Question

How do you solve `3x+ 4y=27` and `-8x+ y=33` using substitution?

Answer 1

The solution is `(-3,9)`.

Explanation:

Solve the system:

`"Equation 1":` `3x+4y=27`

`"Equation 2":` `-8x+y=33`

Both equations are linear equations in standard form. The solution to the system is the point that the two lines have in common. The `x` and `y` coordinates will be determined by substitution.

Solve Equation 1 for `x`.

`3x+4y=27`

`3x=27-4y`

Divide both sides by `3`.

`x=27/3-(4y)/3`

Simplify.

`x=9-(4y)/3`

Substitute `9-4y/3` for `x` in Equation 2. Solve for `y`.

`-8x+y=33`

`-8(9-(4y)/3)+y=33`

`-72+(32y)/3+y=33`

Multiply `y` by `3/3` to create an equivalent fraction with `3` in the denominator.

`-72+(32y)/3+yxx3/3=33`

`-72+(32y)/3+(3y)/3=33`

`-72+(35y)/3=33`

Add `72` to both sides.

`(35y)/3=33+72`

Simplify.

`(35y)/3=105`

Multiply both sides by `3`.

`35y=105xx3`

`35y=315`

Divide both sides by `35`

`y=315/35`

`y=9`

Substitute `9` for `y` in Equation 1. Solve for `x`.

`3x+4(9)=27`

`3x+36=27`

`3x=27-36`

`3x=-9`

`x=(-9)/3`

`x=-3`

The solution is the point of intersection for the two lines, which is `(-3,9)`.

graph{(3x+4y-27)(y-8x-33)=0 [-14.24, 11.07, 1.37, 14.03]}