Question

Solve the following system of equations using any method

10x+2y+7z=2210x+2y+7z=22
6x+6y+5z=−146x+6y+5z=−14
1x+4y+2z=−141x+4y+2z=−14

Answer 1

we notice that 2nd and 3rd line are equal to -14
therefor
6x+6y+5z=−146x+6y+5z=1x+4y+2z=−141x+4y+2z=−14

look at that
6x+6y+5z=-146x+6y+5z
minus (6y+5z) from both sides
6x=-146x
the only value of x that works is if x=0

x=0
cool
now we can get rid of all x

now we have
2y+7z=2y+7z=22
6y+5z=6y+5z=−14
4y+2z=4y+2z=−14

much nicer

lets see
2y+7z=22 and 6y+5z=-14

so multiply 2y+7z=22 by -3 and ad to to other euqaiotn
6y+5z=-14
-6y-21z=-66 +
0y-16z=-80

-16z=-80
divide both sides by -16
z=5

sub back
2y+7z=22
2y+7(5)=22
2y+35=22
minus 35 both sides
2y=-13
divide both sides by 2
y=-6.5



x=0
y=-6.5
z=5

solution
(0,-6.5,5)