How do you solve the system #y = 2x - 3# and #y = -x + 3#?
2 Answers
You may set the two
Explanation:
Plug this into both equations (one for check!):
(they agree).
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]}