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Q. 54

Expert-verifiedFound in: Page 714

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 inconsistent

x + 4y - 3z = -8

3x - y + 3z = 12

x + y + 6z = 1

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

We are given a system of equations

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

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,

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

Also

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

Hence we get

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

We get,

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

hence x=3

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

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

The solution set is $\left\{\right(x,y,z)=(-3,-\frac{8}{3},\frac{1}{9}\left)\right\}$

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