How do you solve #y = -x + 5# and #2x - 3y = -6# using substitution?

1 Answer
Mar 17, 2016

Answer:

#(9/5,16/5)#

Explanation:

As given by the first equation, we see that #color(red)(y=-x+5)#. Because of this, we can plug #color(red)(-x+5# in for #color(red)y# in the second equation.

#2x-3color(red)y=-6" "=>" "2x-3(color(red)(-x+5))=-6#

When distributing #-3# into #(-x+5)#, note that the negative will change the signs of the terms in the parentheses, so:

#-3(color(blue)-xcolor(blue)+5)=color(green)+3xcolor(green)-15#

So, we have

#2x+3x-15=-6#

#5x-15=-6#

#5x=9#

#color(purple)(x=9/5#

With this, we can plug this value for #x# into either equation. I'll choose the first, since it's simpler:

#y=-color(purple)x+5" "=>" "y=-color(purple)(9/5)+5#

Find a common denominator:

#y=-9/5+25/5#

#color(brown)(y=16/5)#

Since we have #color(purple)(x=9/5# and #color(brown)(y=16/5)#, our final answer is the ordered pair #(color(purple)(9/5),color(brown)(16/5))#.