## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 7 - Section 7.2 - Systems of Linear Equations in Three Variables - Exercise Set - Page 829: 14

#### Answer

The solution of the system is $\left( 0,0,4 \right)$.

#### Work Step by Step

Let us consider the given equation: $x+3y+5z=20$ ...... (I) $y-4z=-16$ ...... (II) $3x-2y+9z=36$ ...... (III) Now, multiply equation (I) by 3 and subtract from equation (III): \begin{align} & 3x+9y+15z=6 \\ & 3x-2y+9z=36 \\ \end{align} Solve and we get: $11y+6z=24$ ...... (IV) Multiply equation (II) by 11 and subtract from equation (IV): \begin{align} & 11y+6z=24 \\ & 11y-44z=-176 \\ \end{align} Solve and we get: \begin{align} & 50z\ =\ 200 \\ & z=4 \end{align} Now, putting the value of $z=4$ in equation (II): \begin{align} & y-4\times 4=-16 \\ & y=0 \end{align} Putting the value of $y=0$ and $z=4$ in equation (I) \begin{align} & x+3\left( 0 \right)+5\left( 4 \right)=20 \\ & x+20=20 \\ & x=0 \end{align} Thus, the ordered triple $\left( 0,0,4 \right)$ satisfies the three systems.

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