# Solving systems of equations using substitution

#### ~Flippy

##### OM NOM NOM NOM
I was doing fine up until the problem below, and I am unable to get ahold of a tutor for now so I'd like some more help. Would be greatly appreciated; rep in return for help. Please detail how it was solved so i can apply the method better and see the whole picture.

3x - y - 9 = 0
8x + 5y - 1 = 0

Thanks again!

3x - y - 9 = 0 ----(1)
8x + 5y - 1 = 0 -----(2)

From eq. (1), y = 3x - 9 ----(3)

Substitute (3) into (2),
8x + 5(3x - 9) - 1 = 0
23x - 46 = 0
x = 2
y = 3(2) - 9 = -3

3x - y - 9 = 0
8x + 5y - 1 = 0

you basically have to resolve the system to a single variable let's choose x

3x= y+9 x=y+9/3
8x= -5y+1 x= -5y+1/8

y+9/3= -5y+1/8 solve this then substitute into the original equation.

Use simultaneous eqns. or the substitution method.

yeah like they said. either first use only equation and solve for either y or x, utilizing the other variable as part of that answer. then plug that into the other equation and so on.

OR try to multiply one or both equations by whatever constants so that when you add each equation together ONE of the variables is cancelled out, then proceed.