How do you solve the following system?: `-x+4y=8 , x=2y+1`

Step 1) Because the second equation is already solved for `x` we can substitute `(2y + 1)` for `x` in the first equation and solve for `y`:

`-x + 4y = 8` becomes:

`-(2y + 1) + 4y = 8`

`-2y - 1 + 4y = 8`

`4y - 2y - 1 = 8`

`(4 - 2)y - 1 = 8`

`2y - 1 = 8`

`2y - 1 + color(red)(1) = 8 + color(red)(1)`

`2y - 0 = 9`

`2y = 9`

`(2y)/color(red)(2) = 9/color(red)(2)`

`(color(red)(cancel(color(black)(2)))y)/cancel(color(red)(2)) = 9/2`

`y = 9/2`

Step 2) Substitute `9/2` for `y` in the second equation and calculate `x`:

`x = 2y + 1` becomes:

`x = (2 * 9/2) + 1`

`x = (color(red)(cancel(color(black)(2))) * 9/color(red)(cancel(color(black)(2)))) + 1`

`x = 9 + 1`

`x = 10`

The Solution Is:

`x = 10` and `y = 9/2` or `(10, 9/2)`

`-color(red)(x)+4y=8to(1)`

`color(red)(x)=2y+1to(2)`

`"substitute "color(red)(x)=2y+1" into "(1)`

`rArr-(2y+1)+4y=8`

`rArr-2y-1+4y=8`

`rArr2y=9rArry=9/2`

`"substitute this value into "(2)`

`rArrx=(2xx9/2)+1=9+1=10`

`color(blue)"As a check"`

`"substitute these values into "(1)`

`-10+(4xx9/2)=-10+18=8larr" True"`

`rArr"point of intersection "=(10,9/2)`

Notice that the `x`-terms are ADDITIVE INVERSES. That means that they will add together to give `0`.

Change the second equation into the same form so we have:

`color(white)(xxxxxx)color(blue)(-x)+4y =8" ".......A`
`color(white)(xxxxxx)color(blue)(+x)-2y =1" ".........B`

`A+B:" "color(blue)(0x)+2y=9`

`color(white)(xxxxxxxxxxx)y =4.5`

Substitute this value for `y` into the original equation for `x`

`x =2y+1`

`x = 2(4.5)+1`

`x = 10`

Check in the other equation:`" "-x+4y =8`

`-(10)+4(4.5)`

`-10+18`

`=8" "larr` the answer is correct.