Question

How do you solve `x=y+4` and `x=2y+8` using substitution?

Answer 1

See a solution process below:

Explanation:

Step 1) Because both equations are solve solve for `x`, we substitute `y + 4` from the first equation for `x` in the second equation and solve for `y`:

`y + 4 = 2y + 8`

`-color(red)(y) + y + 4 - color(blue)(8) = -color(red)(y) + 2y + 8 - color(blue)(8)`

`0 - 4 = -color(red)(1y) + 2y + 0`

`-4 = (-color(red)(1) + 2)y`

`-4 = 1y`

`-4 = y`

`y = -4`

Step 2) Substitute `-4` for `y` in either of the original equations and calculate `x`. I will substitute it into the first equation:

`x = y + 4` becomes:

`x = -4 + 4`

`x = 0`

The solution is: `x = 0` and `y = -4` or `(0, -4)`

Answer 2

Place the value `y+4` into the equation `x = 2y + 8` and solve for y, then use the value of y to solve for x.

Explanation:

`x = y + 4` so putting this into the second equation gives.

`y + 4 = 2y +8` solve for y by subtracting y and 8 from both sides.

`y -y + 4 -8 = 2y -y + 8 -8` Which gives

`-4 = y` now put -4 into the first equation for solve for x

`x = -4 +4` so

`x = 0`

Answer 3

`y=-4,x=0`

Explanation:

`x=y+4----(1)`

`x=2y+8----(2)`

substitute `color(magenta)(x=y+4` in (1)

`:.(color(magenta)(y+4))=2y+8`

`:.y-2y=8-4`

`:.-y=4`

`:.color(magenta)(y=-4`

substitute `color(magenta)(y=-4` in (1)

`:.x=(color(magenta)(-4))+4`

`:.color(magenta)(x=0`

substitute `y=-4`and`x=0` in (2)

`:.0=2(-4)+8`

`:.0=-8+8`

`:.color(magenta)(0=0`