Question

Solve each system of linear equations.

2x+4yz=1x+7y+2z=43x2y+3z=3\begin{array}{r} -2 x+4 y-z=-1 \\ x+7 y+2 z=-4 \\ 3 x-2 y+3 z=-3 \end{array}

Solution

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2x+4yz=1x+7y+2z=43x2y+3z=3\begin{align*} -2x+4y-z=-1\\ x+7y+2z=-4\\ 3x-2y+3z=-3 \end{align*}

We will multiply the second equation by 2 and add it to the first equation.

Now the first equation becomes:

2x+4yz+2(x+7y+2z)=1+2(4)-2x+4y-z+2(x+7y+2z)=-1+2\cdot (-4)

2x+4yz+2x+14y+4z=18-2x+4y-z+2x+14y+4z=-1-8

18y+3z=918y+3z=-9

Now we will multiply the second equation by -3 and add it to the third equation.

The third equation becomes:

3x2y+3z3(x+7y+2z)=34(3)3x-2y+3z-3(x+7y+2z)=-3-4\cdot (-3)

3x2y+3z3x21y6z=3+123x-2y+3z-3x-21y-6z=-3+12

23y3z=9-23y-3z=9

The two equations obtained are:

18y+3z=918y+3z=-9

23y3z=9-23y-3z=9

Now we will add the equation 18y+3z=918y+3z=-9 to the equation 23y3z=9-23y-3z=9

18y+3z23y3z=9+918y+3z-23y-3z=-9+9

5y=0-5y=0

y=0\boxed{y=0}

From the equation 18y+3z=918y+3z=-9 it yields

180+3z=918\cdot 0+3z=-9

3z=93z=-9

z=3\boxed{z=-3}

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