Efficient Algebra Question Solutions: Ratios and Equations

  • Thread starter Miike012
  • Start date
  • Tags
    Algebra
In summary, the two questions are asking how to solve a equations in which one variable is over the other by a certain number. The Attempt at a Solution was unable to solve the equation. The Final Solution was to solve the equation for x/y and find that x = 1/3 and y = -1/10.
  • #1
Miike012
1,009
0

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?
 
Physics news on Phys.org
  • #2
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?
 
  • #3
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.
 
  • #4
How would you express the ratio of X and Y?
 
  • #5
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.
 
  • #6
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)
 
  • #7
then i would get 12x = 7y
 
  • #8
Miike012 said:
then i would get 12x = 7y
Good! Now combine that with your second equation.
 
  • #9
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.
 

FAQ: Efficient Algebra Question Solutions: Ratios and Equations

What is a ratio?

A ratio is a comparison between two quantities, typically expressed as a fraction or in the form of a colon (e.g. 2:3). It shows the relationship between the two quantities and can be used to solve various algebra problems.

How do I simplify ratios?

To simplify a ratio, you must find the greatest common factor (GCF) of the two numbers in the ratio. Divide both numbers by the GCF to get the simplified ratio. For example, if the ratio is 12:24, the GCF is 12, so the simplified ratio is 1:2.

Can I solve algebraic equations using ratios?

Yes, ratios can be used to solve algebra problems involving equations. For example, if you have an equation with a ratio on one side, you can multiply both sides by the denominator of the ratio to isolate the variable and solve for it.

What is the difference between a proportion and an equation?

A proportion is an equation that compares two ratios, while an equation compares two expressions or values. Proportions are used to solve problems involving ratios, while equations can involve various mathematical operations.

How can I check my solution to a ratio and equation problem?

To check your solution, you can plug in the values you found into the original equation or ratio and see if it makes the equation true. You can also solve the problem using a different method and see if you get the same answer.

Similar threads

Back
Top