Question

Solve for x, y, and z

Answer 1

ANSWER

$x=-3,y=-1,z=-2$

EXPLANATION

The given equations are:

$3x+4y-5z=-3--(1)$

$-5x-2y+6z=5--(2)$

and

$4x-8y+3z=-10$

$3z=8y-4x-10$

$z=\frac{8y-4x-10}{3}--(3)$

We substitute equation (3) into equation (2) and (1).

This implies that;

$3x+4y-5\frac{(8y-4x-10)}{3}=-3$

We multiply through by 3 to get,

$9x+12y-5(8y-4x-10)=-9$

We expand now to get,

$9x+12y-40y+20x+50=-9$

$29x-28y=-59--(4)$

We now put equation 3 in 2

$-5x-2y+6\frac{(8y-4x-10)}{3}=5$

This gives us,

$-5x-2y+2(8y-4x-10)=5$

We expand to obtain;

$-5x-2y+16y-8x-20=5$

$-13x+14y=25---(5)$

We nw solve equation (5) an (4) simultaneously.

Let us multiply equation (5) by 2 to get,

$-26x+28y=50---(6)$

We add equation (6) and (4) to obtain,

$3x=-9$

$\Rightarrow x=-3$

We substitute the value of $x$ in to equation (5)

$-13(-3)+14y=25$

$39+14y=25$

$14y=25-39$

$14y=-14$

$y=-1$

This implies that,

$z=\frac{8(-1)-4(-3)-10}{3}$

$z=\frac{-8+12-10}{3}$

$z=\frac{-6}{3}$

$z=-2$

$x=-3,y=-1,z=-2$

The correct answer is B.

Alternatively

$3x+4y-5z=-3--(1)$

$-5x-2y+6z=5--(2)$

$4x-8y+3z=-10--(3)$

We multiply equation (1) by 2 to get,

$6x+8y-10z=-6 --(4)$

We multiply equation (2) by 4,

$-20x-8y+24z=20--(5)$

Equation (3)-(5)

$24x-21z=-30--(6)$

Equation (4) + (3)

$10x-7z=-16--(7)$

We solve (7) and (60 simultaneously.

Multiply equation 7 by (3)

$30x-21z=-48--(8)$

Equation (6) -(8)

$-6x=18$

$x=-3$

This implies,

$10(-3)-7z=-16$

$-30-7z=-16$

$-7z=-16+30$

$-7z=14$

$z=-2$

$3(-3)+4y-5(-2)=-3$

$-9+4y+10=-3$

$4y+1=-3$

$4y=-4$

$y=-1$

The solution is

$x=-3,y=-1,z=-2$

Answer 2

I would do linear combinations

looking at the coeffients, I would cancel y first

multiply 2nd equation by 2 and add the 1st equation to it to get

-7x+0y+7z=7

-7x+7z=7

multiply first equation by 2 and add it to the last equation to get

10x+0y-7z=-16

10x-7z=-16

we now consider the 2 equations

-7x+7z=7 and

10x-7z=-16

add them together

3x+0z=-9

3x=-9

x=-3

sub back

-7x+7z=7

-7(-3)+7z=7

21+7z=7

7z=-14

z=-2

sub back

3x+4y-5z=-3

3(-3)+4y-5(-2)=-3

-9+4y+10=-3

4y=-4

y=-1

(x,y,z)=(-3,-1,-2)