Example

Determine if (2,1,3)(-2,-1,3) and (4,3,4)(-4,-3,4) are Solutions to a System of Three Equations

Given the system of equations: {xy+z=22xyz=62x+2y+z=3\begin{cases} x - y + z = 2 \\ 2x - y - z = -6 \\ 2x + 2y + z = -3 \end{cases} To determine if (2,1,3)(-2,-1,3) is a solution, substitute the values into all three equations:

  1. 2(1)+3=2-2 - (-1) + 3 = 2
  2. 2(2)(1)3=62(-2) - (-1) - 3 = -6
  3. 2(2)+2(1)+3=32(-2) + 2(-1) + 3 = -3 Since it makes all equations true, (2,1,3)(-2,-1,3) is a solution. To determine if (4,3,4)(-4,-3,4) is a solution, substitute the values into the equations:
  4. 4(3)+4=32-4 - (-3) + 4 = 3 \neq 2 Since it does not make the first equation true, (4,3,4)(-4,-3,4) is not a solution.

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Updated 2026-05-25

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