# Thread: Math Exam tomorrow (elementary algebra) help :(

1. ## Math Exam tomorrow (elementary algebra) help :(

Hi guys,

I have a Math final worth 100% of my grade and have been knee deep in my text book for the past 2 weeks. I went to the copy centre beside my University and they had old copies of finals, and my friend who did the Math final last year said she used the old exams to study and they're almost identical each year (except obviously the values of the numbers changed etc).

I scanned the most recent exam from last term and would really appreciate if someone could outline the basic way to deal with the questions. Anyway this is the exam PDF, any form of help is appreciated.

Exam url: Exampdf.pdf

Appreciate any advice/help. Thank you!

2. do it by yourself first. when you get stuck, ask here

3. Yes, post any queries here so we can help you. You have to take responsibility..

4. Thanks guys. Right now I'm studying up on factorization of polynomials but for example, I know that d) and e) in Part 3 can be cross multiplied cause of the equal sign, but when the operations are addition/multiplication like in Part 4, what are the steps for that? For now will focus on understanding part 1. Thanks again

edit: ok so I get to Part 2 and on a) I get as far as this "(-8y^3)(-y^6)" but then the exponent outside the bracket is ^-2, do I inverse it and make it the denominator? ugh my brain is so scrambled

5. K guys haha, I'm slowly getting these. In part 4, however, I'm stumped. I know that the fractional equations in Part 3 can be cross multiplied but in Part 4 I have no idea what steps to take. I'm gonna keep trying to learn it but any clarification would be awesome. Thanks

6. I'll do some of the first one for you.

$\displaystyle \frac{1}{3x+y} + \frac{3(x+y)}{9x^2-y^2}$

Now we have to find a common factor in the denominator. Factorise $\displaystyle 9x^2-y^2$ to make it more simple.

$\displaystyle 9x^2-y^2 = (3x+y)(3x-y)$

Difference of Perfect Squares

$\displaystyle \frac{1}{3x+y} + \frac{3(x+y)}{(3x+y)(3x-y)}$

The common factors are $\displaystyle (3x+y) (3x-y)$

$\displaystyle \frac {3x-y+3(x+y)}{(3x+y)(3x-y)}$

$\displaystyle \frac {6x+2y}{(3x+y)(3x-y)}$

$\displaystyle \frac {2(3x+y)}{(3x+y)(3x-y)}$

Cross out $\displaystyle 3x+y$

You are left with:

$\displaystyle \frac {2}{(3x-y)}$

hope i did this right.

So basically find common denominator by factorising if you need to then simplify away!

7. Originally Posted by DannyMath
Hi guys,

I have a Math final worth 100% of my grade and have been knee deep in my text book for the past 2 weeks. I went to the copy centre beside my University and they had old copies of finals, and my friend who did the Math final last year said she used the old exams to study and they're almost identical each year (except obviously the values of the numbers changed etc).

I scanned the most recent exam from last term and would really appreciate if someone could outline the basic way to deal with the questions. Anyway this is the exam PDF, any form of help is appreciated.

Exam url: Exampdf.pdf

Appreciate any advice/help. Thank you!
Do not expect people to provide exam solutions for you. Type out specific questions you cannot do, after you have made a genuine attempt at doing them yourself. Show all your working and say exactly where you get stuck.

Thread closed.