How do you solve x=8-2y and x=-3y+12?

2 Answers
Jun 12, 2018

Answer:

First solve for #y# using substitution, then solve for #x#, you will find that #y=4# and #x=0#

Explanation:

We're very lucky for this problem, as both expressions leave #x# by itself on one side. If you wanted to write this as one long expression:

#8-2y=x=-3y+12#

Now, we can ignore #x# and write the expression as follows (BECAUSE both expressions equal #x#):

#8-2y=-3y+12#

Since #x# is not in either equation, we can solve for #y# easily:

#cancel(8)-2ycolor(red)(+3y)color(blue)(cancel(-8))=cancel(-3y)+12color(red)(cancel(+3y))color(blue)(-8)#

#color(green)(y=4#

Finally, we can solve for #x#:

#x=8-2y#

#x=8-2(4)#

#x=8-8#

#color(green)(x=0)#

Jun 12, 2018

Answer:

#(x,y)to(0,4)#

Explanation:

#x=8-2yto(1)#

#x=-3y+12to(2)#

#"since both equations express x in terms of y , equate"#
#"the right sides"#

#8-2y=-3y+12#

#"add "3y" to both sides"#

#8-2y+3y=12#

#"subtract 8 from both sides"#

#y=12-8=4#

#"substitute "y=4" into either of the 2 equations"#

#(1)tox=8-8=0#

#"the point of intersection "=(0,4)#
graph{(y+1/2x-4)(y+1/3x-4)=0 [-10, 10, -5, 5]}