Question

How do you solve the system `x+2y=13` and `3x-5y=6` using substitution?

Answer 1

`x=7`
`y=3`

Explanation:

So you can pick either one of the 2 equations listed to get 1 term alone. In explanation I will be using the second one `(3x-5y=6)`

So we have `3x-5y=6`

First we add `5y` to both sides giving us
`3x=5y+6`

Next we divide `3` on both sides to get `x` alone, which gives us
`x=5/3y+6/3`

`x=5/3y+2`

Now that have `x` we can plug this into the first equation, which is `(x+2y=13)`

`5/3y+2+2y=13`

Now we subtract the `2` on both sides giving us
`5/3y+2y=11`

Now we add the `y` values by finding common denominators
`5/3y+"(3)2"/"(3)1"y=11`

Then add the `y` values
`5/3y+6/3y=11`

`(5y+6y)/3=11`

`(11y)/3=11`

Now we multiply by `3` on both sides
`11y=33`

Next divide by `11`
`y=3`

So now that we have the value for `y` we can find `x` easily by plugging this into any equation that has `y`. For this part I'm going to use the first equation which was, `x+2y=13`

`y=3`
`x+2y=13`

`x+2(3)=13`

`x+6=13`

Subtracting 6 from both sides gives us
`x=7`

There fore
`x=7`
`y=3`