Question 309474
well once you see how to solve one theres no need for me to solve the other 2 as they are using the same method to solve so all you need is one example.

{{{5x+4y-z=1}}}
{{{2x-2y+z=1}}}
{{{-x-y+z=2}}}
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{{{z=(2+x+y)}}}
{{{5x+4y-(2+x+y)=1}}} 1st equation with z substitute
{{{2x-2y+(2+x+y)=1}}} 2nd equation with z substitute
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{{{5x+4y-2-x-y=1}}}
{{{2x-2y+(2+x+y)=1}}}add like terms
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{{{4x+3y=3}}}
{{{3x-y=-1}}}multiply this line by 3
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{{{4x+3y=3}}}
{{{9x-3y=-3}}}add the 2 lines together
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{{{13x=0}}}
{{{highlight(x=0)}}}
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{{{4x+3y=3}}}multiply this line by 3
{{{3x-y=-1}}}multiply this line by 4
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{{{12x+9y=9}}}
{{{12x-4y=-4}}}subtract these two lines
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{{{13y=13}}}
{{{highlight(y=1)}}}
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{{{z=(2+x+y)}}} solve for z
{{{z=(2+0+1)}}}
{{{highlight(z=3)}}}