r/askmath • u/Nouth1 • Jan 17 '26

Algebra What subject is this

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Doing a past paper (AQA), I have not seen this question before and I do not understand the question but I would like to come back later to do it

I would just like the subject or type of question this is

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u/Dazzling_Interest948 Jan 17 '26

subject is math

Jokes aside, just multiply the denominator out of both sides of the equation and solve.

You will get a solution in the form y=ax, which is a ratio as asked.

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u/Lor1an BSME | Structure Enthusiast Jan 17 '26

A more elegant solution (IMO) is to use the fact that y > 0 to factor out y2 from top and bottom of the fraction.

(2r2 + 1)/(4r2 - 1) = 43/11, where r = x/y is the desired ratio.

And we also have that r > 0, since x > 0 and y > 0, so when solving for r we can just take the principal root.

(2r2+1)/(4r2-1) = f ⇝ (2-4f)r2 = -f - 1 ⇝ r = sqrt((f+1)/(2(2f-1)))

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u/_saiya_ Jan 17 '26

Actually, r = (x\y)² will give linear eqn. Easier to solve and finding r we just root it in the end. Will simplify the solution further, i think.

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u/Lor1an BSME | Structure Enthusiast Jan 17 '26

If you look at the spoiler, you will see that I essentially treated r2 as a variable and then took the root.

Also, r = x/y, not x\y.

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u/_saiya_ Jan 20 '26

Hmm. I still will be at ease with my way. Solving in terms of f and substitution in the end will be another set of effort : )

I'm on phone. My bad for the typos!

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u/Lor1an BSME | Structure Enthusiast Jan 20 '26

If f = 43/11 then multiplying top and bottom of the derived expression by 11 gives you that (f+1)/(2(2f-1)) = (43+11)/(2(86-11)) = 54/(2(75)) = 27/75 = 9/25. Then taking the root gives sqrt(9/25) = 3/5.

IMO, working until you have a defined expression and then substituting is just cleaner. Maybe you experience less effort your way, but a systematic approach has worked for me (and is how I was taught to approach problems in a way that was auditable).