x-axis represents the number of containers. The y-&is represents the price per container.

Answer: The x-axis represents the number of hours spent watching tv. The y-axis represents the number of TVs owned.

FOCUS ON HIGHER ORDER THINKING

Question 12. Multiple Representations Describe what you might see in a table of bivariate data that would lead you to conclude that the scatter plot of the data would show a cluster. Type below: _____________

Answer: A cluster in a scatter plot is when there are a lot of points all grouped around the same location. Look for points that have the same input and output values. If there are a lot of points together, you must have a cluster in your scatter plot.

Question 13. Justify Reasoning Is it possible for a scatter plot to have a positive or negative association that is not linear? Explain. Type below: _____________

Answer: Yes

Explanation: Yes; it is possible for a scatter plot to have a positive or negative association that is not linear. The data points may have a falling or rising curve that will exhibit a nonlinear association.

Question 14. Critical Thinking To try to increase profits, a theater owner increases the price of a ticket by $25 every month. Describe what a scatter plot might look like if x represents the number of months and y represents the profits. Explain your reasoning. Type below: _____________

Answer: Initially, the number of tickets sold might decline a little, but the price increase would offset the loss in sales. That means that profits would increase, showing a positive association. When the price would get too high, ticket sales would decline rapidly, so profits would fall giving a negative association.

Guided Practice – Trend Lines and Predictions – Page No. 442

Angela recorded the price of different weights of several bulk grains. She made a scatter plot of her data. Use the scatter plot for 1–4.

Question 2. How do you know whether your trend line is a good fit for the data? Type below: _____________

Answer: Most of the data points are close to the trend line. The trend line has about the same number of points above and below it.

Scatter Plots and Trend Lines Quiz 1 Answer Key Question 3. Write an equation for your trend line. Type below: _____________

Answer: y = 0.09x

Explanation: The trend line passes through (0, 0) and (19, 1.80). Find the slope by using the slope formula. slope = m = (y2 – y1)/(x2 – x1) = 1.80/19 = 0.09 The line passes through the origin. So, the y-intercept is 0. From an equation for the trend line substituting the slope value for m and the value of the y-intercept b in the slope-intercept formula. y = mx + b y = 0.09x + 0 y = 0.09x

Question 4. Use the equation for your trend line to interpolate the price of 7 ounces and extrapolate the price of 50 ounces. Type below: _____________

Answer: The price for 7 and 50 ounces is $0.63 and $4.50

Explanation: Use the equation for the trend line (y = 0.09x) to interpolate the price of 7 ounces by substituting 7 for x (y= 0.09 • 7) and solving for y. Use the equation for the trend line (y = 0.09x) to interpolate the price of 50 ounces by substituting 50 for x (y= 0.09 • 50) and solving for y.

Question 5. A trend line passes through two points on a scatter plot. How can you use the trend line to make a prediction between or outside the given data points? Type below: _____________

Answer: Use two points on the line. rind the slope and y-intercept. Substitute the values of the slope (m) and y-intercept (b) to form an equation using y = mx + b. Substitute the value of x for which you want to make a prediction and solve for y OR substitute your prediction for y and solve to find its value.

14.2 Independent Practice – Trend Lines and Predictions – Page No. 443

Question 7. What type of association does the trend line show? Type below: _____________

Answer: Negative Association

Explanation: One data set increases – Wind Speed and the other – Wind Chill decreases. So, the trend line shows a Negative Association.

Scatter Plots and Trend Lines Answer Key Question 8. Write an equation for your trend line. Type below: _____________

Answer: y = -0.25x + 2.5

Explanation: Find the slope using the Slope Formula m = (y2 – y1)/(x2 – x1) = ((-10) – 5)/(50 – 30) = -5/20 = -0.25 Find the y-intercept using the Slope-Intercept Formula y = mx + b -5 = -0.25(30) + b -5 = -7.5 + b 2.5 = b Substitute the value of m and b into the Slope-Intercept Formula to form an equation for the trend line. y = -0.25x + 2.5

Question 9. Make a Prediction Use the trend line to predict the wind chill at these wind speeds. a. 36 mi/h _________ °F

Answer: -6.5°F

Explanation: Use the trend line to predict the wind chill at 36mi/h y = -0.25x + 2.5 y = -0.25(36) + 2.5 y = -9 + 2.5 y = -6.5 The wind chill at 36mi/h is -6.5ºF

Question 9. b. 100 mi/h _________ °F

Answer: -22.5°F

Explanation: Use the trend line to predict the wind chill at 100mi/h y = -0.25x + 2.5 y = -0.25(100) + 2.5 y = -25 + 2.5 y = -22.5 The wind chill at 100mi/h is -22.5ºF

Question 10. What is the meaning of the slope of the line? Type below: _____________

Answer: The slope means that the wind chill falls about 1°F for every 4 mph increase in wind speed.

Problem Solving with Trend Lines Worksheet Answers Question 12. Write an equation for your trend line. Type below: _____________

Answer: y = -(2/15)x + 64

Explanation: Find the slope using the Slope Formula m = (y2 – y1)/(x2 – x1) = (72 – 64)/(60 – 0) = 8/60 = -2/15 Find the y-intercept using the Slope-Intercept Formula at (0, 64) y = mx + b b = 64 Substitute the value of m and b into the Slope-Intercept Formula to form an equation for the trend line. y = -2/15x + 64

Question 13. Make a Prediction Use the trend line to predict the apparent temperature at 70% humidity. Type below: _____________

Answer: 73.3º F

Explanation: Use the equation of the trend line. Substitute 70(for 70%) into the equation for x. y = -(2/15)x + 64 y = -(2/15)(70) + 64 y = -140/15 + 64 y = -9.3 + 64 y = 73.3 The apparent temperature is 73.3º F

Question 14. What is the meaning of the y-intercept of the line? Type below: _____________

Answer: The y-intercept explains that at 0% humidity, the apparent temperature is 64ºF

FOCUS ON HIGHER ORDER THINKING – Trend Lines and Predictions – Page No. 444

Question 15. Communicate Mathematical Ideas Is it possible to draw a trend line on a scatter plot that shows no association? Explain. _____________

Explanation: It is not possible to draw a trend line on a scatter plot that shows no association. If the scatter plot shows no association, the data points have no relationships with one another. You can draw a trend line if a linear association is available.

Question 16. Critique Reasoning Sam drew a trend line that had about the same number of data points above it as below it, but did not pass through any data points. He then picked two data points to write the equation for the line. Is this a correct way to write the equation? Explain. _____________

Explanation: Sam did not use the correct way to write an equation. Sam may have drawn a correct trend line but using the data points that are not on the trend line may have an incorrect equation for the line. He should use two points on that trend line to write the equation.

Problem Solving with Trend Lines Homework 4 Answer Key Question 17. Marlene wanted to find a relationship between the areas and populations of counties in Texas. She plotted x (area in square miles) and y (population) for two counties on a scatter plot: Kent County (903, 808)                                Edwards County (2118, 2002) She concluded that the population of Texas counties is approximately equal to their area in square miles and drew a trend line through her points. a. Critique Reasoning Do you agree with Marlene’s method of creating a scatter plot and a trend line? Explain why or why not. _____________

Answer: I do not agree with Marlene’s method of creating a scatter plot and a trend line. She did not have enough data. Marlene should have collected and plotted data for many more counties.

Question 17. b. Counterexamples Harris County has an area of 1778 square miles and a population of about 4.3 million people. Dallas County has an area of 908 square miles and a population of about 2.5 million people. What does this data show about Marlene’s conjecture that the population of Texas counties is approximately equal to their area? Type below: _____________

Answer: The data collected are only of two counties whose populations are nearly equal to their area. The fact that the populations of Harris and Dallas counties are in the millions, Marlene’s conjecture about the population of Texas counties being equivalent to their area is invalid.

Ready to Go On? – Model Quiz – Page No. 445

14.1 Scatter Plots and Association

Unit Scatter Plots and Data Homework 1 Answer Key Question 2. Describe the association you see between the number of quarts purchased and the price per quart. Explain. Type below: _____________

Answer: Negative nonlinear association

Explanation: The association seen between the number of quarts purchased and the price per quart is negative and nonlinear. As the number of quarts rises, the price per quart decreases but you can see a data curve.

14.2 Trend Lines and Predictions

The scatter plot below shows data comparing wind speed and wind chill for an air temperature of 20 °F. Use the scatter plot for Exs. 3–5.

Question 4. Write an equation for your trend line. Type below: _____________

Answer: y = -0.35x + 12.25

Explanation: The line passes through (10, 8.75) and (35, 0) so we can use these points to find the slope. The slope of the line is : Slope = m = (y2 – y1)/(x2 – x1) = (0 – 8.75)/(35 – 10) = -8.75/25 = -0.35 Find the y-intercept using the slope-intercept formula : y = mx + b 0 = -0.35 . 35 + b 0 = -12.25 + b b = 12.25 Substitute the slope m and the y-intercept b in the slope-intercept formula. The equation for the trend line is : y = mx + b y = -0.35x + 12.25

Problem-Solving with Trend Lines Homework 4 Answers Question 5. Use your equation to predict the wind chill to the nearest degree for a wind speed of 60 mi/h. ________ °F

Answer: 9°F

Explanation: y = −0.35x + 12.25 y = -0.35(60) + 12.25 y = -21 + 12.25 y = -8.75 The wind chill to the nearest degree for a wind speed of 60 mi/h is 9°F.

ESSENTIAL QUESTION

Question 6. How can you use scatter plots to solve real-world problems? Type below: _____________

Answer: Using a scatter plot, you can see positive and negative trends such as prices over time. You can also make predictions such as height at a certain age.

Selected Response – Mixed Review – Page No. 446

Answer: b. B

Question 2. What type of association would you expect between a person’s age and hair length? Options: a. linear b. negative c. none d. positive

Answer: c. none

Explanation: The length of their hair reduces. This is because the length of hair changes with the growth phase of the hair follicles. When one is young, the cells of the papilla divide more rapidly, and hence the length of the hair is long before reaching the transitional phase and then shedding off in the telogen phase. The older one gets, the papilla cells do not divide as rapidly and the length of the hair shortens with age.

Answer: d. positive association

Explanation: The scatter plot shows a cluster, some outliers, and a negative association. It does not show a positive association.

Unit Scatter Plots and Data Homework 3 Answer Key Question 4. A restaurant claims to have served 352,000,000 hamburgers. What is this number in scientific notation? Options: a. 3.52 × 10 6 b. 3.52 × 10 8 c. 35.2 × 10 7 d. 352 × 10 6

Answer: b. 3.52 × 10 8

Explanation: 100,000,000 So, 3.52 × 10 8

Answer: b. y = −\(\frac{1}{4}\)x

Explanation: In order to find out the relationship between x and y, we have to use the values in the question and substitute them into the solution options. So, y = -1/4x

Question 6. b. Which data point is an outlier? Type below: ______________

Answer: The outlier is the point (92, 135).

Question 6. c. Predict the number of visitors on a day when the high temperature is 102 °F. Type below: ______________

Answer: Based on the cluster around 100°F, I would expect that on a day with a temperature of 102 °F, the pool would have between 350 and 400 visitors.

Conclusion:

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