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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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