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

Jun 21, 2017

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

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

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

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

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

$x + \left(4 + x\right) + \left(x + 9\right) = 73$

$3 x + 13 = 73$

$3 x = 60$

$x = 20$

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

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

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

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