Question #dfe7b

1 Answer
Mar 24, 2017

Answer:

#(3,0)#

Explanation:

Label the equations.

#color(red)(y)=-2x+6to(1)#

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

Since both equation have y as the subject we can equate the right sides.

#rArrx-3=-2x+6#

add 2x to both sides.

#x+2x-3=cancel(-2x)cancel(+2x)+6#

#rArr3x-3=6#

add 3 to both sides.

#3xcancel(-3)cancel(+3)=6+3#

#rArr3x=9#

divide both sides by 3

#(cancel(3) x)/cancel(3)=9/3#

#rArrx=3#

Substitute this value into either of the 2 equations.

#"Substitute "x=3" in " (2)#

#x=3toy=3-3=0#

#color(blue)"As a check"#

#"Substitute " x=3" in " (1)#

#x=3to(-2xx3)+6=-6+6=0to" true"#

#rArr(3,0)" is the point of intersection"#
graph{(y+2x-6)(y-x+3)=0 [-10, 10, -5, 5]}