How do you solve #2x-3y=1# and #5x+4y=14# using substitution?

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Answer 1 · 2018-06-30T15:59:17

#x=2,y=1#

solving the first equation for #x#:
substracting #3y#

#2x=1+3y#
dividing by #2# we get
#x=1/2+3/2y#
plugging this in the second equation

#5(1/2+3/2y)+4y=14#

multiplying by #2#

#5(1+3y)+8y=28#
#5+15y+8y=28#

collecting like Terms

#23y+5=28#

substracting #5# and dividing by #23#

#y=1#

so we get

#x=1/2+3/2=4/2=2#