6x - 4y = 36
2x - 8y = 32
I assume you meant "substitution." To solve this system by substitution, first isolate one of the variables in either of the equations. I'll isolate y in the second equation.
2x - 8y = 32 (given)
-8y = 32 - 2x (subtract 2x from both sides)
y = (32 - 2x)/-8 (divide both sides by -8)
y = -4 + 1/4x (simplify)
Substitute -4 + 1/4x for y into the first equation and solve algebraically for x.
6x - 4y = 36 (given)
6x - 4(-4 + 1/4x) = 36 (substitute)
6x + 16 - x = 36 (distribute)
5x + 16 = 36 (collect like terms)
5x = 20 (subtract 16 from both sides)
x = 4 (divide both sides by 5)
Substitute 4 for x into either of the original equations to find y.
6x - 4y = 36 (given)
6(4) - 4y = 36 (substitute)
24 - 4y = 36 (multiply)
-4y = 12 (subtract 24 from both sides)
y = -3 (divide both sides by -4)
Lastly, plug x- and y-values into original equations to check work.
6x - 4y = 36
6(4) - 4(-3) = 36
24 + 12 = 36
36 = 36
2x - 8y = 32
2(4) - 8(-3) = 32
8 + 24 = 32
32 = 32
Answer:
x = 4 and y = -3; (4, -3).
