SOLUTION: how do I solve -x-5y+z=17 -x-y+z=1 2x+5y-3z=-10

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Question 1116288: how do I solve -x-5y+z=17
-x-y+z=1
2x+5y-3z=-10
Found 2 solutions by josgarithmetic, MathTherapy:
Answer by josgarithmetic(39620) About Me  (Show Source):
You can put this solution on YOUR website!
How, depends what you know.

system%28x%2B5y-z=-17%2Cx%2By-z=-1%2C2x%2B5y-3z=-10%29

R2-R1 for new R2;
R3-2R1 for new R3;

system%28-4y=16%2C-5y-z=-8%29-------two linear equations, two unknown variables

system%28y=-4%2Cz%2B5y=8%29

system%28y=-4%2Cz=28%29--------two of the variables solved - and you can finish the remaining variable however you want.



(not the only way to solve)

Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!
how do I solve -x-5y+z=17
-x-y+z=1
2x+5y-3z=-10 
- x - 5y + z = 17 ------- eq (i)

- x - y + z = 1 --------- eq (ii)

2x + 5y - 3z = - 10 ----- eq (iii)

4y = - 16  -------- Subtracting eq (i) from eq (ii)

Right away you can see that highlight_green%28matrix%281%2C5%2C+y%2C+%22=%22%2C+%28-+16%29%2F4%2C+%22=%22%2C+-+4%29%29

- x - 5(- 4) + z = 17 ------- Substituting - 4 for y in eq (i)

- x + 20 + z = 17

- x + z = - 3 ------- eq (iv)



2x + 5(- 4) - 3z = - 10 ------- Substituting - 4 for y in eq (iii)

2x - 20 - 3z = - 10

2x - 3z = 10 ------- eq (v)

- 3x + 3z = - 9 ------- Multiplying eq (iv) by 3 ------ eq (vi)

- x = 1 ------- Adding eqs (v) & (vi)

highlight_green%28matrix%281%2C5%2C+x%2C+%22=%22%2C+1%2F%28-+1%29%2C+%22=%22%2C+-+1%29%29

I leave it to you to find z

By the way. z CAN NEVER = 28, as YOU-KNOW-WHO will have you believe!