How do you solve $x + y + z = 73, - 2 x + 2 y = 8, x - z = - 9$ $x+y+z=73, -2x+2y=8, x-z=-9$ ?

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Answer 1 · 2017-06-21T04:31:16

This is a system of linear equations.Let's label them as

eq1 : $x + y + z = 73$ $x+y+z=73$
eq2 : $- 2 x + 2 y = 8$ $-2x+2y=8$
eq3 : $x - z = - 9$ $x-z=-9$

Now eq2 can be written as $- x + y = 4$ $-x+y=4$ or $y = 4 + x$ $y=4+x$ (divide by $2$ $2$ )

and eq3 can be written as $z = x + 9$ $z=x+9$

Now we replace the values of $y$ $y$ and $z$ $z$ in eq1 hence

$x + (4 + x) + (x + 9) = 73$ $x+(4+x)+(x+9)=73$

$3 x + 13 = 73$ $3x+13=73$

$3 x = 60$ $3x=60$

$x = 20$ $x=20$

Now replace the value of $x$ $x$ in eq2 and eq3 to get the values of $y$ $y$ and $z$ $z$ hence

$y = 4 + x = 4 + 20 = 24$ $y=4+x=4+20=24$

$z = x + 9 = 20 + 9 = 29$ $z=x+9=20+9=29$

Finally the solution of the system is $(x, y, z) = (20, 24, 29)$ $(x,y,z)=(20,24,29)$

How do you solve x+y+z=73,−2x+2y=8,x−z=−9x+y+z=73, -2x+2y=8, x-z=-9?