How do you solve the following system: #11y + 3x = 4, -3y + x = -3 #?

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Answer 1 · 2018-03-24T15:03:20

#color(blue)(y=13/20)#

#color(blue)(x=-21/20)#

#11y+3x=4 \ \ \ \ \ \ \ \ \ [1]#

#-3y+x=-3 \ \ \ \ \[2]#

Multiply equation #[2]# by 3 and subtract it from equation #[1]#

#11y+3x=4 \ \ \ \ \ \ \ \ \ [1]#

#-9y+3x=-9 \ \ \ \ \[2]#

# \ \ \ \ \ \ \ \ \ \ \ \ \20y=13#

Divide by 20:

#color(blue)(y=13/20)#

We now substitute this into one of the original equations, it dosen't matter which one we choose.

In #[2]#

#-3(13/20)+x=-3#

Add #-3(13/20)# to both sides:

#x=-3+3(13/20)#

#color(blue)(x=-21/20)#