How do you solve by substitution #x + 3y = 5# and #4x + 5y = 13#?

1 Answer
Jun 8, 2015

#x+3y=5#
#color(blue)(x=5-3y# ( we substract #3y# on each side )

Now that we have #color(blue)x#, we can substitute it in the second equation :

#4color(blue)x+5y=13#
#4*color(blue)((5-3y))+5y=13#
#20-12y+5y=13#
#20-7y=13#
#20=13+7y# ( we add #7y# on each side )
#20-13=7y#
#7y=7#
#color(red)(y=1# ( we divide by #7# on each side )

Now that we have #color(red)y#, we can find #color(blue)x# :

#color(blue)x=5-3color(red)y#
#color(blue)x=5-3*1#
#color(blue)x=5-3#
#color(blue)(x=2)#