# Algebra (probably easy) problem

• Spirochete
In summary, the conversation discusses two equations with three variables and the task of solving for a numerical value of a third expression. The equations are P/R=3/5 and P/T=9/10, and the expression to be evaluated is (R+T)/R. The correct answer is 5/3, but the person is unsure how to get there. The solution involves solving for R and T separately and substituting their values into the expression. However, since P is unknown, there will be a free variable in the final answer. The person believes this is a special case where the P's happen to cancel, allowing for a numerical answer.
Spirochete

## Homework Statement

There are two equations and 3 variables and I'm asked to solve for the numerical value of a third expression:

We are told that

P/R=3/5

P/T=9/10

The question asks for the value of (R+T)/R

## Homework Equations

The correct answer is 5/3 but I have no idea how to get there. This is from a GRE practice test.

## The Attempt at a Solution

I tried addin' em together. Also solving for "P" to get rid of it doesn't work right away.

Spirochete said:

## Homework Statement

There are two equations and 3 variables and I'm asked to solve for the numerical value of a third expression:

We are told that

P/R=3/5

P/T=9/10

The question asks for the value of (R+T)/R

## Homework Equations

The correct answer is 5/3 but I have no idea how to get there. This is from a GRE practice test.

## The Attempt at a Solution

I tried addin' em together. Also solving for "P" to get rid of it doesn't work right away.

Try solving the first equation for R and the second for T. Then substitute those values into the expression you need to evaluate.

oh I did not see in advance that the P's would cancel out like that and I'd get a number. I never would've thought to do that. Could you explain how you saw that?

When you have equations that share variables, you can always do that setup to eliminate the common variable.

Yes but you couldn't give a numerical value for just R+T, in this case you'd be left with 25P/9

Spirochete said:
Yes but you couldn't give a numerical value for just R+T, in this case you'd be left with 25P/9
This is to be expected. You can't get a numerical value for R unless you know P, and you can't get a numerical value for T unless you know P. Since you don't know P, you are not going to be able to get a numerical value for (R + T)/R.

You have 2 equations and 3 unknowns. Chances are, you will have a free variable.

Mark44 said:
This is to be expected. You can't get a numerical value for R unless you know P, and you can't get a numerical value for T unless you know P. Since you don't know P, you are not going to be able to get a numerical value for (R + T)/R.

Actually I believe this is a special case where you can get a numerical answer just because the P's happen to cancel.

## 1. What is Algebra and why is it important?

Algebra is a branch of mathematics that deals with symbols and the rules for manipulating these symbols to solve equations. It is important because it is the foundation for more advanced math concepts and is used in many fields, including science, engineering, and economics.

## 2. How do I solve an Algebra problem step-by-step?

To solve an Algebra problem, you first need to identify the known values and unknown variables. Then, use the appropriate rules and operations to manipulate the symbols and solve for the unknown variable. Finally, check your solution by plugging it back into the original problem to see if it makes sense.

## 3. What are the basic rules of Algebra?

The basic rules of Algebra include the commutative property, associative property, distributive property, and the rules for combining like terms and solving equations. These rules help us to manipulate symbols and solve equations in a consistent and logical manner.

## 4. How can I improve my Algebra skills?

To improve your Algebra skills, it is important to practice regularly and familiarize yourself with the basic rules and operations. You can also seek help from a tutor or use online resources and practice problems to strengthen your understanding of Algebra concepts.

## 5. Can Algebra be used in real-life situations?

Yes, Algebra can be used in many real-life situations, such as calculating expenses and budgets, determining the growth rate of investments, and solving problems in engineering and science. It is a valuable tool for problem-solving in various fields and everyday life.

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