# How do you solve the system x+y+z=6, x-2y=-7, and 4x+3y+z=7?

$\left(x , y , z\right) = \left(- \frac{3}{2} , \frac{11}{4} , \frac{19}{4}\right)$

#### Explanation:

I'm going to label these equations:

$E 1 : x + y + z = 6$
$E 2 : x - 2 y = - 7$
$E 3 : 4 x + 3 y + z = 7$

There's no $z$ term in E2, so let's solve it for $x$ and then substitute into E1 and E3:

$E 2 : x = 2 y - 7$

$E 1 : \left(2 y - 7\right) + y + z = 6$
$E 3 : 4 \left(2 y - 7\right) + 3 y + z = 7$

I'm going to simplify them:

$E 1 : 3 y + z = 13$
$E 3 : 11 y + z = 35$

We can subtract E1 from E3 to eliminate the $z$ terms:

$E 3 - E 1 : \left(11 y - 3 y\right) + \left(z - z\right) = \left(35 - 13\right)$

E3-E1: 8y=22=>y=22/8=>color(blue)(ul(bar(abs(color(black)(y=11/4))))

I'm going to plug this into E2 for to solve for $x$, and then will substitute the $x , y$ values into E1 and E3 to solve for $z$ and verify my work.

$E 2 : x - 2 y = - 7$

$E 2 : x - 2 \left(\frac{11}{4}\right) = - 7$

$E 2 : x - \left(\frac{11}{2}\right) = - \frac{14}{2}$

E2: x=-14/2+11/2=-3/2=>color(blue)(ul(bar(abs(color(black)(x=-3/2))))

~~~~~

color(brown)(E1: x+y+z=6
color(red)(E3:4x+3y+z=7

color(brown)(E1: -3/2+11/4+z=6
color(red)(E3:4(-3/2)+3(11/4)+z=7

color(brown)(E1: z=24/4+6/4-11/4=19/4
color(red)(E3:z=28/4+24/4-33/4=19/4

And so this shows we've got it right and that z=38/8=>color(blue)(ul(bar(abs(color(black)(z=19/4))))