Algebraic fractions help!

Okay this is the question I have.

The product of these three expressions is 8. Use this information to find the value of y.

1) y^2+10y+21/2y+8. 2) y^2-16/15. 3) 60/y^2-y-12

Okay I checked in the back of my text book and the answer is -3 but I'm not exactly sure how it got there. Can someone please explain this to me?

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Reply 1

Matt85259

Could you try clarifying the equations with brackets where appropriate if you're not able to use the latex functions? It's hard to read what's what.

EDIT: Are they:

\frac{y^2+10y+21}{2y+8}

\frac{y^2-16}{15}

\frac{60}{y^2-y-12}

EDIT2:

Ok just checked that out and they are. Ok, so our aim is to look for cancellations - is there a way we could make these possible cancellations easier to see when we have all these things multiplying together? What do we generally like to do when we have quadratics? When you have these cancellations it should leave you with a simple equation to solve!

(edited 10 years ago)

Reply 2

TinM

Okay and sorry it's because I'm using my phone.

The first expression is:

y(squared)+10y+21/ 2y+8

The second is:

y(squared)-16/ 15

And the final one:

60/ y(squared)-y-12

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Reply 3

Matt85259

I managed to figure them out! Check my first post and see if the hint helps you!

Reply 4

TinM

Okay, hold on I'll try it.

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Reply 5

TinM

I've managed to factorise the expressions but I'm not how to individually cancel each one out.

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Reply 6

Matt85259

Okay good, so now you have these we can just multiply like normal fractions and set them equal to 8. Do you see some common terms top and bottom when all 3 equations are combined?

Reply 7

TinM

y+4?

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Reply 8

Matt85259

That's one of them yes, are you able to show your working so far? You should be able to cancel down to just one bracket containing y multiplied by a constant.

Reply 9

TinM

Yeah, I've got that down so far.

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Reply 10

Matt85259

So what are you left with in your equation now? Can you post it so I can see what stage you've got to?

Reply 11

TinM

Wait so would divide 60/15 and leave the y+4 that is on the denominator of the first fraction or would I cancel that out as well?

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Reply 12

Matt85259

Can you post your big fraction? If I can see what working you've done it'll be much easier to guide you. Just do it like

Numerator = 5(x+1)(x+2)
Denominator = 10(x+2)

Reply 13

TinM

Numerator= (y+7)(y+3)(y-4)x4
Denominator=2y^2+2y+24

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Reply 14

Matt85259

Okay, check for a sign errors in your denominator I think you may have inadvertantly changed them. Can you see a common factor in each term of the denominator? After correcting the signs can you do more with the denominator?

Reply 15

TinM

Would it be -24 and 2?

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Reply 16

Matt85259

okay so your denominator should be 2y^2 - 2y - 24. (check the denominator of your equation 3). Can you see a common factor? Can you factorise?

Reply 17

TinM

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So who it be y^2-y+12?

Reply 18

Matt85259

Why did you change from -24 to +12? Don't forget you need to keep the 2, as we're working with the denominator but you can move it to the outside and bracket the equation right?

so we'd have:

denominator = 2(y^2 - y - 12)

Can we factorise the bit in the bracket? Can we cancel the 2 with anything in the denominator?