How do you solve the system #2x+7y=8# and #x+5y=7# using substitution?

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Answer 1 · 2015-06-07T01:44:41

#x=-3# and #y=2#.

Solve #x+5y=7# for #x#.

#x=7-5y#

Substitute #7-5y# for #x# into the other equation.

#2(7-5y)+7y=8# =

#14-10y+7y=8# '

#14-3y=8#

Subtract #14# from both sides.

#-3y=-6#

Divide both sides by #-3#.

#y=2#

Substitute #2# for #y# in the first equation.

#x+5(2)=7# =

#x+10=7#

Subtract #10# from both sides.

#x=-3#

Check by substituting #-3# for #x# and #2# for #y# into both equations.

#2x+7y=8# =

#2(-3)+7(2)=8# =

#-6+14=8# =

#8=8# Check.

#x+5y=7# =

#-3+5(2)=7# =

#-3+10=7# =

#7=7# Check.