By LearningExpress Editors
Updated on Oct 3, 2011
To review these concepts, go to Systems of Equations—Solving with Elimination Study Guide.
Systems of Equations—Solving with Elimination Practice Questions
Problems
Practice 1
Solve each system of equations.
 8y + 3x = –2, –8y – 14 = 5x
 x – 2y = 20, 4x + 7y = 5
 –4x + 3y = –4, 6x – 5y = 2
Practice 2
Find the solution to this system of equations:
 –x + y + 3z = 1
 2y + 9x + 4z = –1
 –8z – 4x = 3y
Solutions
Practice 1
1.  Choose a variable to eliminate. The first equation contains an 8y term and the second equation contains a –8y term. Add these equations to eliminate y: 
Solve for x:  
3x – 14 = 5x – 2  
–14 = 2x – 2  
–12 = 2x  
–6 = x  
Substitute –6 for x in either equation and solve for y:  
8y + 3x = –2  
8y + 3(–6) = –2  
8y – 18 = –2  
8y = 16  
y = 2  
The solution to this system of equations is x = –6, y = 2.  
2.  Choose a variable to eliminate. Adding the two equations will not eliminate a variable, and subtracting one equation from the other will not eliminate a variable, either. To eliminate y, we would have to multiply both equations by constants before adding. To eliminate x, we can multiply just the first equation by –4 before adding: 
–4(x – 2y) = –4(20)  
–4x + 8y = –80  
The first equation now contains a –4x term and the second equation contains a 4x term. Add these equations to eliminate x:  
Divide by 15 to solve for y:  
15y = –75  
y = –5  
Substitute –5 for y in either equation and solve for x:  
x – 2y = 20  
x – 2(–5) = 20  
x + 10 = 20  
x = 10  
The solution to this system of equations is x = 10, y = –5.  
3.  Choose a variable to eliminate. Adding the two equations will not eliminate a variable, and subtracting one equation from the other will not eliminate a variable, either. To eliminate x, we can multiply the first equation by 3 and multiply the second equation by 2 before adding: 
3(–4x + 3y) = 3(–4) 2(6x – 5y) = 2(2)  
–12x + 9y = –12 12x – 10y = 4  
The first equation now contains a –12x term and the second equation contains a 12x term. Add these equations to eliminate x:  
Divide by –1 to solve for y:  
–y = –8  
y = 8  
Substitute 8 for y in either equation and solve for x:  
–4x + 3y = –4  
–4x + 3(8) = –4  
–4x + 24 = –4  
–4x = –28  
x = 7  
The solution to this system of equations is x = 7, y = 8. 

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From Algebra in 15 Minutues A Day. Copyright © 2009 by LearningExpress, LLC. All Rights Reserved.
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