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 830: 26


The numbers are $x=-3$ ; $y=0$ ; $z=2$

Work Step by Step

Let us consider the given system of equations and mark the equations as shown below: $\frac{x+3}{2}-\frac{y-1}{2}+\frac{z+2}{4}=\frac{3}{2}$ (A) $\frac{x-5}{2}+\frac{y+1}{3}-\frac{z}{4}=-\frac{25}{6}$ (B) $\frac{x-3}{4}-\frac{y+1}{2}+\frac{z-3}{2}=-\frac{5}{2}$ (C) Simplify the first equation by taking the L.C.M and the equation becomes: $\begin{align} & 2\left( x+3 \right)-2\left( y-1 \right)+z+2=6 \\ & 2x+6-2y+2+z+2=6 \\ & 2x-2y+z+10=6 \end{align}$ Now, subtract 10 from both sides: $\begin{align} & 2x-2y+z+10-10=6-10 \\ & 2x-2y+z=-4 \end{align}$ Simplify the second equation by taking the L.C.M and the equation becomes: $\begin{align} & 6\left( x-5 \right)+4\left( y+1 \right)-3z=-50 \\ & 6x-30+4y+4-3z=-50 \\ & 6x+4y-3z-26=-50 \end{align}$ By adding 26 to both sides we get: $\begin{align} & 6x+4y-3z-26+26=-50+26 \\ & 6x+4y-3z=-24 \end{align}$ Simplify the third equation by taking the L.C.M and the equation becomes: $\begin{align} & x-3-2\left( y+1 \right)+2\left( z-3 \right)=-10 \\ & x-3-2y-2+2z-6=-10 \\ & x-2y+2z-11=-10 \end{align}$ Add 11 to both sides to get: $\begin{align} & x-2y+2z-11+11=-10+11 \\ & x-2y+2z=1 \end{align}$ The system of equations now becomes: $2x-2y+z=-4$ $6x+4y-3z=-24$ $x-2y+2z=1$ Mark equations as shown below: $2x-2y+z=-4$ (I) $6x+4y-3z=-24$ (II) $x-2y+2z=1$ (III) By multiplying equation (III) by −2 and eliminate x equations (I) and (II): we get, $\begin{align} & \text{ }2x-2y+\text{ }z=-4 \\ & \underline{-2x+4y-4z=-2} \\ & \text{ 2}y-3z\text{ }=-6 \\ \end{align}$ (IV) By multiplying equation (I) by −6 and eliminating x from equations (I) and (II): we get $\begin{align} & \text{ }6x+\text{ }4y-\text{ }3z=\text{ }-24 \\ & \underline{-6x+12y-12z=\text{ }-\text{ }6} \\ & \text{ }16y\text{ }-\text{ 15}z=-\text{ }30 \\ \end{align}$ (V) Multiply equation (IV) by −8 and then add equations (IV) and (VI) to get the value of z: $\begin{align} & \text{ }16y-15z=-30 \\ & \underline{-16y+24y=\text{ }48} \\ & \text{ }9z=18 \\ \end{align}$ (VI) By dividing equation (V) by 9 we get: $\begin{align} & \frac{9z}{9}=\frac{18}{9} \\ & \text{ }z=2 \end{align}$ Substitute the value of z in equation (IV): $\begin{align} & 2y-3\left( 2 \right)=-6 \\ & 2y-6=-6 \end{align}$ Now, adding 6 to both sides, we get $\begin{align} & 2y-6+6=-6+6 \\ & 2y=0 \end{align}$ It implies $y=0.$ Now, putting the values of y and z in equation (III) to get the value of x: $\begin{align} & x-2\left( 0 \right)+2\left( 2 \right)=1 \\ & x+4=1 \end{align}$ Now, subtract 4 from both sides of the equation: $\begin{align} & x+4-4=1-4 \\ & x=-3 \end{align}$ Hence, the values are $x=-3$, $y=0,$ and $z=2$.
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