Example

Determine if (1,3,2)(1,-3,2) and (4,1,5)(4,-1,-5) are Solutions to a System of Three Equations

Given the system of equations: {3x+y+z=2x+2y+z=33x+y+2z=4\begin{cases} 3x + y + z = 2 \\ x + 2y + z = -3 \\ 3x + y + 2z = 4 \end{cases} To determine if (1,3,2)(1,-3,2) is a solution, substitute the values into all three equations:

  1. 3(1)+(3)+2=23(1) + (-3) + 2 = 2
  2. 1+2(3)+2=31 + 2(-3) + 2 = -3
  3. 3(1)+(3)+2(2)=43(1) + (-3) + 2(2) = 4 Since it makes all equations true, (1,3,2)(1,-3,2) is a solution. To determine if (4,1,5)(4,-1,-5) is a solution, substitute the values into the equations:
  4. 3(4)+(1)+(5)=623(4) + (-1) + (-5) = 6 \neq 2 Since it does not make the first equation true, (4,1,5)(4,-1,-5) is not a solution.

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Updated 2026-05-26

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