Question

How do you solve this system of equations: `4x - 3y = 39 and 3x + y = 39`?

Answer 1

See a solution process below:

Explanation:

Step 1) Solve the second equation for `y`:

`3x + y = 39`

`-color(red)(3x) + 3x + y = -color(red)(3x) + 39`

`0 + y = -3x + 39`

`y = -3x + 39`

Step 2) Substitute `(-3x + 39)` for `y` in the second equation and solve for `x`:

`4x - 3y = 39` becomes:

`4x - 3(-3x + 39) = 39`

`4x + (-3 xx -3x) + (-3 xx 39) = 39`

`4x + 9x + (-117) = 39`

`4x + 9x - 117 = 39`

`(4 + 9)x - 117 = 39`

`13x - 117 = 39`

`13x - 117 + color(red)(117) = 39 + color(red)(117)`

`13x - 0 = 156`

`13x = 156`

`(13x)/color(red)(13) = 156/color(red)(13)`

`(color(red)(cancel(color(black)(13)))x)/cancel(color(red)(13)) = 12`

`x = 12`

Step 3) Substitute `12` for `x` in the solution to the second equation at the end of Step 1 and calculate `y`:

`y = -3x + 39` becomes:

`y = (-3 xx 12) + 39`

`y = -36 + 39`

`y = 3`

The Solution Is: `x = 12` and `y = 3` or `(12, 3)`

Answer 2

`x=12`, `y=3`

Explanation:

Let's start by solving for a single variable. `3x+color(red)(y)=39` looks easy, since `color(red)(y)` doesn't have a coefficient to make things messy. So, let's rewrite that as

`color(red)(y)=39-3x`

Now let's substitute

`4x-3color(red)(y)=39`

`4x-3(color(red)(39-3x))=39`

`4x-117+9x = 39`

`13x = 156`

`color(blue)(x=12)`

`color(white)(0)`

Now that we have `x`, let's solve for `y`:

`color(white)(0)`

`3(color(blue)(x))+y=39`

`3(color(blue)(12))+y=39`

`color(red)(y=3)`

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Now let's double check our math here:

We say that `x=12` and that `y=3`

`4x-3y=39`

`4(12)-3(3)` should equal `39`, if our math is right

`48-9`

`39`

We were right! `39` is `39`, and so `x` must equal `12` and that `y` does equal `3`.

Another way we could have checked that is graphed the two functions and see where they intercepted, which is at the point `(12, 3)`.

So either way, we can see that we are correct! Nice work

Answer 3

`x=12`
`y=3`

Here's how I did it!

Explanation:

First you want to pick one of your two equations and a variable you want to isolate in it. I'm going to pick `3x+y=39` because it'll be pretty easy to solve for `y`.

`3x+y=39`

`y=-3x+39`

Perfect! Now we have a value for `y`. We can plug this into our second equation, `4x-3y=39`, and solve for a value of our other variable, `x`.

`4x-3y=39`

`4x-3(-3x+39)=39`

`4x+9x-117=39`

`13x-117=39`

`13x=156`

`x=12`

Great! Now that we have a numerical value for `x`, we can plug that back into our first equation, `3x+y=39`.

`3x+y=39`

`3(12)+y=39`

`36+y=39`

`y=3`

There we go!