Question #1762a

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Answer 1 · 2018-02-18T12:56:58

#x=2, y=1#

#[1] 2x+y=5#
#[2] 3x-2y=4#

We will use a method called elimination.

Multiply #[1]# by 2

#4x+2y=10#

We can then add this to #[2]#

#4x+2y=10#
#+3x-2y=4#

#7x=14#
#x=2#

Substitute this into any equation; I will use #[1]#

#2(2)+y=5#
#4+y=5#
#y=1#

graph{(2x+y-5)(3x-2y-4)=0 [-10, 10, -5, 5]}