How do you solve the system #2x+4y=6# and #2x+y= -3# by substitution?

1 Answer
Bhavin B.
Apr 30, 2018

x=-3, y=3

Explanation:

#2x+4y=6 # eq 1
#2x+y=-3# eq 2
From eq 2, make y the subject:
#y=-3+2x#
Substitute y from above in eq 1
#2x+4(-3+2x)=6#
#2x-12+8x=6#
#2x-8x=6+12#
#-6x=18#
#x=-18/6#
#x=-3#
Now, substitute x=-3 in eq 1 or eq 2. Lets use eq 1.
#2(-3)+4y=6#
Now solve for y
#-6+4y=6#
#4y=6+6#
#4y=12#
#y=12/4#
#y=3#
Check for answer:
use eq 1 or eq 2
#2(-3) + 3 = -6+3=-3#
So the answers x=-3 and y=3 are correct

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