Determining Whether an Ordered Triple Is a Solution to a System

We will check each equation by substituting in the values of the ordered triple for $x,y,$ and $z.$

$\begin{array}{ccccc}\begin{array}{r}\hfill x+y+z=2\\ \hfill (3)+(\mathrm{-2})+(1)=2\\ \hfill \text{True}\end{array}& & \begin{array}{r}\hfill \text{<mn>6</mn><mi>x</mi><mn>−4</mn><mi>y</mi><mo>+</mo><mn>5</mn><mi>z</mi><mo>=</mo><mn>31</mn>}\\ \hfill 6(3)\mathrm{-4}(\mathrm{-2})+5(1)=31\\ \hfill 18+8+5=31\\ \hfill \text{True}\end{array}& & \begin{array}{r}\hfill \text{<mn>5</mn><mi>x</mi><mo>+</mo><mn>2</mn><mi>y</mi><mo>+</mo><mn>2</mn><mi>z</mi><mo>=</mo><mn>13</mn>}\\ \hfill 5(3)+2(\mathrm{-2})+2(1)=13\\ \hfill \text{<mn>15</mn><mn>−4</mn><mo>+</mo><mn>2</mn><mo>=</mo><mn>13</mn>}\\ \hfill \text{True}\end{array}\end{array}$

The ordered triple $\left(3,\mathrm{-2},1\right)$ is indeed a solution to the system.