How do you solve 3x+y=43x+y=4 and 2x+y=12x+y=1?

1 Answer
mimi
Apr 4, 2018

(3, -5)(3,−5)

Explanation:

I used substitution.

Isolate yy in one of the equations.
3x+y=43x+y=4
y=4-3xy=4−3x

Substitute yy in 2x+y=12x+y=1 for 4-3x4−3x.
2x+(4-3x)=12x+(4−3x)=1
2x+4-3x=12x+4−3x=1
4-x=14−x=1
-x=-3−x=−3
x=3x=3

Substitute xx back into y=4-3xy=4−3x find the value of yy.
y=4-3(3)y=4−3(3)
y=4-9y=4−9
y=-5y=−5

Answer (Coordinate Form): (3, -5)(3,−5)

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