Efficient Algebra Question Solutions: Ratios and Equations

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Homework Help Overview

The discussion revolves around solving algebraic problems involving ratios and equations. The first problem requires finding two numbers in a specific ratio that differ by a given amount, while the second problem involves determining the ratio of two variables based on a quadratic equation.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore algebraic expressions to represent the relationships in the first problem, questioning how to set up equations based on the given ratio and difference.
  • In the second problem, there is a focus on manipulating the original equation to express it in terms of the ratio x/y, with participants discussing potential substitutions and transformations.

Discussion Status

Participants have made progress in formulating equations for both problems. Some have offered hints and guidance on how to approach solving the equations, while others express uncertainty about their methods and seek clarification on their reasoning.

Contextual Notes

There is a noted lack of confidence among some participants regarding their algebra skills, which may affect their ability to express the problems mathematically. Additionally, there is an emphasis on finding quicker methods rather than relying solely on trial and error.

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Homework Statement


I know the answers to the question however I was wonder if there was a way of figuring out the answer rather than just pluggin in answers and guessing.

first question: Find two numbers in the ratio of 7:12 so that the greater exceeds the less by 275.

second questionif 15(2x^2-y^2)=7xy, find the ratio x:y




The Attempt at a Solution


I aloready know the answers but it was only because I plugged in numbers one after the other... is there a quicker way?
 
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Miike012 said:
first question: Find two numbers in the ratio of 7:12 so that the greater exceeds the less by 275.
How about using algebra? Let the numbers be X and Y. How would you express the two relationships given using algebraic equations?
 
Sorry, I am drawling a blank. I don't have much alg under my belt.
The only thing that I can think of is obviously the denominator is larger than the numerator... therefore the den. must be larger then the num. by 275. And the final answer must be a ratio of 7:12. But I don't know how to represent the information I just gave you into two numbers X and Y.
 
How would you express the ratio of X and Y?
 
This was what I was thinking exactly... I thought of x/y and because we are relating x/y and the ratio 7/12.. i set x/y = 7/12, then I know that y needs to be bigger than x by 275
so I set y = x + 275... but that really wouldn't make sence... and that is where I am stuck... hmmm.
 
Miike012 said:
This was what I was thinking exactly... I thought of x/y and because we are relating x/y and the ratio 7/12.. i set x/y = 7/12, then I know that y needs to be bigger than x by 275
so I set y = x + 275... but that really wouldn't make sence... and that is where I am stuck... hmmm.
Why wouldn't that make sense? It's perfect, so far. Now all you have to do is solve those two equations together.

Hint: Rewrite x/y = 7/12. (Cross multiply)
 
then i would get 12x = 7y
 
Miike012 said:
then i would get 12x = 7y
Good! Now combine that with your second equation.
 
Alright I am back... ok combine 12x = 7y If y = x + 275 then

12x = 7x + 1925
5x = 1925
x = 385
y = 385 + 275 = 660 !
Woohoo! thanks lol.
 
  • #10
Good. Now try the other one. How can you rewrite that equation totally in terms of x/y?
 
  • #11
Would you first set

15(2x^2-y^2) - 7xy = x/y ?
 
  • #12
Miike012 said:
Would you first set

15(2x^2-y^2) - 7xy = x/y ?
No. How did you get that?

Hint: Start with your original equation and divide. Twice.
 
  • #13
The only thing that I can think of is trying to get the variables on one side so I would divide by 15 and xy

then I would get 2x^2/xy - y^2/xy = 7/15

But to be honest I don't know what I am looking for.
 
  • #14
Hint: What do you have to divide x^2 by to get (x/y)^2?
 
  • #15
divide x^2 by y^2
 
  • #16
Miike012 said:
divide x^2 by y^2
Good. So divide both sides of your original equation by y^2 and see what happens.
 
  • #17
30x^2/y^2 -1 = 7x/y
 
  • #18
almost looks kinda like a quadratic but I've never seen a quadratic with x over y...
 
  • #19
unless I substitue x/y for u then I get 30u^2 - 7u -1 = 0
then I get u = 1/3 ; -1/10

then x/y = 1/3 ---- x = 1/3y ; x/y = -1/10 ---- x = -1/10 y

then I plug in x = 1/3y ; x/y = -1/10 ---- x = -1/10 y
to the original equation both seperatly
That wouldn't get my anyhting would it?
 
  • #20
Miike012 said:
30x^2/y^2 -1 = 7x/y
Double check the coefficient of that second term.

Miike012 said:
almost looks kinda like a quadratic but I've never seen a quadratic with x over y...
Well, now you have! Since x/y is what you are asked to find, solve that equation and you're good. (But correct it first.)
 
  • #21
30x^2/y^2 - 15 = 7x/y
 
  • #22
Miike012 said:
30x^2/y^2 - 15 = 7x/y
Good. Now solve that quadratic. (It might be easier to let z = x/y and rewrite it in terms of z.)
 
  • #23
your right... thank you. I hope to be good like that one day lol. I just need a tons more practice.
But thakn you.
 

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