How do you solve `y=4x-9` and `y=x-3` using substitution?

Step 1) Because the first equation is already solve for `y`, substitute `4x - 9` for `y` in the second equation and solve for `x`:

`y = x - 3` becomes:

`4x - 9 = x - 3`

`4x - 9 + color(red)(9) - color(blue)(x) = x - 3 + color(red)(9) - color(blue)(x)`

`4x - color(blue)(1x) - 9 + color(red)(9) = x - color(blue)(x) - 3 + color(red)(9)`

`(4 - color(blue)(1))x - 0 = 0 + 6`

`3x = 6`

`(3x)/color(red)(3) = 6/color(red)(3)`

`(color(red)(cancel(color(black)(3)))x)/cancel(color(red)(3)) = 2`

`x = 2`

Step 2) Substitute `2` for `x` in the first equation and calculate `y`:

`y = 4x - 9` becomes:

`y = (4 xx 2) - 9`

`y = 8 - 9`

`x = 2`

The solution is: `x = 2` and `x = 2` or `(2, -1)`

Since both equations are `y =`, substitute the first equation for the second "y":

`4x - 9 = x - 3`

Now solve for `x`:
`4x -x - 9 = x -x - 3`

`3x - 9 = - 3`

`3x -9 +9 = -3 +9`

`3x = 6`

`3/3x = 6/3 = 2`

Now substitute `x` into either equation to find `y`:

`y = 2 - 3 = -1`

Check by substituting both putting both values into the first equation:
`-1 = 4(2) -9`
`-1 = 8-9`
`-1 = -1` TRUE

So `x = 2, y = -1`

Which is the point of intersection: `(2, -1)`

Labelling the equations.

`color(red)(y)=4x-9to(1)`

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

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

`rArr4x-9=x-3`

subtract x from both sides.

`4x-x-9=cancel(x)cancel(-x)-3`

`rArr3x-9=-3`

add 9 to both sides.

`3xcancel(-9)cancel(+9)=-3+9`

`rArr3x=6`

divide both sides by 3

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

`rArrx=2`

Substitute this value into either of the equations

`"Substitute " x=2" in " (2)`

`rArry=2-3=-1`

`color(blue)"As a check"`

`"Substitute " x=2" in "(1)`

`rArry=(4xx2)-9=8-9=-1to" true"`

`rArr(2.-1)" is the point of intersection"`
graph{(y-4x+9)(y-x+3)=0 [-10, 10, -5, 5]}