# Algebra II Word Problem Involving Fractional Equations

• velox_xox
In summary, the problem is asking for the new size of the Computer Club after 8 new members joined. The club was originally trying to raise $1200 by assessing equal amounts to each member. With the addition of 8 new members, the per-member assessment was reduced by$7.50. The equations 1200=md and 8d- 7.5m= 60 are used to solve the problem, resulting in a quadratic equation of 7.5m^2 +60m- 9600=0. After solving for m, it is found that there are 40 members in the new club.
velox_xox
Hi everyone! This time I have a lesson on word problems, and that is probably my absolute weakest point in math. (In fact, as I write, I'm puzzling over several different problems, but here is the one I'll post for now.)

## Homework Statement

Members of the Computer Club were assessed equal amounts to raise $1200 to buy some software. When 8 new members joined, the per-member assessment was reduced by$7.50. What was the new size of the club?

--

## The Attempt at a Solution

I understand what it is asking for, and the information provided. My problem with word problems is translating the facts into a formula.

I assigned d = dollars and m = the number of members

The original amount of members and dollars would be:

$$\frac{1200}{m} = d$$

Then, with the new members:
$$\frac{1200}{m + 8} = d - 7.5$$

This is my attempt. I'm not confident that I set the formula up correctly or how to solve it if it is correct. I could use some hints right about now. Please and thank you!

The equations are correct. This is just a system of equations. You already have one variable expressed in therms of another. Now you only have to substitute that expression in the second equation and solve it. It will be a quadratic equation, when you clean it up.

Remember that one of the solutions to the quadratic won't make sense and you should throw away that term.

velox_xox said:
Hi everyone! This time I have a lesson on word problems, and that is probably my absolute weakest point in math. (In fact, as I write, I'm puzzling over several different problems, but here is the one I'll post for now.)

## Homework Statement

Members of the Computer Club were assessed equal amounts to raise $1200 to buy some software. When 8 new members joined, the per-member assessment was reduced by$7.50. What was the new size of the club?

--

## The Attempt at a Solution

I understand what it is asking for, and the information provided. My problem with word problems is translating the facts into a formula.

I assigned d = dollars and m = the number of members

The original amount of members and dollars would be:

$$\frac{1200}{m} = d$$
Excellent! And that is the same as 1200= md

Then, with the new members:
$$\frac{1200}{m + 8} = d - 7.5$$
Good. So 1200= (m+8)(d- 7.5)= md+ 8d- 7.5m- 60= 1200+ 8d- 7.5m- 60
which reduces to 8d- 7.5m= 60. You can solve that for d (or m) and put that into 1200= md.

This is my attempt. I'm not confident that I set the formula up correctly or how to solve it if it is correct. I could use some hints right about now. Please and thank you!

Ah ha ha! My math instincts aren't quite there yet. I thought it might be a system of equations, and even got as far as $8d- 7.5m= 60$ but I second guessed it and tossed that idea out. Going with it then, I think I'm doing the math wrong.

So, for starters, if I'm expressing 'd' in terms of 'm' is this correct: $d =\frac {7.5m + 60}{8}$

Or, expressing 'm' in terms of 'd' would be: $d =\frac {-8d + 60}{7.5}$

Are both of those correct? And you plug them back in for $1200 = md$ When I do that, I get a huge quadratic equation that is confusing to solve, such as $8d^2 -60d + 9000$

I also forgot to include the answer. Sorry, it's proof I was frazzled working on all those word problems! (I'll go back and edit it.) The answer is 40 members [of the new club].

First, I can see no point in reducing to an equation for "d" because you don't need to know d to answer the question- it asks for the number of people in the club, not how much each has to pay.
The second equation has a typo- you mean
$$m= \frac{-8d+ 60}{7.5}$$
Yes, both are correct. Now, combine the first, where you solved for d, with 1200= md so you have
$$1200= m\frac{7.5m+ 60}{8}$$

Multiplying both sides by 8, $9600= 7.5m^2+ 60m$ or $7.5m^2+ 60m- 9600= 0$. If you don't like fractions, you can multiply both sides by 2 to get $15m^2+ 120m- 19200= 0$. Dividing through by 5 gives $3m^2+ 24m-3840= 0$ and then dividing by 3, $m^2+ 8m- 1280= 0$.

Solve that using the quadratic formula. Don't forget that the problem asks for the number of people in the new club- m+ 8.

@HallsofIvy: Well, technically, it would be adding an extra step, but I could solve for 'd' first and then use that to find 'm', right?

So factoring out, $7.5m^2 +60m−9600=0$, I got $(1.5m - 48)(5m + 200)$.

One of which, as scurty said, isn't possible since you can't have negative people. And, so that leaves $\frac {48}{1.5}$, which equals 32.

And so, if m = 32, then m + 8 = 40. So, there are 40 people in the new club. And that is the correct answer the book provides.

So, was my method correct?

Yes, that is correct. And I didn't say it was wrong to write the equation of d and then, after finding d, solve another equation for m. I just said that I did not see any point in doing that rather than just set up the equation in terms of m in the first place.

@HallsofIvy: I understand. I was just double-checking.

Thank you hamsterman, scurty, and HallsofIvy! I appreciate all of your help! :)

## 1. What is a fractional equation in Algebra II?

A fractional equation is an equation that contains at least one fraction. This means that the unknown variable in the equation is in the form of a fraction, and the equation must be solved using algebraic methods.

## 2. How do you solve a fractional equation in Algebra II?

To solve a fractional equation, you must first isolate the variable by getting rid of any fractions. This can be done by multiplying both sides of the equation by the denominator of the fraction. Then, solve the resulting equation as you would a regular algebraic equation.

## 3. Can you provide an example of an Algebra II word problem involving fractional equations?

Example: A recipe calls for 2/3 cup of flour and makes 8 cookies. How many cookies can be made with 3 cups of flour?

To solve this problem, we can set up the equation: 2/3 = x/3, where x represents the number of cookies that can be made with 3 cups of flour. By cross-multiplying, we get 2(3) = 3x, which simplifies to 6 = 3x. Solving for x, we get x = 2. Therefore, 2 cups of flour will make 24 cookies.

## 4. What are some common mistakes to avoid when solving fractional equations in Algebra II?

One common mistake is forgetting to multiply both sides of the equation by the denominator when trying to isolate the variable. Another mistake is forgetting to apply the distributive property when necessary. It is also important to carefully check your work and ensure that all steps are accurate.

## 5. How can I practice solving Algebra II word problems involving fractional equations?

There are many resources available for practicing solving fractional equations, including textbooks, online practice problems, and worksheets. It is also helpful to create your own word problems and practice solving them. Additionally, working with a tutor or joining a study group can provide additional support and practice opportunities.

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