Let's solve this by using Substitution method.

Since the equation was solved for `y` , replace all occurrences of `y` in the other equations with the solution `(4x+12)` .

`y=4x+12 `

`3(4x+12)=8-12x `

Multiply `3` by each term inside the parentheses.

`y=4x+12 `

`12x+36=8-12x `

Move all terms not containing `x` to the right-hand side of the equation.

`y=4x+12 `

`24x+36=8 `

Move all terms not containing `x` to the right-hand side of the equation.

`y=4x+12 `

`24x=-28 `

Divide each side by `24`

`x=-28/24`

Then, simplify by dividing the LCD `4`

`x= -7/6`

Since the equation was solved for `x,` replace all occurrences of `x` in the other equations with the solution `(-7/6)`

`y=4(-7/6)+12 `

Multiply `4` by each term inside the parentheses.

`y=-14/3 +12`

Combine `-14/3 +12` into a single expression by finding the least common denominator (LCD). The LCD of `-14/3 +12` is `3` .

`y=(-14+36)/3`

`y=22/3`

This is the solution to the system of equations.

`y=22/3`

`x=-7/6`

`y = 4x+12 `

`3y = 8 - 12x`

Using the substitution method, substitute `4x+12` for `y` into the 2nd equation.

`3(4x+12)=8 - 12x`

`12x + 36 = 8 - 12x`

` ` `24x = -28`

`x = (-28)/24 = (-7)/6`

Now substitute `-7/6` into the first equation to find y.

`y = 4(-7/6)+12`

` ` `y = -14/3+12`

`y = -14/3 + 36/3`

`y = 22/3`

**The solutions is `(-7/6,22/3)`**

**` ` **

y = 4x + 12 -----eq(i)

3y = 8 - 12x -----eq(ii)

We can solve this with the method of substitution. Since the value of y is given, insert that in eq(ii)

3y = 8 - 12x

3(4x + 12) = 8 - 12x

12x + 36 = 8 - 12x

12x + 36 - 36 = 8 - 12x - 36 Subtract 36 from both sides

12x + 12x = -28 - 12x + 12x Add 12x to both sides

24x = -28

24x/24 = -28/24 Divide 24 by both sides

**x = -7/6 **

Insert the value of x in eq(i)

y = 4x + 12

y = 4(-7/6) + 12

y = -14/3 + 12

y = -14/3 + 36/3

**y = -22/3**

**x = -7/6 Answer.**

Insert both values in eq(i) to verify the answer,

y = 4x + 12

-22/3 = 4(-7/6) + 12

-22/3 = -14/3 + 12

-22/3 = -14/3 + 36/3

-22/3 = -22/3

LHS = RHS proved.