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# Systems of Equations—Solving with Elimination Practice Questions (page 2)

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Updated on Oct 3, 2011

#### Practice 2

Choose a variable to eliminate. Adding any two of the three equations will not eliminate a variable, and subtracting one equation from another will not eliminate a variable, either. To eliminate y, we can multiply the first equation by –2 and add it to the second equation:

–2(–x + y + 3z) = –2(1)
2x – 2y – 6z = –2

We can combine another pair of equations so that we have a second equation with just the variables x and z. Multiply the second equation by 3 and third equation by –2:

 3(2y + 9x + 4z) = 3(–1) –2(–8z – 4x) = –2(3y)
 6y + 27x + 12z = –3 16z + 8x = –6y

Add the equations to eliminate the variable y:

Simplify:

27x + 12z = 16z + 8x – 3
19x + 12z = 16z – 3
19x – 4z = –3

We have two equations with x and z:

11x – 2z = –3
19x – 4z = –3

Multiply the first equation by 2 and then subtract the second equation:

2(11x – 2z) = 2(–3)
22x – 4z = –6
x = –1

We have the value of x. Substitute it into one of the two equations with just x and z:

11x – 2z = –3
11(–1) – 2z = –3
–11 – 2z = –3
–2z = 8
z = –4

Substitute the values of x and z into any one of the original equations to find the value of y:

x + y + 3z = 1
–(–1) + y + 3(–4) = 1
1 + y – 12 = 1
y – 11 = 1
y = 12

The solution to this system of equations is x = –1, y = 12, z = –4.

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