unit linear equations homework 5 answer key

4.1 Linear Functions

m = 4 − 3 0 − 2 = 1 − 2 = − 1 2 ; m = 4 − 3 0 − 2 = 1 − 2 = − 1 2 ; decreasing because m < 0. m < 0.

y = − 7 x + 3 y = − 7 x + 3

H ( x ) = 0.5 x + 12.5 H ( x ) = 0.5 x + 12.5

Possible answers include ( − 3 , 7 ) , ( − 3 , 7 ) , ( − 6 , 9 ) , ( − 6 , 9 ) , or ( − 9 , 11 ) . ( − 9 , 11 ) .

( 16 , 0 ) ( 16 , 0 )

• ⓐ f ( x ) = 2 x ; f ( x ) = 2 x ;
• ⓑ g ( x ) = − 1 2 x g ( x ) = − 1 2 x

y = – 1 3 x + 6 y = – 1 3 x + 6

4.2 Modeling with Linear Functions

ⓐ C ( x ) = 0.25 x + 25 , 000 C ( x ) = 0.25 x + 25 , 000 ⓑ The y -intercept is ( 0 , 25 , 000 ) ( 0 , 25 , 000 ) . If the company does not produce a single doughnut, they still incur a cost of $25,000.

ⓐ 41,100 ⓑ 2020

21.57 miles

4.3 Fitting Linear Models to Data

54 ° F 54 ° F

150.871 billion gallons; extrapolation

4.1 Section Exercises

Terry starts at an elevation of 3000 feet and descends 70 feet per second.

d ( t ) = 100 − 10 t d ( t ) = 100 − 10 t

The point of intersection is ( a , a ) . ( a , a ) . This is because for the horizontal line, all of the y y coordinates are a a and for the vertical line, all of the x x coordinates are a . a . The point of intersection is on both lines and therefore will have these two characteristics.

y = 3 5 x − 1 y = 3 5 x − 1

y = 3 x − 2 y = 3 x − 2

y = − 1 3 x + 11 3 y = − 1 3 x + 11 3

y = − 1.5 x − 3 y = − 1.5 x − 3

perpendicular

f ( 0 ) = − ( 0 ) + 2 f ( 0 ) = 2 y − int : ( 0 , 2 ) 0 = − x + 2 x − int : ( 2 , 0 ) f ( 0 ) = − ( 0 ) + 2 f ( 0 ) = 2 y − int : ( 0 , 2 ) 0 = − x + 2 x − int : ( 2 , 0 )

h ( 0 ) = 3 ( 0 ) − 5 h ( 0 ) = − 5 y − int : ( 0 , − 5 ) 0 = 3 x − 5 x − int : ( 5 3 , 0 ) h ( 0 ) = 3 ( 0 ) − 5 h ( 0 ) = − 5 y − int : ( 0 , − 5 ) 0 = 3 x − 5 x − int : ( 5 3 , 0 )

− 2 x + 5 y = 20 − 2 ( 0 ) + 5 y = 20 5 y = 20 y = 4 y − int : ( 0 , 4 ) − 2 x + 5 ( 0 ) = 20 x = − 10 x − int : ( − 10 , 0 ) − 2 x + 5 y = 20 − 2 ( 0 ) + 5 y = 20 5 y = 20 y = 4 y − int : ( 0 , 4 ) − 2 x + 5 ( 0 ) = 20 x = − 10 x − int : ( − 10 , 0 )

Line 1: m = –10 Line 2: m = –10 Parallel

Line 1: m = –2 Line 2: m = 1 Neither

Line 1 :   m = – 2       Line 2 :   m = – 2       Parallel Line 1 :   m = – 2       Line 2 :   m = – 2       Parallel

y = 3 x − 3 y = 3 x − 3

y = − 1 3 t + 2 y = − 1 3 t + 2

y = − 5 4 x + 5 y = − 5 4 x + 5

y = 3 x − 1 y = 3 x − 1

y = − 2.5 y = − 2.5

y = 3 y = 3

x = − 3 x = − 3

Linear, g ( x ) = − 3 x + 5 g ( x ) = − 3 x + 5

Linear, f ( x ) = 5 x − 5 f ( x ) = 5 x − 5

Linear, g ( x ) = − 25 2 x + 6 g ( x ) = − 25 2 x + 6

Linear, f ( x ) = 10 x − 24 f ( x ) = 10 x − 24

f ( x ) = − 58 x + 17.3 f ( x ) = − 58 x + 17.3

• ⓐ a = 11,900 , b = 1000.1 a = 11,900 , b = 1000.1
• ⓑ q ( p ) = 1000 p – 100 q ( p ) = 1000 p – 100

y = − 16 3 y = − 16 3

x = a x = a

y = d c – a x – a d c – a y = d c – a x – a d c – a

y = 100 x – 98 y = 100 x – 98

x < 1999 201 , x > 1999 201 x < 1999 201 , x > 1999 201

$45 per training session.

The rate of change is 0.1. For every additional minute talked, the monthly charge increases by $0.1 or 10 cents. The initial value is 24. When there are no minutes talked, initially the charge is $24.

The slope is –400. this means for every year between 1960 and 1989, the population dropped by 400 per year in the city.

4.2 Section Exercises

Determine the independent variable. This is the variable upon which the output depends.

To determine the initial value, find the output when the input is equal to zero.

6 square units

20.01 square units

P ( t ) = 75 , 000 + 2500 t P ( t ) = 75 , 000 + 2500 t

(–30, 0) Thirty years before the start of this model, the town had no citizens. (0, 75,000) Initially, the town had a population of 75,000.

Ten years after the model began

W ( t ) = 0.5 t + 7.5 W ( t ) = 0.5 t + 7.5

( − 15 , 0 ) ( − 15 , 0 ) : The x -intercept is not a plausible set of data for this model because it means the baby weighed 0 pounds 15 months prior to birth. ( 0 , 7 . 5 ) ( 0 , 7 . 5 ) : The baby weighed 7.5 pounds at birth.

At age 5.8 months

C ( t ) = 12 , 025 − 205 t C ( t ) = 12 , 025 − 205 t

( 58 . 7 , 0 ) : ( 58 . 7 , 0 ) : In roughly 59 years, the number of people inflicted with the common cold would be 0. ( 0 , 12 , 0 25 ) ( 0 , 12 , 0 25 ) Initially there were 12,025 people afflicted by the common cold.

y = − 2 t +180 y = − 2 t +180

In 2070, the company’s profit will be zero.

y = 3 0 t − 3 00 y = 3 0 t − 3 00

(10, 0) In the year 1990, the company’s profits were zero

During the year 1933

• ⓐ 696 people
• ⓒ 174 people per year
• ⓓ 305 people
• ⓔ P(t) = 305 + 174t
• ⓕ 2,219 people
• ⓐ C(x) = 0.15x + 10
• ⓑ The flat monthly fee is $10 and there is a $0.15 fee for each additional minute used

P(t) = 190t + 4,360

• ⓐ R ( t ) = − 2 . 1 t   +   16 R ( t ) = − 2 . 1 t   +   16
• ⓑ 5.5 billion cubic feet
• ⓒ During the year 2017

More than 133 minutes

More than $42,857.14 worth of jewelry

More than $66,666.67 in sales

4.3 Section Exercises

When our model no longer applies, after some value in the domain, the model itself doesn’t hold.

We predict a value outside the domain and range of the data.

The closer the number is to 1, the less scattered the data, the closer the number is to 0, the more scattered the data.

61.966 years

Interpolation. About 60 ° F . 60 ° F .

This value of r indicates a strong negative correlation or slope, so C This value of r indicates a strong negative correlation or slope, so C

This value of r indicates a weak negative correlation, so B This value of r indicates a weak negative correlation, so B

Yes, trend appears linear because r = 0. 985 r = 0. 985 and will exceed 12,000 near midyear, 2016, 24.6 years since 1992.

y = 1 . 64 0 x + 13 . 8 00 , y = 1 . 64 0 x + 13 . 8 00 , r = 0. 987 r = 0. 987

y = − 0.962 x + 26.86 ,       r = − 0.965 y = − 0.962 x + 26.86 ,       r = − 0.965

y = − 1 . 981 x + 6 0. 197; y = − 1 . 981 x + 6 0. 197; r = − 0. 998 r = − 0. 998

y = 0. 121 x − 38.841 , r = 0.998 y = 0. 121 x − 38.841 , r = 0.998

( −2 , −6 ) , ( 1 , −12 ) , ( 5 , −20 ) , ( 6 , −22 ) , ( 9 , −28 ) ; ( −2 , −6 ) , ( 1 , −12 ) , ( 5 , −20 ) , ( 6 , −22 ) , ( 9 , −28 ) ; Yes, the function is a good fit.

( 189 .8 , 0 ) ( 189 .8 , 0 ) If 18,980 units are sold, the company will have a profit of zero dollars.

y = 0.00587 x + 1985 .4 1 y = 0.00587 x + 1985 .4 1

y = 2 0. 25 x − 671 . 5 y = 2 0. 25 x − 671 . 5

y = − 1 0. 75 x + 742 . 5 0 y = − 1 0. 75 x + 742 . 5 0

Review Exercises

y = − 3 x + 26 y = − 3 x + 26

y = 2 x − 2 y = 2 x − 2

Not linear.

( –9 , 0 ) ; ( 0 , –7 ) ( –9 , 0 ) ; ( 0 , –7 )

Line 1: m = − 2 ; m = − 2 ; Line 2: m = − 2 ; m = − 2 ; Parallel

y = − 0.2 x + 21 y = − 0.2 x + 21

More than 250

y = − 3 00 x + 11 , 5 00 y = − 3 00 x + 11 , 5 00

• ⓑ 100 students per year
• ⓒ P ( t ) = 1 00 t + 17 00 P ( t ) = 1 00 t + 17 00

Extrapolation

y = − 1.294 x + 49.412 ;   r = − 0.974 y = − 1.294 x + 49.412 ;   r = − 0.974

Practice Test

y = −1.5x − 6

y = −2x − 1

Perpendicular

(−7, 0); (0, −2)

y = −0.25x + 12

Slope = −1 and y-intercept = 6

y = 875x + 10,625

• ⓑ dropped an average of 46.875, or about 47 people per year
• ⓒ y = −46.875t + 1250

In early 2018

y = 0.00455x + 1979.5

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• Book title: College Algebra
• Publication date: Feb 13, 2015
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• Section URL: https://openstax.org/books/college-algebra/pages/chapter-4

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UNIT 5 - LINEAR RELATIONS & EQUATIONS

Part 1 - Graphing Linear Relations (Ch 6)

Part 2 - Solving Linear Equations (Ch 6)

• Unit 5 Notes Package •

6.0 - ALGEBRA (REVIEW)

6.1 - representing patterns.

PG 183   #1-5, 8, 11, 13, 16 , 19

6.2 (1) - GRAPHING LINEAR RELATIONS

6.2 (2) - graphing linear relations (cont...).

PG 194  #1-9, 12, 15

6.2 (3) - GRAPHING LINEAR RELATIONS (CONT...)

6.3 (1) - modelling equations to solve.

PG 204   #1- 4

6.3 ( 2 ) - USING ALGEBRA TO SOLVE EQUATIONS

6. 4 (1) - solving equations with more than 1 step, 6.4 (2) - the distributive property, 6.4 (2) - cont..., 6.4 (3) - variables on both sides, 6.4 (4) - word problems, unit 5 review.

GRAPHING REVIEW

WORKSHEET     ANSWER KEY

SOLVING EQUATIONS REVIEW

GENERAL REVIEW

Textbook - PG 215

UNIT 5 TEST

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Chapter 3: Graphing

3.4 Graphing Linear Equations

There are two common procedures that are used to draw the line represented by a linear equation. The first one is called the slope-intercept method and involves using the slope and intercept given in the equation.

If the equation is given in the form [latex]y = mx + b[/latex], then [latex]m[/latex] gives the rise over run value and the value [latex]b[/latex] gives the point where the line crosses the [latex]y[/latex]-axis, also known as the [latex]y[/latex]-intercept.

Example 3.4.1

Given the following equations, identify the slope and the [latex]y[/latex]-intercept.

• [latex]\begin{array}{lll} y = 2x - 3\hspace{0.14in} & \text{Slope }(m)=2\hspace{0.1in}&y\text{-intercept } (b)=-3 \end{array}[/latex]
• [latex]\begin{array}{lll} y = \dfrac{1}{2}x - 1\hspace{0.08in} & \text{Slope }(m)=\dfrac{1}{2}\hspace{0.1in}&y\text{-intercept } (b)=-1 \end{array}[/latex]
• [latex]\begin{array}{lll} y = -3x + 4 & \text{Slope }(m)=-3 &y\text{-intercept } (b)=4 \end{array}[/latex]
• [latex]\begin{array}{lll} y = \dfrac{2}{3}x\hspace{0.34in} & \text{Slope }(m)=\dfrac{2}{3}\hspace{0.1in} &y\text{-intercept } (b)=0 \end{array}[/latex]

When graphing a linear equation using the slope-intercept method, start by using the value given for the [latex]y[/latex]-intercept. After this point is marked, then identify other points using the slope.

This is shown in the following example.

Example 3.4.2

Graph the equation [latex]y = 2x - 3[/latex].

First, place a dot on the [latex]y[/latex]-intercept, [latex]y = -3[/latex], which is placed on the coordinate [latex](0, -3).[/latex]

Now, place the next dot using the slope of 2.

A slope of 2 means that the line rises 2 for every 1 across.

Simply, [latex]m = 2[/latex] is the same as [latex]m = \dfrac{2}{1}[/latex], where [latex]\Delta y = 2[/latex] and [latex]\Delta x = 1[/latex].

Placing these points on the graph becomes a simple counting exercise, which is done as follows:

Once several dots have been drawn, draw a line through them, like so:

Note that dots can also be drawn in the reverse of what has been drawn here.

Slope is 2 when rise over run is [latex]\dfrac{2}{1}[/latex] or [latex]\dfrac{-2}{-1}[/latex], which would be drawn as follows:

Example 3.4.3

Graph the equation [latex]y = \dfrac{2}{3}x[/latex].

First, place a dot on the [latex]y[/latex]-intercept, [latex](0, 0)[/latex].

Now, place the dots according to the slope, [latex]\dfrac{2}{3}[/latex].

This will generate the following set of dots on the graph. All that remains is to draw a line through the dots.

The second method of drawing lines represented by linear equations and functions is to identify the two intercepts of the linear equation. Specifically, find [latex]x[/latex] when [latex]y = 0[/latex] and find [latex]y[/latex] when [latex]x = 0[/latex].

Example 3.4.4

Graph the equation [latex]2x + y = 6[/latex].

To find the first coordinate, choose [latex]x = 0[/latex].

This yields:

[latex]\begin{array}{lllll} 2(0)&+&y&=&6 \\ &&y&=&6 \end{array}[/latex]

Coordinate is [latex](0, 6)[/latex].

Now choose [latex]y = 0[/latex].

[latex]\begin{array}{llrll} 2x&+&0&=&6 \\ &&2x&=&6 \\ &&x&=&\frac{6}{2} \text{ or } 3 \end{array}[/latex]

Coordinate is [latex](3, 0)[/latex].

Draw these coordinates on the graph and draw a line through them.

Example 3.4.5

Graph the equation [latex]x + 2y = 4[/latex].

[latex]\begin{array}{llrll} (0)&+&2y&=&4 \\ &&y&=&\frac{4}{2} \text{ or } 2 \end{array}[/latex]

Coordinate is [latex](0, 2)[/latex].

[latex]\begin{array}{llrll} x&+&2(0)&=&4 \\ &&x&=&4 \end{array}[/latex]

Coordinate is [latex](4, 0)[/latex].

Example 3.4.6

Graph the equation [latex]2x + y = 0[/latex].

[latex]\begin{array}{llrll} 2(0)&+&y&=&0 \\ &&y&=&0 \end{array}[/latex]

Coordinate is [latex](0, 0)[/latex].

Since the intercept is [latex](0, 0)[/latex], finding the other intercept yields the same coordinate. In this case, choose any value of convenience.

Choose [latex]x = 2[/latex].

[latex]\begin{array}{rlrlr} 2(2)&+&y&=&0 \\ 4&+&y&=&0 \\ -4&&&&-4 \\ \hline &&y&=&-4 \end{array}[/latex]

Coordinate is [latex](2, -4)[/latex].

For questions 1 to 10, sketch each linear equation using the slope-intercept method.

• [latex]y = -\dfrac{1}{4}x - 3[/latex]
• [latex]y = \dfrac{3}{2}x - 1[/latex]
• [latex]y = -\dfrac{5}{4}x - 4[/latex]
• [latex]y = -\dfrac{3}{5}x + 1[/latex]
• [latex]y = -\dfrac{4}{3}x + 2[/latex]
• [latex]y = \dfrac{5}{3}x + 4[/latex]
• [latex]y = \dfrac{3}{2}x - 5[/latex]
• [latex]y = -\dfrac{2}{3}x - 2[/latex]
• [latex]y = -\dfrac{4}{5}x - 3[/latex]
• [latex]y = \dfrac{1}{2}x[/latex]

For questions 11 to 20, sketch each linear equation using the [latex]x\text{-}[/latex] and [latex]y[/latex]-intercepts.

• [latex]x + 4y = -4[/latex]
• [latex]2x - y = 2[/latex]
• [latex]2x + y = 4[/latex]
• [latex]3x + 4y = 12[/latex]
• [latex]4x + 3y = -12[/latex]
• [latex]x + y = -5[/latex]
• [latex]3x + 2y = 6[/latex]
• [latex]x - y = -2[/latex]
• [latex]4x - y = -4[/latex]

For questions 21 to 28, sketch each linear equation using any method.

• [latex]y = -\dfrac{1}{2}x + 3[/latex]
• [latex]y = 2x - 1[/latex]
• [latex]y = -\dfrac{5}{4}x[/latex]
• [latex]y = -3x + 2[/latex]
• [latex]y = -\dfrac{3}{2}x + 1[/latex]
• [latex]y = \dfrac{1}{3}x - 3[/latex]
• [latex]y = \dfrac{3}{2}x + 2[/latex]
• [latex]y = 2x - 2[/latex]

For questions 29 to 40, reduce and sketch each linear equation using any method.

• [latex]y + 3 = -\dfrac{4}{5}x + 3[/latex]
• [latex]y - 4 = \dfrac{1}{2}x[/latex]
• [latex]x + 5y = -3 + 2y[/latex]
• [latex]3x - y = 4 + x - 2y[/latex]
• [latex]4x + 3y = 5 (x + y)[/latex]
• [latex]3x + 4y = 12 - 2y[/latex]
• [latex]2x - y = 2 - y \text{ (tricky)}[/latex]
• [latex]7x + 3y = 2(2x + 2y) + 6[/latex]
• [latex]x + y = -2x + 3[/latex]
• [latex]3x + 4y = 3y + 6[/latex]
• [latex]2(x + y) = -3(x + y) + 5[/latex]
• [latex]9x - y = 4x + 5[/latex]

Answer Key 3.4

Intermediate Algebra Copyright © 2020 by Terrance Berg is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License , except where otherwise noted.

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2.4: Graphing Linear Equations- Answers to the Homework Exercises

• Last updated
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• Page ID 45036

• Darlene Diaz
• Santiago Canyon College via ASCCC Open Educational Resources Initiative

Graphing and Slope

• \(\frac{1}{3}\)
• \(\frac{4}{3}\)
• \(\frac{1}{2}\)
• \(-\frac{1}{3}\)
• \(\frac{16}{7}\)
• \(-\frac{7}{17}\)
• \(\frac{1}{16}\)
• \(\frac{24}{11}\)
• \(x=\frac{23}{6}\)
• \(y=-\frac{29}{6}\)

Equations of Lines

• \(y=-\frac{3}{4}x-1\)
• \(y = −6x + 4\)
• \(y = − \frac{1}{4} x + 3\)
• \(y = \frac{1}{3} x + 3\)
• \(y = −3x + 5\)
• \(y = − \frac{1}{10} x − \frac{37}{10}\)
• \(y = \frac{7x}{3} − 8\)
• \(y = −4x + 3\)
• \(y = \frac{1}{10} x − \frac{3}{10}\)
• \(y = − \frac{4}{7} x + 4\)
• \(y=\frac{5}{2}x\)

• \(y − (−5) = 9(x − (−1))\)
• \(y − (−2) = −3(x − 0)\)
• \(y − (−3) = \frac{1}{5} (x − (−5))\)
• \(y − 2 = 0(x − 1)\)
• \(y − (−2) = −2(x − 2)\)
• \(y − 1 = 4(x − (−1))\)
• \(y − (−4) = − \frac{2}{3} (x − (−1))\)
• \(y = − \frac{3}{5} x + 2\)
• \(y = − \frac{3}{2} x + 4\)
• \(y = x − 4\)
• \(y = − \frac{1}{2} x\)
• \(y = − \frac{2}{3} x − \frac{10}{3}\)
• \(y = − \frac{5}{2} x − 5\)
• \(y = −3\)
• \(y − 3 = −2(x + 4)\)
• \(y + 2 = \frac{3}{2} (x + 4)\)
• \(y + 3 = − \frac{8}{7} (x − 3)\)
• \(y − 5 = − \frac{1}{8} (x + 4)\)
• \(y + 4 = −(x + 1)\)
• \(y = − \frac{8}{7} x − \frac{5}{7}\)
• \(y = −x + 2\)
• \(y = − \frac{1}{10} x − \frac{3}{2}\)
• \(y=\frac{1}{3}x+1\)

Parallel and Perpendicular Lines

• \(m_{||} = 2\)
• \(m_{||} = 1\)
• \(m_{||} = − \frac{2}{3}\)
• \(m_{||} = \frac{6}{5}\)
• \(m_{⊥} = 0\)
• \(m_{⊥} = −3\)
• \(m_{⊥} = 2\)
• \(m_{⊥} = − \frac{1}{3}\)
• \(y − 4 = \frac{9}{2} (x − 3)\)
• \(y − 3 = \frac{7}{5} (x − 2)\)
• \(y + 5 = −(x − 1)\)
• \(y − 2 = \frac{1}{5} (x − 5)\)
• \(y − 2 = − \frac{1}{4} (x − 4)\)
• \(y + 2 = −3(x − 2)\)
• \(y = −2x + 5\)
• \(y = − \frac{4}{3} x − 3\)
• \(y = − \frac{1}{2} x − 3\)
• \(y = − \frac{1}{2} x − 2\)
• \(y = x − 1\)
• \(y=-2x+5\)

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Unit 3: Linear relationships

Lesson 3: representing proportional relationships.

• Graphing proportional relationships: unit rate (Opens a modal)
• Graphing proportional relationships from a table (Opens a modal)
• Graphing proportional relationships from an equation (Opens a modal)
• Graphing proportional relationships Get 3 of 4 questions to level up!

Lesson 4: Comparing proportional relationships

• Rates & proportional relationships example (Opens a modal)
• Rates & proportional relationships: gas mileage (Opens a modal)
• Rates & proportional relationships Get 5 of 7 questions to level up!

Lesson 7: Representations of linear relationships

• Linear & nonlinear functions: missing value (Opens a modal)

Lesson 8: Translating to y=mx+b

• Intro to slope-intercept form (Opens a modal)
• Graph from slope-intercept equation (Opens a modal)

Lesson 9: Slopes don't have to be positive

• Intro to intercepts (Opens a modal)
• Slope-intercept equation from slope & point (Opens a modal)
• Linear & nonlinear functions: word problem (Opens a modal)
• Intercepts from a graph Get 3 of 4 questions to level up!
• Slope from graph Get 3 of 4 questions to level up!
• Slope-intercept intro Get 3 of 4 questions to level up!
• Graph from slope-intercept form Get 3 of 4 questions to level up!
• Slope-intercept equation from graph Get 3 of 4 questions to level up!

Lesson 10: Calculating slope

• No videos or articles available in this lesson
• Slope from two points Get 3 of 4 questions to level up!

Lesson 11: Equations of all kinds of lines

• Converting to slope-intercept form (Opens a modal)

Extra practice: Slope

• Intro to slope (Opens a modal)
• Worked examples: slope-intercept intro (Opens a modal)
• Graphing slope-intercept form (Opens a modal)
• Writing slope-intercept equations (Opens a modal)
• Slope-intercept form review (Opens a modal)
• Slope-intercept from two points Get 3 of 4 questions to level up!

Lesson 12: Solutions to linear equations

• Solutions to 2-variable equations (Opens a modal)
• Worked example: solutions to 2-variable equations (Opens a modal)
• Solutions to 2-variable equations Get 3 of 4 questions to level up!

Lesson 13: More solutions to linear equations

• Completing solutions to 2-variable equations (Opens a modal)
• Complete solutions to 2-variable equations Get 3 of 4 questions to level up!

Extra practice: Intercepts

• x-intercept of a line (Opens a modal)
• Intercepts from an equation (Opens a modal)
• Worked example: intercepts from an equation (Opens a modal)
• Intercepts of lines review (x-intercepts and y-intercepts) (Opens a modal)
• Intercepts from an equation Get 3 of 4 questions to level up!

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Homework 5 Slope Graphing Lines Review

Homework 5 Slope Graphing Lines Review - Displaying top 8 worksheets found for this concept.

Some of the worksheets for this concept are Graphing lines, Ws3, Unit 4 analyze and graph linear equations functions and, Graphing lines in slope intercept form, Graphing linear equations work answer key, Slope date period, Work, Table of contents chapter 3 representations of a line 4.

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1. Graphing Lines SI.ks-ia1

3. unit 4: analyze and graph linear equations, functions and ..., 4. graphing lines in slope-intercept form -, 5. graphing linear equations worksheet answer key, 6. slope date period -, 7. worksheet, 8. table of contents chapter 3: representations of a line (4 ....

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Linear Equations Activity Ideas

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Linear equations require lots of practice as the skill progresses to include more steps and increases in complexity. Any of these activities can be used from the basics of simplifying expressions to solving equations with variables on both sides. 

Model by Using Algebra Tiles

Are you surprised I am starting with Algebra tiles? They are foundational for concrete understanding! You can learn more about using Algebra tiles in your classroom by grabbing our free Getting Started with Algebra Tiles guide, which can be found by checking out our post on Solving Equations . This Modeling Equations with Variables on Both Sides activity is a great way to practice solving equations with Algebra tiles. You just need some Algebra tiles (or your students can draw them!).

Student Teacher

Put students into pairs and show an equation on the board. Have one student instruct the other on how to solve as the student listening writes each step and solution. Then, show a new equation and have students switch roles. This gives students a chance to teach and reinforce what they remember about linear equations. I love this activity because it is simple and it makes every student explain their thinking. You, as the teacher, can circulate listening to each pair.

Round Table

Give students individual white boards and have them work in teams of 2-4.  With one equation written on the board, the first person will solve step one.  The second person will complete the second step in solving and the third will complete the next step. Keep rotating until the problem is solved and the last person checks the solution.  Have groups hold up their boards when they are finished.  If you want something like this, we have this Solving Equations with Variables on Both Sides Round Table available! 

Board Races

After students have had time to practice, implement “Board Races.” Two students will come up to the board and race to solve an equation shown on the board. The person who solves it correctly first stays up at the board for the next equation with a new competitor. I like to have the students who aren’t at the board working the equations on notebook paper to help check the solutions. An element of competition makes repetitive practice more fun! For race type activities, I like to have teams compete (boys v. girls or one side of the room v. the other side of the room).

Digital Activities 

I love our digital activities! This linear equations set of digital activities includes simplifying expressions, solving one and two-step equations, solving multi-step equations, and equations with variables on both sides, making it the perfect review before a linear equations test. 

What are some fun ways that students practice solving linear equations in your class? You can check out how to turn any worksheet into an activity which can easily be used for linear equations too!

Getting Started with Algebra Tiles

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