208k views
2 votes
Algebra 2 - Solving a system of three equations with three variables.. PLEASE HELP!

I've attached an image here. I've been trying to solve this problem for over 30 minutes and neither the class or textbook makes sense to me. I am looking for an explanation. Thank you in advance.

Algebra 2 - Solving a system of three equations with three variables.. PLEASE HELP-example-1

2 Answers

3 votes
ya need to eliminate 1 variable from 2 equations then eliminate one variable from those 2

ok

we see the last equation already has y eliminated

so we need to eliminate y from one of the other equation

ok we got
x+3y-z=2 and
4x+2y+5z=1
let's elimiate from top equation

multiply top equation by 2 and 2nd by -3
2x+6y-2z=4
-12x-6y-15z=-3
add them

2x+6y-2z=4
-12x-6y-15z=-3 +
-10x+0y-17z=1

-10x-17z=1

use that equation and the last one

3x+z=12 and
-10x-17z=1

eliminate z in first equation
multiply top eqquation by 17
51x+17z=204

add to other equation
51x+17z=204
-10x-17z=1 +
41x+0z=205

41x=205
divide both sides by 41
x=5

sub back in other equation

3x+z=12
3(5)+z=12
15+z=12
z=-3

sub back again

x+3y-z=2
5+3y-(-3)=2
5+3y+3=2
8+3y=2
3y=-6
y=-2



x=5
y=-2
z=-3

(x,y,z)
(5,-2,-3)

answer is 2nd option
User Kkpoon
by
6.7k points
0 votes
The answer is X=5 Y=-2 and Z=-3 :)

User Zhu Shengqi
by
6.1k points