confusedchildren
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9. Show that 1/(〖6x〗^2+7x-5)÷1/(〖4x〗^2-1) simplifies to (ax+b)/(cx+d), where a, b¸ c and d are integers.
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student10005555
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when you divide a fraction by another fraction, flip the second fraction and then multiply the 2 fractions together. then factorise and simplify.
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RDKGames
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(Original post by confusedchildren)
9. Show that 1/(〖6x〗^2+7x-5)÷1/(〖4x〗^2-1) simplifies to (ax+b)/(cx+d), where a, b¸ c and d are integers.
Rewrite the expression as a single fraction first.
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confusedchildren
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(Original post by amaraub)
when you divide a fraction by another fraction, flip the second fraction and then multiply the 2 fractions together. then factorise and simplify.
i did the first part and got 4x^2 - 1 / 6x^2 + 7x -5 but idl how to factorise that
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confusedchildren
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(Original post by RDKGames)
Rewrite the expression as a single fraction first.
i got 4x^2 - 1 / 6x^2 + 7x -5
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(Original post by confusedchildren)
i got 4x^2 - 1 / 6x^2 + 7x -5
Great, now factorise the numerator and denominator.
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(Original post by confusedchildren)
i did the first part and got 4x^2 - 1 / 6x^2 + 7x -5 but idl how to factorise that
i'd suggest you brush up on factorising quadratics then, there are plenty of videos on youtube etc
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confusedchildren
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(Original post by RDKGames)
Great, now factorise the numerator and denominator.
im stuck i dont know how to do it
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(Original post by confusedchildren)
im stuck i dont know how to do it
The numerator is a difference of two squares (something I recommend knowing by heart) whereby a^2 - b^2 = (a+b)(a-b) hence 4x^2 -1 = (2x+1)(2x-1).

To factorise the denominator 6x^2+7x-5, do you have any methods you went over that you can use?

If not, then a nice way is to think of two numbers which multiply to make 6\times -5 = -30 and add to make 7. What are they?? Then split the middle 7 as a sum of these two numbers with coefficient of x. Then factorise.
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confusedchildren
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(Original post by RDKGames)
The numerator is a difference of two squares (something I recommend knowing by heart) whereby a^2 - b^2 = (a+b)(a-b) hence 4x^2 -1 = (2x+1)(2x-1).

To factorise the denominator 6x^2+7x-5, do you have any methods you went over that you can use?

If not, then a nice way is to think of two numbers which multiply to make 6\times -5 = -30 and add to make 7. What are they?? Then split the middle 7 as a sum of these two numbers with coefficient of x. Then factorise.
so i got -3 and 10 as the two numbers so it was 6x^2 -3x +10x -5 and i factorised it to get (2x-1)(3x+5) so i canceled (2x-1) from the numerator and denominator to get (2x+1)/ (3x+5) what do i do next?
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(Original post by confusedchildren)
what do i do next?
Celebrate.
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confusedchildren
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(Original post by RDKGames)
Celebrate.
Cant celebrate if ive still got 3 more questions left ugh XD but thanks again
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