Thread: 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.

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

...

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???

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:



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