Question

Solve each system of linear equations.$\begin{array}{r} -2 x+4 y-z=-1 \\ x+7 y+2 z=-4 \\ 3 x-2 y+3 z=-3 \end{array}$

Solution

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Step 1
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\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:

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

$-2x+4y-z+2x+14y+4z=-1-8$

$18y+3z=-9$

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

The third equation becomes:

$3x-2y+3z-3(x+7y+2z)=-3-4\cdot (-3)$

$3x-2y+3z-3x-21y-6z=-3+12$

$-23y-3z=9$

The two equations obtained are:

$18y+3z=-9$

$-23y-3z=9$

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

$18y+3z-23y-3z=-9+9$

$-5y=0$

$\boxed{y=0}$

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

$18\cdot 0+3z=-9$

$3z=-9$

$\boxed{z=-3}$

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