Question

Can figure out this one. a+6 over4-a^2 subtract a+3 over a+2 plus a-3 over 2-a. Please help

Answer 1

factorise (4-a^2) = (2+a)(2-a) , spot the common denominators and make all of the denominators the same and add/substract as per normal

Answer 2

\[a+6-(2-a)(a+3)+(2+a)(a-3) \over (2-a)(2+a)\]

Answer 3

I dont think i did that right, I got -5a+6 over (2-a)(2+a)

Answer 4

The numerator calculates out as follows:
a + 6 -2a - 6 + a^2 + 3a + 2a - 6 + a^2 - 3a = 2a^2 + a -6 =(2a - 3)(a + 2)
Therefore putting this over the denominator:
((2a - 3)(a + 2))/(2 - a)(2 + a) = (2a - 3)/(2 - a)
So the answer is:
2a - 3
--------
2 - a

Answer 5

thank you so much, I always screw up with a negative or two

Answer 6

You're not the only one :) Take care :)