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

Question

1788 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-23T10:56:12

$(x, y, z)=(-3/2,11/4,19/4)$

I'm going to label these equations:

$E1: x+y+z=6$
$E2: x-2y=-7$
$E3:4x+3y+z=7$

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

$E2: x=2y-7$

$E1: (2y-7)+y+z=6$
$E3: 4(2y-7)+3y+z=7$

I'm going to simplify them:

$E1: 3y+z=13$
$E3: 11y+z=35$

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

$E3-E1: (11y-3y)+(z-z)=(35-13)$

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

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.

$E2: x-2y=-7$

$E2: x-2(11/4)=-7$

$E2: x-(11/2)=-14/2$

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

~~~~~

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

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

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

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

How do you solve the system x+y+z=6x+y+z=6, x-2y=-7x-2y=-7, and 4x+3y+z=74x+3y+z=7?