Question

How do you solve `9x+6y=3` and `-3x+y=14` using substitution?

Answer 1

First, you take 1 equation and you make it so that the `y` is the main thing like `y =`something... Then you take that and you substitute it into your other equation to get `x =`something...You then take that and you substitute it into your first equation and done.

Explanation:

NOTE:
color `color(gold)(gold` and the bolded text is the main stuff
Other writing is explanations

So, to start you got these 2 equations...
`color(gold)(9x + 6y = 3`
`color(gold)(-3x + y = 14`

Let's call equation 1 A and equation 2 B. So...
`9x + 6y = 3`: `color(red)(A`
`-3x + y = 14`: `color(blue)(B`

So to start substituting, you first take `color(blue)(B` because it has the lowest number of '`y`'s...
`-3x +y = 14` `color(blue)(B`
`9x + 6y = 3` `color(red)(A`
Equation `color(blue)(B` has the lowest number of `y`.

Now, you take equation `color(blue)(B` and you twist and turn it to make the `y` in the equation the main letter. Like so...

`color(gold)(-3x + y = 14` | This is your starting equation
`y = 14 + 3x` | you move over the '`-3x`' to the other side of the '`=`' sign. Since it was a 'negative' number before, going over to the other side makes '`-3x`' a positive number, '`3x`'.
`color(gold)(y = 14 + 3x` | now '`y`' is the main letter.

Now you have this, you get your other equation (equation `color(red)(A`) and you 'combine' it with your new equation `color(blue)(B`. Like so...

`9x + 6y = 3`: `color(red)(A`
`y = 14 + 3x`: NEW `color(blue)(B`

You substitute your NEW `color(blue)(B` into `color(red)(A`...
`9x+ 6color(gold)((14 + 3x)) = 3` | In the NEW `color(blue)(B`, it says the `y =` 'that'. Therefore, you can make the `y` in `color(red)(A` (`6y`) also equal 'that'.

Now you just have to do some switch-a-roos and stuff to get your answers: x = something and y = something, like this...

`color(gold)(9x + 6(14 + 3x) = 3` | what you got now
`9x + 84 + 18x = 3` | expanded the brackets
`27x + 84 = 3` | added the similar numbers
`27x = 3 - 84` | moved `84` to the other side
`27x = -81`
`x = -81/27` | moved `27` to the other side.
`color(gold)(x = -3` | your first answer.

Now...
`color(gold)(-3x + y = 14`: `color(blue)(B` | The old equation `color(blue)(B`
Since `x = -3` so...
`-3(-3) + y = 14` | substituting `x = -3`
`9 + y = 14`
`y = 14 - 9` | moving 9 to the other side
`color(gold)(y = 5` | your second answer.

That's how you do it.