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

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Answer 1 · 2016-12-23T07:19:50

#x=-4 1/5# and #y=-10 3/5#

#y=3x+2#
#y=4/3x-5#

In the second equation, substitute #y# with #color(red)((3x+2))#. This equivalence is taken from the first equation.

#color(red)(3x+2)=4/3x-5#

Multiply all terms by #3#.

#9x+6=4x-15#

Subtract #4x# from each side.

#5x+6=-15#

Subtract #6# from each side.

#5x=-21#

Divide both sides by #5#.

#x=-21/5# or #x=-4 1/5#

In the first equation, substitute #x# with #color(blue)(-21/5)#.

#y=3x+2#

#y=3color(blue)((-21/5))+2#

#y=-63/5+2#

#y=(-63+10)/5#

#y=-53/5# or #y=-10 3/5#