# Finding the two numbers... (check my work)

• Oct 20th 2012, 09:28 PM
Tall Jessica
Finding the two numbers... (check my work)
Hey, everyone!

I'm having a little trouble with this one... I'm pretty sure I'm missing something, though. Here's the problem:

"The product of two numbers is 407. One of the numbers is three times the other number, augmented by four. Find the sum of the two numbers."

So, I chose $\displaystyle x$ and $\displaystyle y$ as my numbers and this is what I got, based on the description:

$\displaystyle x$
$\displaystyle y=3x+4$

When multiplied, you get 407

$\displaystyle xy=407$

So, I substitute and simplify:

$\displaystyle x(3x+4)=407$
$\displaystyle 3x^2+4x=407$
$\displaystyle 3x^2+4x-407=0$
$\displaystyle (3x+37)(x-11)=0$

$\displaystyle x=-\frac{3}{37}, 11$

I'm assuming the solution I should choose is the 11, right?

Now, substituting for $\displaystyle x=11$

$\displaystyle y=3x+4$
$\displaystyle y=3(11)+4$
$\displaystyle y=37$

$\displaystyle 37 + 11 = 48$

Thoughts?

• Oct 20th 2012, 09:29 PM
Prove It
Re: Finding the two numbers... (check my work)
You will have to show BOTH possible sets of solutions.
• Oct 20th 2012, 09:41 PM
Tall Jessica
Re: Finding the two numbers... (check my work)
Quote:

Originally Posted by Prove It
You will have to show BOTH possible sets of solutions.

So, it's just a matter of plugging in the other solution, too? Is what I've done correct, then?

Thank you so much!!