Question

How do you solve the system `y=2x` and `x+3y=-14` using substitution?

Answer 1

`-2,-4)`

Explanation:

`"Label the equations"`

`color(white)(xxxx)color(red)(y)=2xto(1)`

`x+3color(red)(y)=-14to(2)`

`"Substitute " color(red)(y)=2x" in " (2)`

`rArrx+(3xx2x)=-14`

`rArr7x=-14`

`"divide both sides by 7"`

`(cancel(7) x)/cancel(7)=(-14)/7`

`rArrx=-2`

`"Substitute this value into " (1)`

`y=(2xx-2)=-4`

`color(blue)"As a check"`

`"substitute values for x and y in " (2)`

`-2+(3xx-4)=-2-12=-14to" true"`

graph{(y-2x)(y+1/3x+14/3)=0 [-10, 10, -5, 5]}
`rArr" point of intersection is " (-2,-4)`