(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

(a) [tex]\frac{3}{x}+\frac{3}{2x} = 2 [/tex]

that is 4x = 9 so x = 2.25

(b) I can do the first question (a) easy peasy , however the secnond question says "using your answer to part (a), or otherwise solve:

[tex]\frac{3}{(y-1)^{2}}+\frac{3}{2(y-1)^{2}} = 2[/tex]

What part of (a) do I use to solve part (b)

3. The attempt at a solution

for (b)

1) Expand (y-1)² = y² - 2y + 1

2) Expand 2(y-1)² = 2(y² - 2y + 1 ) = 2y² - 4y + 2

3) Double 1'st fraction to get equal denominators then add fractions

[tex]\frac{6}{2y^{2} - 4y + 2}[/tex]

4) Crossmultiply so you get [tex]3 = 4y^{2} - 8y + 4[/tex]

then complete the square

(4y - 4)² - 16 + 4 = 3

y = [tex]\frac{4 +- \sqrt{15}}{4}[/tex]

but that GETS a bit complicated for NON CALCULATOR level so i tried using the formula which gave me y = 1 or 1 (when at the bottom of the question there is 2 answers). Have I made a mistake in my calculations, and what about using part (a) to help answer part (b), I havn't done that, how do they link in together?

Thx

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# Homework Help: Question 18 GCSE higher tier

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