# Math Help - Struggling with several questions

1. ## Struggling with several questions

Howdy all,

Currently doing a practice test for my upcoming assignment and have flat out hit the wall on the following!! Any assistance would be greatly appreciated:

1) Subtract the fractions and simplify: (x+4)/(x+5)(x+9) - (x+4)/(x+9)
2) 4/t + 1/2t + 1/2+t
3) find p in terms of the other unknowns 2p+4s+pt = q-s
4) Solve 6/A^2-1 = 1/a+1 - 1/a-1
5) The area of a rectangular vegetable garden is 384 square metres, if its length is 20m more than its width, what are the dimensions.

2. ## Re: Struggling with several questions

Originally Posted by robbyc1981
Howdy all,

Currently doing a practice test for my upcoming assignment and have flat out hit the wall on the following!! Any assistance would be greatly appreciated:

1) Subtract the fractions and simplify: (x+4)/(x+5)(x+9) - (x+4)/(x+9)

common denominator ...

(x+4)/[(x+5)(x+9)] - (x+4)(x+5)/[(x+9)(x+5)] = [(x+4) - (x+4)(x+5)]/[(x+5)(x+9)] , expand the numerator and combine like terms

2) 4/t + 1/2t + 1/2+t

once again, common denominator ... you do it.

3) find p in terms of the other unknowns 2p+4s+pt = q-s

2p + pt = q - 5s

p(2 + t) = q - 5s ... finish it

4) Solve 6/A^2-1 = 1/a+1 - 1/a-1

common denominator again ... then solve the equation formed by the numerators

5) The area of a rectangular vegetable garden is 384 square metres, if its length is 20m more than its width, what are the dimensions.

hint ... L = 20+W ... and, of course, you know the area formula for a rectangle
...

3. ## Re: Struggling with several questions

Thanks for that. The determining the common denom is giving me real head aches at the moment. Just to clarify with 5), i got to that part and understand the need to use A = L x W but hit a wall thereafter as to how we determine the dimensions algebraically???

4. ## Re: Struggling with several questions

Originally Posted by robbyc1981
Thanks for that. The determining the common denom is giving me real head aches at the moment. Just to clarify with 5), i got to that part and understand the need to use A = L x W but hit a wall thereafter as to how we determine the dimensions algebraically???
As skeeter pointed out you know

[1]: A = 384
[2]: L = W + 20

Plug in these terms and values into the equation of the area of a rectangle:

$A = L \cdot W~\implies~384=(W+20)\cdot W$

Solve for W and consequently for L using [2].