SOLUTION: hi, i an usually good with systems but this one is giving me trouble and i have a test tomorrow and could really use some help! 3x+2y=12 -4x+5z=12 y-4z=-13 thanx!!!

Algebra ->  Algebra  -> Coordinate Systems and Linear Equations -> SOLUTION: hi, i an usually good with systems but this one is giving me trouble and i have a test tomorrow and could really use some help! 3x+2y=12 -4x+5z=12 y-4z=-13 thanx!!!      Log On

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Question 13893: hi, i an usually good with systems but this one is giving me trouble and i have a test tomorrow and could really use some help!
3x+2y=12
-4x+5z=12
y-4z=-13
thanx!!!
Answer by Earlsdon(6103) About Me  (Show Source):
You can put this solution on YOUR website!
1) 3x%2B2y=12
2) -4x%2B5z=12
3) y-4z=-13 Solve this for y: y+=+4z-13 and substitute into equation 1)
1) 3x+%2B+2%284z-13%29+=+12 Simplify.
3x+%2B+8z+-+26+=+12
1a)3x+%2B+8z+=+38 Now take equation 2)
2) -4x+%2B+5z+=+12 Multiply equ. 1a) by 4 and equ. 2) by 3.
1b) 12x+%2B+32z+=+152
2a) -12x+%2B+15z+=+36 Now add these two to eliminate the x-term.
47z+=+188 Divide both sides by 47.
z+=+4 Substitute this into y+=+4z+-+13 and solve for y.
y+=+4%284%29+-+13
y+=+3 Substitute this into equation 1) and solve for x.
3x+%2B+2%283%29+=+12
3x+=+6
x+=+2
x = 2
y = 3
z = 4