How do you solve the system #y = 2x - 3# and #y = -x + 3#?

2 Answers
Apr 3, 2017

Answer:

You may set the two #y#'s equal to each other:

Explanation:

#2x-3=-x+3#. Now add #3# to both sides:

#2x-cancel3+cancel3=-x+3+3#. Now add #x#:

#2x+x=cancelx-cancelx+6->3x=6->x=2#

Plug this into both equations (one for check!):
#y=2*2-3=1#
#y=-2+3=1#
(they agree).

Apr 3, 2017

Answer:

#(2,1)#

Explanation:

Label the equations.

#color(red)(y)=2x-3to(1)#

#color(red)(y)=-x+3to(2)#

Since both equations give #color(red)(y)# expressed in terms of x we can equate the right sides.

#rArr2x-3=-x+3#

add x to both sides.

#2x+x-3=cancel(-x)cancel(+x)+3#

#rArr3x-3=3#

add 3 to both sides.

#3xcancel(-3)cancel(+3)=3+3#

#rArr3x=6#

divide both sides by 3

#(cancel(3) x)/cancel(3)=6/3#

#rArrx=2#

Substitute this value into either ( 1 ) or ( 2 ) and evaluate for y

#"Substituting in " (1)#

#x=2toy=(2xx2)-3=4-3=1#

#color(blue)"As a check"#

#"Substituting in " (2)#

#x=2toy=-2+3=1to" true"#

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