Question

How do you solve the system `2x-3y=-24` and `x+6y=18` using substitution?

Answer 1

See a solution process below:

Explanation:

Step 1) Solve the second equation for `x`:

`x + 6y = 18`

`x + 6y - color(red)(6y) = 18 - color(red)(6y)`

`x + 0 = 18 - 6y`

`x = 18 - 6y`

Step 2) Substitute `(18 - 6y)` for `x` in the first equation and solve for `y`:

`2x - 3y = -24` becomes:

`2(18 - 6y) - 3y = -24`

`(2 xx 18) - (2 xx 6y) - 3y = -24`

`36 - 12y - 3y = -24`

`36 + (-12 - 3)y = -24`

`36 + (-15y) = -24`

`36 - 15y = -24`

`-color(red)(36) + 36 - 15y = -color(red)(36) - 24`

`0 - 15y = -60`

`-15y = -60`

`(-15y)/color(red)(-15) = (-60)/color(red)(-15)`

`(color(red)(cancel(color(black)(-15)))y)/cancel(color(red)(-15)) = 4`

`y = 4`

Step 3) Substitute `4` for `y` in the solution to the second equation at the end of Step 1 and calculate `x`:

`x = 18 - 6y` becomes:

`x = 18 - (6 * 4)`

`x = 18 - 24`

`x = -6`

The solution is: `x = -6` and `y = 4` or `(-6, 4)`

Answer 2

`color(blue)(y=4,x=-6`

Explanation:

`:.2x-3y=-24----(1)`

`x+6y=18----(2)`
:.
in (2) `x=18-6y`

substitute `color(blue)(x=18-6y` in (1)

`:.2(color(blue)(18-6y))-3y=-24`

`:.36-12y-3y=-24`

`:.-15y=-24-36`

multiply both sides by`-1`

`:.15y=24+36`

`:.15y=60`

`:.color(blue)(y=4`

substitute `color(blue)(y=4` in (2)

`:.x+6(color(blue)4)=18`

`:.x+24=18`

`:.x=18-24`

`:.color(blue)(x=-6`

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

`color(blue)(check:`

substitute `color(blue)(y=4` and`color(blue)(x=-6` in (2)

`:.(color(blue)(-6))+6(color(blue)(4))=18))`

`:.-6+24=18`

`:.color(blue)(18=18`