# How do you solve the system of equations algebraically 36x-15y+50z=-10, 2x+25y=40, 54x-5y+30z=-160?

May 24, 2017

Usually, we’d move on to Cramer’s Rule with three or more equations. But, they can be done by ‘elimination’ yet.
x = -5 ; y = 2 ; z = 4

#### Explanation:

Set up the equations with all of the variables in columns. Use multiplication/division on a set of two equations followed by subtraction to ‘eliminate’ one of the variables.

Repeat with another two, and then combine them to eliminate a second variable. Solve for the final variable, then substitute the value back into one of the previous equations to find the second variable, and then finally into one of the originals to find the third variable. Then use all three variables in one of the original equations to check for correctness.
1) 36x − 15y + 50z = −10
2) $2 x + 25 y = 40$ (no ‘z’ term in this one)
3) 54x − 5y + 30z = −160

In this case the second equation also makes it easy to define x in terms of y for a direct substitution.
2) $2 x + 25 y = 40$ ; 2x = 40 – 25y ; $x = 20 - 12.5 y$

1) 36x − 15y + 50z = −10
3) 54x − 5y + 30z = −160

1) 36*(20 - 12.5y) − 15y + 50z = −10
3) 54*(20 - 12.5y) − 5y + 30z = −160

1) 720 - 450y − 15y + 50z = −10
3) 1080 - 675y − 5y + 30z = −160

Multiply by inverse multiples to get equal 'z' coefficients:
1) 720 - 465y + 50z = −10 $\cdot 3$
3) 1080 - 680y + 30z = −160 $\cdot 5$

1) 2160 - 1395y + 150z = −30 $\cdot 3$
3) 5400 - 3400y + 150z = −800 $\cdot 5$ Subtract 3) from 1)

$- 3240 + 2005 y = 770$; $y = 2$

2) $2 x + 25 y = 40$
2) $2 x + 25 \cdot 2 = 40$ ; $x = - 5$

3) 54x − 5y + 30z = −160
3) 54*(-5) − 5*(2) + 30z = −160
 -270 – 10 + 30z = -160
$- 280 + 30 z = - 160$ ; $z = 4$

Then CHECK :
1) 36x − 15y + 50z = −10
1) 36*(-5) − 15*(2) + 50*(4) = −10
-180 − 30 + 200 = −10 ; $10 = 10$ CORRECT!