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

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Answer 1 · 2018-06-23T09:40:32

#x=1# and #y=0#

From second equation, #y=3-3x#

After putting value into first one,

#2x+5*(3-3x)=2#

#2x+15-15x=2#

#15-13x=2#

#-13x=-13#, so #x=(-13)/(-13)=1#

Thus #y=3-3*1=0#