Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 7 - Section 7.2 - Systems of Linear Equations in Three Variables - Exercise Set - Page 829: 18


The solution of the system is $\left( 0.5,3,-2 \right)$.

Work Step by Step

It is given that: $7z-3=2\left( x-3y \right)$ ...... (I) $5y-3z-7=4x$ ...... (II) $4+5z=3\left( 2x-y \right)$ ...... (III) From equation (I) $7z-3=2\left( x-3y \right)$: $2x-6y-7z=-3$ ...... (IV) From equation (II) $5y-3z-7=4x$: $4x-5y-3z=-7$ ...... (V) From equation (III) $4+5z=3\left( 2x-y \right)$: $6x-3y-5z=4$ ...... (VI) Now, multiply the equation (IV) by 2 and subtract from equation (V): $\begin{align} & 4x-12y-14z=-6 \\ & 4x-5y-3z=-7 \\ \end{align}$ Solve and we get: $-7y-11z=1$ ...... (VII) Multiply equation (IV) by 3 and subtract from equation (VI): $\begin{align} & 6x-18y-21z=-9 \\ & 6x-3y-5z=4 \\ \end{align}$ Solve and we get: $-15y-16z=-13$ ...... (VIII) From equation (VIII) and, putting the value of z in equation (VII): $\begin{align} & -7y-11\left( \frac{13-15y}{16} \right)=1 \\ & -7y-\frac{143}{16}+\frac{165y}{16}=1 \\ & -112y-143+165y=16 \\ & 53y=159 \end{align}$ So, $\begin{align} & y=\frac{159}{53} \\ & y=3 \\ \end{align}$ Putting the value of $y=3$ in equation (VII): $\begin{align} & -7\left( 3 \right)-11z=1 \\ & -21-11z=1 \\ & z=-2 \end{align}$ Putting the values of $y=3$ and $z=-2$ in equation (I): $\begin{align} & 2x-6\left( 3 \right)-7\left( -2 \right)=-3 \\ & x=0.5 \end{align}$ Thus, the ordered triple $\left( 0.5,3,-2 \right)$ satisfies the three systems.
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