# Thread: How to solve this problem algebraically?

1. ## How to solve this problem algebraically?

Qs: One number is 6 more than another. Their product is 520. Find the numbers.

-- Let the numbers be: x, x+6
So, x(x+6) = 520
or, x^2+6x=520
then what..?

The numbers are 20 & 26.
And if I put them in the above formula, it works..
20(20+6) = 520
or, 20x26=520

2. ## Re: How to solve this problem algebraically?

$x^2+6x-520=0$

Are you aware of any of the methods available for solving a quadratic equation?

3. ## Re: How to solve this problem algebraically?

One number is 6 more than another. Their product is 520. Find the numbers.

Let any letter of choice be the number we are searching for. I will use n for number but you can use x, y, z, or whatever letter you like.

Let n = one of the numbers

Let n + 6 = the other number

The words "Their product" should tell you right away that multiplication is what the question is leading to.

n times (n + 6) = 250

n(n + 6) = 250

n^2 + 6n -250 = 0

In the quadratic formula, let a = 1, b = 6 and c = -250.

If after visiting the site you have no clue what's going on, write back. I think it is important for you to try this one on your own. Keep in mind that x is typically used in the formula. Just replace my letter n with the typical variable x and follow the steps given in the link above.

4. ## Re: How to solve this problem algebraically?

Thanks for Both of the above posts. I've solved it using the factoring method e.g.
factors of 520 that will match to x, x+6 are 20 & 26
Since, x^2 + 6x -520 = 0
(-20)(26) = -520
(-20)+(26)= 6
So, (x-20)(x+26)=0
x-20 = 0, or x=20
x+26 = 0, or x=-26
x=20 ... this matches the equation, so the #s are 20 & 26.
Will try the formula way also. Thanks again...