Algebra Word Problem

by Abby Hope
(Henderson, NC)

The product of two fractions is 1/9. The larger fraction divided by the smaller fraction is 4. What is the sum of the two fractions? Express your answer as a common fraction.

Comments for
Algebra Word Problem

Average Rating starstarstarstarstar

Click here to add your own comments

Sep 29, 2010
Rating
starstarstarstarstar
Algebra Word Problem
by: Karin

Hi Abbey,

This is a great problem!

Let's first identify the variables. You know that you are dealing with 2 numbers and one is larger than the other.

Let x = the larger number
Let y = the smaller number

Now we can write two equations:

The product of the two numbers is 1/9. Product means multiply.

xy = 1/9

The larger number divided by the smaller number is 4.

x/y = 4

So, we have: xy = 1/9 and x/y = 4

Since we have two equations, we actually have a system of equations. We need to solve the system.

I am going to use the substitution method.

Step 1: Solve one of the equations for a variable. I will solve x/y = 4 for x.

Multiply both sides by y.

y(x/y)= 4y

x = 4y

Now, we'll substitute 4y for x into the equation:
xy = 1/9

4y(y) = 1/9

4y^2 = 1/9

Now divide by 4 on both sides.

4y^2/4 = 1/9 / 4
y^2 = 1/36

Now take the square root of both sides and you end up with:

y = 1/6

Now substitute 1/6 for y into the equation x = 4y

x = 4y
x = 4(1/6)
x = 4/6
x = 2/3

So, the two numbers are 1/6, & 2/3.

Now let's check before we finish the problem:

xy = 1/9
(1/6)(2/3) = 1/9
2/18 = 1/9
1/9 = 1/9

AND

x/y = 4

2/3 / 1/6 = 4
2/3 * 6/1 = 4
12/3 = 4
4 = 4

Yes, our answers are correct. So, now the question asked for the sum of the numbers, so we must add the numbers together.

1/6 + 2/3 = 5/6

The correct answer is 5/6

Hope this helps,
Karin
xy = 1/9
(1/6)(2/3) =

Click here to add your own comments


Having Trouble with Your Homework?

algebrator




[ ?] Subscribe To This Site

XML RSS
Add to Google
Add to My Yahoo!
Add to My MSN
Add to Newsgator
Subscribe with Bloglines


Enjoy This Site?
Then why not use the button below, to add us to your favorite bookmarking service?

| Homepage | Contact Me | Privacy Policy | Disclaimer |Affiliates



Return to top
Copyright© Algebra-class.com 2009.