How do you solve the system #2x+5y=4# and #x+y=-1# by substitution?

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Answer 1 · 2015-05-26T03:14:40

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

Solve #x+y=-1# for #x#.

#x=-y-1#

Substitute #-y-1# for #x# into the other equation and solve for #y#.

#2x+5y=4# =

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

#-2y-2+5y=4# =

#3y-2=4# =

#3y=6#

Divide both sides by #3#.

#y=2#

Substitute #2# for #y# into the first equation, and solve for #x#.

#x+y=-1# =

#x+2=-1# =

#x=-3#

Check by substituting the #x# and #y# values into both equations.

#x+y=-1# =

#-3+2=-1# =

#-1=-1#

#2x+5y=4# =

#2(-3)+5(2)=4

#-6+10=4# =

#4=4#