Efficient Algebra Question Solutions: Ratios and Equations

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The discussion focuses on finding efficient algebraic solutions for two problems involving ratios and equations. The first problem requires finding two numbers in a 7:12 ratio, where the greater number exceeds the lesser by 275. Participants suggest using algebraic equations to express the relationships, leading to the realization that setting y = x + 275 and using the ratio gives a solvable system. The second problem involves manipulating an equation to find the ratio x:y, with hints provided to rewrite it in a more manageable form. Ultimately, the conversation emphasizes the importance of algebraic manipulation for solving ratio problems efficiently.
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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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