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# Systems of Equations—Solving with Elimination Study Guide (page 2)

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Updated on Oct 3, 2011

## Systems With Three Variables

We can also solve systems of equations with three variables using the elimination method. Just as with the substitution method, we begin by reducing three equations with three variables to two equations with two variables. In fact, with the elimination method, sometimes we can eliminate two variables at once.

xy + 2z = 1
3x + 1 = 7y – 2z
x + 2y – 2z = 3

The first equation has an x term and the third equation has an –x term. Adding these equations will eliminate the variable x. But look again—the first equation has a –2z term and the third equation has a –2z term. When we add, we will eliminate the variable z, too:

We have the value of y, but we still need to find the values of the other two variables. Replace y with 4 in each equation:

 x – y + 2z = 1 3x + 1 = 7y – 2z –x + 2y – 2z = 3 x – 4 + 2z = 1 3x + 1 = 7(4) – 2z –x + 2(4) – 2z = 3 x + 2z = 5 3x + 1 = 28 – 2z –x + 8 – 2z = 3 3x + 2z = 27 –x – 2z = –5 x + 2z = 5

We have three equations, but the first and third equations are the same. Combine the first two equations. Both have a 2z term, so we can subtract the second equation from the first:

Divide both sides of the equation by –2, and we find that x = 11. Now that we have the values of x and y, we can substitute them into any of the original three equations to find the value of z:

xy + 2z = 1
(11) – (4) + 2z = 1
7 + 2z = 1
2z = –6
z = –3

The solution to this system of equations is x = 11, y = 4, z = –3.

Find practice problems and solutions for these concepts at Systems of Equations—Solving with Elimination Practice Questions.

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