How do you solve #3x+y=2# and #2x+3y=13# using substitution?

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Answer 1 · 2016-03-04T17:44:06

#"Point of intersection "(x,y)" "->" "(-1,5)#

Given:
#3x+y=2#............................(1)
#2x+3y=13#.......................(2)
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Consider equation (1)

Subtract #3x# form both sides

#3x-3x+y=2-3x#

#0+y=-3x+2" "..................(1_a)#

Using equation #1_a# substitute for #y# in equation (2)

#2x+3(-3x+2)=13................(2_a)#

#2x-9x+6=13#

#-7x+6=13#

Subtract 6 from both sides

#-7x+6-6=13-6#

#-7x+0=7#

Divide bot sides by -7

#(-7)/(-7)xx x=(+7)/(-7)#

#(+1)xx x=-1#

#color(red)(x=-1)#
'~~~~~~~~~~~~~~~~~~~~~~~~~
Substitute #x=-1# into equation (1)

#3x+y=2" "->" "3(-1)+y=2 #

#-3+y=2#

Add 3 to both sides

#-3+3+y=2+3#

#color(red)(y=5)#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
To prove this:
write equation (1) as: #y=-3x+2#
write equation (2) as: #y=-2/3 x+13/3#
Tony B