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Can figure out this one. a+6 over4-a^2 subtract a+3 over a+2 plus a-3 over 2-a. Please help
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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
\[a+6-(2-a)(a+3)+(2+a)(a-3) \over (2-a)(2+a)\]
I dont think i did that right, I got -5a+6 over (2-a)(2+a)
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
thank you so much, I always screw up with a negative or two
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