How do you solve #5x+y=5# and #3x+2y=3# using substitution?

Question

2370 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-21T19:24:20

#x=1 and y=2#

#" "5x+y+(−5x)=5+(−5x)" "#(Add #-5x# to both sides)

#" "y=−5x+5#.

Substitute #color(blue)((−5x+5))# for #color(blue)(y) " in " 3x+2y=3#:

#" "3x+2color(blue)(y)=3#

# 3x+2color(blue)((−5x+5))=3#

#3x-10x+10 =3#

# −7x+10=3#

# −7x+10+ (−10)=3+(−10)" "# (Add #-10# to both sides)

# −7x=−7#

#(7x)/-7=(-7)/-7" "# (Divide both sides by # -7#)

# x =1.#

Substitute #1 # for #x# in #y=−5x+5#

# y=−5x+5#

# y=(−5)(1)+5#

#y=0" " # (Simplify both sides of the equation).