Question

How do you solve the following system: `-2x + 5y = 20, x +4y = 16`?

Answer 1

Let's use substitution:

`-2x + 5y = 20`

`x + 4y = 16`

We need to solve for `x` in the second equation

`x = 16 - 4y`

Now we substitute `(16 - 4y)` for `x` in the first equation

`-2(16 - 4y) + 5y = 20`

distribute the `-2`

`-32 + 8y + 5y = 20`

Solve for `y`. Add `32` to both sides

`13y = 52`

Divide by `13` on both sides

`y = 4`

`* * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * *`

We have `y`, let's find `x`:

`x = 16 - 4y`

`x = 16 - 4(4)`

`x = 16 - 16`

`x = 0`

`* * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * *`

To check our work, let's plug our values for `x` and `y` into the first equation and see if it equals `20`:

`-2(0) + 5(4)`

`0 + 20`

`20` EQUALS `20`! We were right