Question

How do you solve the following linear system: `y = 3x - 2 , 14x - 3y = 0`?

Answer 1

The solution is `(-6/5,-28/5)` or `(-1.2,-5.6)`.

Explanation:

Solve the linear system:

`"Equation 1":` `y=3x-2`

`"Equation 2":` `14x-3y=0`

The solution is the point `(x,y)` that the two lines have in common, which is the point of intersection. I'm going to use substitution to solve the system.

Equation 1 is already solved for `y`. Substitute `3x-2` for `y` in Equation 2 and solve for `x`.

`14x-3(3x-2)=0`

Expand.

`14x-9x+6=0`

Simplify.

`5x+6=0`

Subtract `6` from both sides.

`5x=-6`

Divide both sides by `5`.

`x=-6/5` or `-1.2`

Substitute `-6/5` for `x` in Equation 1. Solve for `y`.

`y=3(-6/5)-2`

Expand.

`y=-18/5-2`

Multiply `2` by `5/5` to get an equivalent fraction with `5` as the denominator.

`y=-18/5-2xx5/5`

`y=-18/5-10/5`

Simplify.

`y=-28/5` or `-5.6`

The solution is `(-6/5,-28/5)` or `(-1.2,-5.6)`.

graph{(y-3x+2)(14x-3y+0)=0 [-6.366, 4.73, -8.243, -2.696]}