Question

How do you solve `x = y - 8` and `-x - y = 0` using substitution?

Answer 1

x= -4 , y= 4

Explanation:

From the second equation -x-y=0, it is y=-x. Now substitute this in the first equation ,

x= -x-8 `->` 2x =-8 `->` x= -4

Hence y=4

Answer 2

`(-4,4)`

Explanation:

`"Given "color(red)(x=y-8)` we can `color(blue)"substitute"` this directly into the other equation, and solve for y

`rArr-(color(red)(y-8))-y=0`

distributing gives.

`-y+8-y=0`

simplifying.

`-2y+8=0`

subtract 8 from both sides of the equation.

`-2ycancel(+8)cancel(-8)=0-8`

`rArr-2y=-8`

To solve for y, divide both sides by - 2

`(cancel(-2) y)/cancel(-2)=(-8)/(-2)`

`rArry=4`

To find x, substitute y = 4 into `x=y-8`

`y=4tox=4-8=-4`

`rArr(-4,4)" is the solution"`

Answer 3

Replace `x` with `(y-8)` in `"–"x-y=0`; solve for `y`.
Use this `y`-value in either equation to solve for `x`.

`(x,y)=("–"4,4)`.

Explanation:

Each of these equations represents a line in 2D-space. Solving the system of these two equations means finding all the `(x,y)` points where the lines cross.

We are given the equations `x=y-8` and `"–"x-y=0`. From the first equation, we have a value for `x` in terms of `y`. That is, for the first line, `x` is always `y-8`. To see if this line has any points in common with the other line `"–"x-y=0`, we want to check if an `x` value of `y-8` works for any point on that other line.

So, we substitute `x` out for `y-8` in the second equation:

`"      ""–"x"    "-y=0`
`"–"(y-8)-y=0`
`"  ""–"y+8" "-y=0`
`"               –"2y="–"8`
`"                   "y=4`

So yes—there is a point on the second line where `x` will be `y-8`, and that point occurs when `y=4`.

The only thing left to do is to find the `x`-value for this point. To do that, we can plug `y=4` into either of our line equations and solve for `x`. (Since `y=4` is where the lines cross, both equations will have the same `x`-value for that `y`).

Using the first equation, we get:

`x=y-8`
`x=4-8`
`color(white)x="–"4`

(Or, using the second equation, we get

`"–"x-y=0`
`"–"x-4=0`
`"–"x"       "=4`
`"         "x="–"4`

which gives the same `x`-value, as we'd expect.)

So our solution for the system is `(x,y)=("–"4, 4)`.

graph{(x-y+8)(x+y)=0 [-12.17, 7.83, -2.76, 7.24]}