Question

Question #dfe7b

Answer 1

`(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]}