Question

How do you solve the system of equations `6x - 2y = - 30` and `3x + 8y = 39`?

Answer 1

`(x,y)to(-3,6)`

Explanation:

`6x-2y=-30to(1)`

`3x+8y=39to(2)`

`"we can eliminate y by multiplying equation "(1)" by 4"`

`24x-8y=-120to(3)`

`"add "(2)" and "(3)" term by term"`

`(3x+24x)+cancel((8y-8y))^0=(120-39)`

`rArr27x=-81`

`"divide both sides by 27"`

`(cancel(27) x)/cancel(27)=(-81)/27`

`rArrx=-3`

`"substitute "x=-3" into either equation "(1)" or "(2)`

`(2)to-9+8y=39`

`"add 9 to both sides"`

`rArr8y=48`

`"divide both sides by 8"`

`(cancel(8) y)/cancel(8)=48/8`

`rArry=6`

`"the point of intersection "=(-3,6)`

Answer 2

The point of intersection is `(-3,6)`.

Explanation:

Solve the system of equations:

The equations are linear equations in standard form: `Ax+By="C"`. The resulting `x`- and `y`-values make up the point of intersection between the two equations. There are several possibilities for solving a system of equation. I am going to use substitution to solve this system.

Equation 1: `6x-2y=-30`

Equation 2: `3x+8y=39`

Solve Equation 1 for `y`.

`6x-2y=-30`

Subtract `6x` from both sides.

`-2y=-6x-30`

Divide both sides by `-2`

`y=(-6x)/(-2)-30/(-2)`

Simplify.

`y=3x+15`

Substitute `3x+15` for `y` in Equation 2 and solve for `x`.

`3x+8(3x+15)=39`

Expand.

`3x+24x+120=39`

Simplify.

`27x+120=39`

Subtract `120` from both sides.

`27x=39-120`

Simplify.

`27x=-81`

Divide both sides by `27`.

`x=(-81)/27`

Simplify.

`x=-3`

Substitute `-3` for `x` in Equation 1 and solve for `y`.

`6(-3)-2y=-30`

Simplify.

`-18-2y=-30`

Add `18` to both sides.

`-2y=-30+18`

Simplify.

`-2y=-12`

Divide both sides by `-2`.

`y=(-12)/(-2)`

Simplify.

`y=6`

Point of intersection: `(-3,6)`