Mathematics
OpenStudy (anonymous):

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!

OpenStudy (anonymous):

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.

hero (hero):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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?

hero (hero):

It's correct

hero (hero):

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

OpenStudy (anonymous):

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

hero (hero):

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

OpenStudy (anonymous):

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

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