SOLUTION: 3x+5y-z=1 4x+7y+z=2

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Question 1070318: 3x+5y-z=1
4x+7y+z=2
Found 2 solutions by josgarithmetic, ikleyn:
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
system%283x%2B5y=1%2Bz%2C4x%2B7y=2-z%29

Make z as some constant: k=z.

system%283x%2B5y=1%2Bk%2C4x%2B7y=2-k%29

Prepare to use Elimination.
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system%2812x%2B60y=12%2B12k%2C12x%2B21y=6-6k%29
Subtract second from first.
39y=6%2B18k
13y=2%2B6k
highlight%28y=%282%2B6k%29%2F13%29
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system%2821x%2B35y=7%2B7k%2C20x%2B35y=10-5k%29
subtract second from first.
highlight%28x=-3%2B12k%29

k and z were made the same, so you can pick any real value for k (or z) and find the corresponding x and y values. Infinitely many solutions to the three-variable, two-equation system.

Answer by ikleyn(52781) About Me  (Show Source):
You can put this solution on YOUR website!
.
All calculations made by "josgarithmetic" are wrong.
Below I redo it by the correct way. 

3x + 5y = 1 + z,        (1)
4x + 7y = 2 - z.        (2)


I will solve the system (1), (2) for x using Elimination. 
For it, multiply eq.(1) by 4 (both sides). Multiply eq.(2) by 3 (both sides). You will get an equivalent system

12x + 20y = 4 + 4z,     (3)
12x + 21y = 6 - 3z      (4)

Subtract (3) from (4). You will get

 y = -7z + 2            (5)


Now, I will solve the system (1), (2) for y using Elimination. 
For it, multiply eq.(1) by 7 (both sides). Multiply eq.(2) by 5 (both sides). You will get an equivalent system

21x + 35y =  7 + 7z,    (6)
20x + 35y = 10 - 5z     (7)

Subtract (7) from (6). You will get

 x = 12z - 3            (8)


The given system of two equations in three unknowns has infinitely many solutions.
You can take z as an arbitrary real number and calculate x and y according to the formulas (5) and (8):

 x =  12z - 3,
 y =  -7z + 2.

Again, all calculations made by "josgarithmetic" are wrong.