Question 191226
Three equations would normally have three variables.
2^5 + 5 + 5*3 = 155
5   - 2*5 + 0 = -20
-5  + 3*5 - 5 = -5
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Now put x, y, and z in place of the 5's:
2x + y + 3z = 155
x  - 2y + 0 = -20
-x + 3y - z = -5
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The solutions should be x = 5, y=5, z=5
Cheers,
Stan H.