Question

Solve the system by elimination.
-2x+2y+3z=0
-2x-y+z=-3
2x+3y+3z=5

Could you also show me how you got the solutions? Thanks!

Answer 1

eq1+eq3: 5y+6z=5 (4)
eq1-eq2: 3y+2z=3 (5)
eq4-3*eq5: -4y=-4
y=1
Substitute into equations to get your other values.

Answer 2

Pair equations one and three together:
-2x+2y+3z=0
2x+3y+3z=5

Pair equations two and three together:
-2x-y+z=-3
2x+3y+3z=5

For both pairs, eliminate the z variable:

For equations one and three, subtract the 1st equation from the 2nd to get:
4x + y = 5

For equations two and three, multiply the 1st equation by 3 to get:
-6x - 3y +3z = -9

Then subtract that from 2x + 3y +3z = 5 to get:
8x + 6y = 14

Reduce that by dividing both sides by 2:
4x + 3y = 7
Now you have two equations in variables x and y:

4x + y = 5
4x + 3y = 7

Subtract the 1st equation from the second to get:

2y = 2

and y = 1

Now use back substitution to find x and z

Answer 3

Thanks you guys. I appreciate the help, I've been stuck on this one for awhile now. The step-by-step stuff really helped. (:

Answer 4

I think I did something wrong. Is this right?

4x+1=5
4x=4
x=1

And then:

-2(1)+2(1)+3z=0
3z=0
z=0

Something looks wrong, and I don't know what I did wrong. Did I not plug in the y value in the correct equation?

Answer 5

It's correct

Answer 6

(x,y,z) = (1,1,0)

Answer 7

I guess I just wasn't expecting it to be (1,1, 0). Thanks. (:

Answer 8

If you had any doubts you could test out another equation:

2x + 3y + 3z = 5
2(1) + 3(1) + 3(0) = 5
2 + 3 + 0 = 5
5 = 5

Answer 9

I'll definitely test it out next time and save you the trouble lol. Thank you for all the help.

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