Question

How do you solve the system of equations `-8x - 2y = - 1` and `3x - 12y = 4`?

Answer 1

`x=10/51` & `y=-29/102`

Explanation:

Given system of linear equations:

`-8x-2y=-1`

or `8x+2y=1\ ...... .......(1)`

`3x-12y=4\ ...... .......(2)`

Multiplying (1) by `6` & adding to (2) as follows

`3x-12y+6(8x+2y)=4+6\times 1`

`51x=10`

`x=10/51`

setting `x=10/51` in (1) we get

`y=\frac{1-8x}{2}=\frac{1-8\times 10/51}{2}=-29/102`

hence, the solution is `x=10/51` & `y=-29/102`

Answer 2

`x=10/51`

`y=-29/102`

Explanation:

`color(blue)("Equation A: ") -8x-2y=-1`

`color(green)("Equation B: ") 3x-12y=4`

Multiply `color(blue)("Equation A")` by 6, then use elimination to first eliminate `y` from each equation.

`color(blue)("Equation A: ") 6*(-8x-2y) = 6*(-1)`

`color(blue)("Equation A: ") -48x-12y=-6`

`color(green)("Equation B: ") 3x-12y=4`

Subtract `color(green)("Equation B")` from `color(blue)("Equation A")`:

`-51x = -10`

`x = 10/51`

Substitute this `x` value into one of the original equations to find `y`:

`-8(10/51) - 2y =-1`

`-2y = -1 + 80/51`

`y = 1/2 - 40/51`

`y = frac{51-80}{102}`

`y = -29/102`