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Systems of Linear Equations
11/11/12 lntaylor ©
Table of Contents
Learning Objectives
Solving Systems by Graphing
Solving Systems by Substitution
Solving Systems by Elimination
Word Problems
Practice
11/11/12 lntaylor ©
Learning Objectives
TOC11/11/12 lntaylor ©
Solving Systems of Linear Equations
• In these sections you will learn/review how to:
– Meet or exceed proficiency in AF 9.0
– Pass questions regarding these standards on District or State Tests
– Solve Systems of Equations using three (3) different methods
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1st MethodGraphing
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Systems of Linear Equations
• Graphing Method
– Graph two different linear equations– Determine the point where the lines crosses– Prove that point works with both equations
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Find the solution to:
y = 2x + 1 and
y = - x + 4
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0,0
Find the solution to y = 2x + 1 and y = - x + 4
Step 1 – Graph the 1st lineIdentify slope m and bLocate 0,0Go up or down b Put a dot for the 1st pointGo up or down ΔyGo right ΔxPut a dot for 2nd pointConnect dotsLabel the line
y = 2x + 1
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Step 2 – Graph the 2nd lineIdentify slope m and bLocate 0,0Go up or down b Put a dot for the 1st pointGo up or down ΔyGo right ΔxPut a dot for 2nd pointConnect dotsLabel the lineFind the point where the lines crossPlug into each equationThe point (1,3) is the solution for both equations
y = - x + 4
y = 2x + 1 and y = -x + 43 = 2(1) + 1 and 3 = -1 + 43 = 3 and 3 = 3
(1,3)
Now you try!
y = 1/2x + 1 and
y = - x + 4
TOC11/11/12 lntaylor ©
0,0
Find the solution to y = 1/2x + 1 and y = - x + 4
Step 1 – Graph the 1st lineIdentify slope m and bLocate 0,0Go up or down b Put a dot for the 1st pointGo up or down ΔyGo right ΔxPut a dot for 2nd pointConnect dotsLabel the line
y = 1/2x + 1
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Step 2 – Graph the 2nd lineIdentify slope m and bLocate 0,0Go up or down b Put a dot for the 1st pointGo up or down ΔyGo right ΔxPut a dot for 2nd pointConnect dotsLabel the lineFind the point where the lines crossPlug into each equationThe point (1,3) is the solution for both equations
y = - x + 4
y = 1/2x + 1 and y = - x + 42 = 1/2(2) + 1 and 2 = - 2 + 42 = 2 and 2 = 2
(2,2)
Now you try!
y = 1/3x – 4and
y = - x
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0,0
Find the solution to y = 1/3x - 4 and y = - x
Step 1 – Graph the 1st lineIdentify slope m and bLocate 0,0Go up or down b Put a dot for the 1st pointGo up or down ΔyGo right ΔxPut a dot for 2nd pointConnect dotsLabel the line
y = 1/3x - 4
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Step 2 – Graph the 2nd lineIdentify slope m and bLocate 0,0Go up or down b Put a dot for the 1st pointGo up or down ΔyGo right ΔxPut a dot for 2nd pointConnect dotsLabel the lineFind the point where the lines crossPlug into each equationThe point (1,3) is the solution for both equations
y = - x + 4
y = 1/3x – 4 and y = - x-3 = 1/3(3) – 4 and - 3 = - 3- 3 = - 3 and - 3 = - 3
(3, - 3)
2nd MethodSubstitution
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Systems of Linear Equations
• Substitution Method
– Set one equal to y – Substitute the 2nd equation into the 1st equation
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Find the solution to:
y = 2x + 1 and
y = - x + 4
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Step 4
Step 5
Rewrite without the y and solve
Plug the answer into both equations
Step 6 The 1st number is x; the 2nd number is y
Step 7 The solution is (x,y) or in this equation (1,3)
Step 2
Step 1 Write the 1st equation as y =
Write the 2nd equation as = y
Step 3 They are both = y
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Find the solution to y = 2x + 1 and y = - x + 4y = 2x + 1 - x + 4 = y
- x + 4 = 2x + 1 4 – 1 = 2x + x 3 = 3x x = 1
- x + 4 = 2x + 1- (1) + 4 = 2(1) + 1 - 1 + 4 = 2 + 1 3 = 3
(1,3)
Now you try!
y = 1/2x + 1 and
y = - x + 4
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Step 4
Step 5
Rewrite without the y and solve
Plug the answer into both equations
Step 6 The 1st number is x; the 2nd number is y
Step 7 The solution is (x,y) or in this equation (2,2)
Step 2
Step 1 Write the 1st equation as y =
Write the 2nd equation as = y
Step 3 They are both = y
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Find the solution to y = 1/2x + 1 and y = - x + 4y = 1/2x + 1 - x + 4 = y
- x + 4 = 1/2x + 1 4 – 1 = 1/2x + x 3 = 1½ x 6 = 3x x = 2
- x + 4 = 1/2x + 1- (2) + 4 = ½(2) + 1 - 2 + 4 = 1 + 1 2 = 2(2,2)
Now you try!
y = 1/3x – 4and
y = - x
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Step 4
Step 5
Rewrite without the y and solve
Plug the answer into both equations
Step 6 The 1st number is x; the 2nd number is y
Step 7 The solution is (x,y) or in this equation (3,- 3)
Step 2
Step 1 Write the 1st equation as y =
Write the 2nd equation as = y
Step 3 They are both = y
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Find the solution to y = 1/3x - 4 and y = - x y = 1/3x - 4 - x = y
- x = 1/3x - 4 4 = 1/3x + x 4 = 1⅓ x 12 = 4x x = 3
- x = 1/3x - 4- (3) = ⅓(3) - 4 - 3 = 1 - 4 - 3 = - 3(3,- 3)
3rd MethodElimination
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Systems of Linear Equations
• Elimination Method
– Write the equations underneath each other – Choose a multiplier for one equation that eliminates a
variable– Plug in the answer and solve for the 2nd variable
TOC11/11/12 lntaylor ©
Find the solution to:
y = 2x + 1 and
y = - x + 4
TOC11/11/12 lntaylor ©
Step 4
Step 5
Add the equations and solve for the variable
Plug the answer into an equation and solve
Step 6 Make sure you put the numbers in the right order!
Step 7 The solution is (x,y) or in this equation (1,3)
Step 2
Step 1 Write the equations underneath each other
Pick a multiplier that will eliminate 1 variable
Step 3 Rewrite the equations
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Find the solution to y = 2x + 1 and y = - x + 4y = 2x + 1 y = - x + 4
(2) y = (2)(-x + 4) 2y = - 2x + 82y = - 2x + 8
3y = 9 y = 3 (x,3)
y = 2x + 13 = 2x + 12 = 2xx = 1
(1,3)
Now you try!
y = 1/2x + 1 and
y = - x + 4
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Step 4
Step 5
Add the equations and solve for the variable
Plug the answer into an equation and solve
Step 6 Make sure you put the numbers in the right order!
Step 7 The solution is (x,y) or in this equation (2,2)
Step 2
Step 1 Write the equations underneath each other
Pick a multiplier that will eliminate 1 variable
Step 3 Rewrite the equations
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Find the solution to y = ½x + 1 and y = - x + 4y = ½x + 1 y = - x + 4
(2) y = (2)(½x + 1) 2y = x + 22y = x + 2
3y = 6 y = 2 (x,2)
y = ½x + 12 = ½x + 11 = ½x2 = x
(2,2)
Now you try!
y = 1/3x – 4and
y = - x
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Step 4
Step 5
Add the equations and solve for the variable
Plug the answer into an equation and solve
Step 6 Make sure you put the numbers in the right order!
Step 7 The solution is (x,y) or in this equation (3,- 3)
Step 2
Step 1 Write the equations underneath each other
Pick a multiplier that will eliminate 1 variable
Step 3 Rewrite the equations
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Find the solution to y = ⅓x - 4 and y = - xy = ⅓x - 4 y = - x
(3) y = (3)(⅓x - 4) 3y = x - 123y = x - 12
4y = - 12 y = - 3 (x,- 3)
y = ⅓x - 4- 3 = ⅓x - 4 1 = ⅓x 3 = x
(3,- 3)
Word Problems
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Step 4
Step 5
Pick a multiplier that will eliminate 1 variable
Add the equations and solve
Step 6 Plug in the answer and solve
Step 2
Step 1 Highlight the first equation
Highlight the second equation
Step 3 Write the equations underneath each other
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4A + 9C = $824A + 1C = $34 – 1(4A + 1C) = – 1($34)
– 4A – 1C = - $34
4A + 9C = $82– 4A – 1C = – $34
8C = $48 C = $6
4A + 9C = $824A + 9($6) = $824A + $54 = $82 4A = $28 A = $7
Adult Tickets were $7Child Tickets were $6
Step 4
Step 5
Pick multiplier(s) that will eliminate 1 variable
Add the equations and solve
Step 6 Plug into the original equation and solve
Step 2
Step 1 Highlight the first equation
Highlight the second equation
Step 3 Write the equations underneath each other
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5A + 7C = $1296A + 10C = $174
6(5A + 7C) = 6($129) or 30A + 42C = $774 30A + 42C = $774– 30A – 50C = – $870
– 8C = – $96 C = $12
5A + 7C = $1295A + 7($12) = $1295A + $84 = $129 5A = $45 A = $9
Adult Tickets were $9Child Tickets were $12
– 5(6A + 10C) = – 5($174) or – 30A – 50C = – $870
Step 4
Step 5
Pick multiplier(s) that will eliminate 1 variable
Add the equations and solve
Step 6 Plug into the original equation and solve
Step 2
Step 1 Highlight the first equation
Highlight the second equation
Step 3 Write the equations underneath each other
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3V + 6B = 19210V + 5B = 205
10(3V + 6B) = 10(192) or 30V + 60B = 1920 30V + 60B = 1920– 30V – 15B = – 615
45B = 1305 B = 29
3V + 6B = 192 3V + 6(29) = 192 3V + 174 = 192 3V = 18 V = 6
6 students in each van29 students in each bus
– 3(10V + 5B) = – 3(205) or – 30V – 15B = – 615
Step 4
Step 5
No multiplier(s) is needed here!
Add the equations and solve
Step 6 Plug into the original equation and solve
Step 2
Step 1 Highlight the first equation
Highlight the second equation
Step 3 Write the equations underneath each other
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A + B = 16 A – B = 2
2A = 18 A = 9
A + B = 16 9 + B = 16 B = 7 A is 9
B is 7
Practice
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11/11/12 lntaylor ©TOC
Problem Answer
y = 2x + 5 and y = – 5x – 16
y = – 4x + 5 and y = – 2x + 3
y = 5x + 10 and 7x – 7y = - 14
y = – 6x + 20 and – x – y = - 10
y = 5x – 5y = – 20 and x + 2y = 17
x – 4y = 16 and – 3x – 3y= 12
– 5x – 4y = – 22 and – 3x – y = – 16
– x – 5y = – 3 and – 8x – 6y = 10
2x – 7y = 6 and – 7x + 5y = – 21
> (-3, -1) >
>
>
>
>
(1, 1)(- 2, 0)
(2, 8)
(3, 7)
(0, - 4)
> (6, - 2)
> (- 2, 1)
> (3, 0)
clear answers