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Answers without the blur. Sign up and see all textbooks for free! Q. 54

Expert-verified Found in: Page 714 ### Precalculus Enhanced with Graphing Utilities

Book edition 6th
Author(s) Sullivan
Pages 1200 pages
ISBN 9780321795465 # solve each system of equations. If the system has no solution, say that it is inconsistentx + 4y - 3z = -83x - y + 3z = 12x + y + 6z = 1

The solution of the system can be given as $\left\{\left(x,y,z\right)=\left(-3,-\frac{8}{3},\frac{1}{9}\right)\right\}$

See the step by step solution

## Step 1: Given information

We are given a system of equations

$x+4y-3z=-8\left(1\right)\phantom{\rule{0ex}{0ex}}3x-y+3z=12\left(2\right)\phantom{\rule{0ex}{0ex}}x+y+6z=1\left(3\right)$

## Step 2: Add equation 1 and 2

We get,

$x+4y-3z=-8+3x-y+3z=124x+3y=4$

Hence we get, $4x+3y=4$ (4)

Now Multiply equation 1 by 2 and add it to equation 3

We get,

$2\left(x+4y-3z=-8\right)\phantom{\rule{0ex}{0ex}}2x+8y-6z=-16$

Also

$2x+8x-6z=-16+x+y+6z=13x+9y=-15$

Hence we get

$3x+9y=-15\phantom{\rule{0ex}{0ex}}x+3y=-5\left(5\right)$

## Step 3: Subtract equation 4 and 5

We get,

$4x+3y=4-x-3y=53x=9$

hence x=3

## Step 4: Find the values of y and z

We get,

$x+3y=-5\phantom{\rule{0ex}{0ex}}3+3y=-5\phantom{\rule{0ex}{0ex}}3y=-8\phantom{\rule{0ex}{0ex}}y=-\frac{8}{3}$

Now also we have

$x+4y-3z=-8\phantom{\rule{0ex}{0ex}}-3+4\left(-\frac{8}{3}\right)-3z=-8\phantom{\rule{0ex}{0ex}}z=\frac{1}{9}$

The solution set is $\left\{\left(x,y,z\right)=\left(-3,-\frac{8}{3},\frac{1}{9}\right)\right\}$ ### Want to see more solutions like these? 