Hi, Omid.

It looks like you're learning about solving systems of equations. The easiest way to approach this particular problem is to use substitution. In order to do that, you need to get one of the variables by itself. I think it is easier to work with the first equation so that we can avoid mistakes made by using fractions.

-2x + 3y - 12 = 0

Let's get y alone.

-2x + 3y - 12 = 0

+2x +2x

3y - 12 = 2x

3y - 12 = 2x

+ 12 +12

3y = 2x + 12

Now divide each term on both sides of the equation by 3

3y = 2x + 12

3 3 3

y = 2/3 x + 4

Now that we have y alone in the first equation, we can substitute its expression in the other equation. Here we go:

x * y = 90

x * (2/3 x + 4) = 90

Distribute the x...

2/3 x^2 + 4x = 90

Let's multiply all the terms (on both sides) by 3 to clear the fractions:

2 x^2 + 12x = 270

Do you see that we now have a quadratic equation to solve?

Let's get all the terms on one side first:

2 x^2 + 12x = 270

-270 -270

2 x^2 + 12x - 270 = 0

It's easiest to factor if the coefficient of the x^2 term is a 1. Since all four terms are divisible by 2, let's divide both sides by 2 (all the terms):

2 x^2 + 12x - 270 = 0

2 2 2 2

x^2 + 6x - 135 = 0

Do you know how to factor this?

When you get your two answers, be sure to test them in the original two equations to ensure they are solutions.

## Comments

Roman, this was clever, and it cleaned up the problem very quickly.

Nice job.