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What you do is find the common denomonator which is simply (x-3)(x^2-9). Then you multiply that by each fraction.
(7 over x-3) times (x-3)(x^2-9) is 7x^2-63 because the (x-3)'s cancel out and you distribute the 7 to the (x^2-9).
(4 over x^2-9) times (x-3)(x^2-9) is 4x-12 (the (x^2-9)'s cancel out and distribute the 4 to the x-3)
so then you have 7x^2-63-4x-12 which simplifies to 7x^2-4x-75!
To perform the operation,
(7 over x-3) - (4 over x^2 - 9):
7 over x-3 = 7/x -3: The reason is a o b - c means subtract c from the result a o b. Operation 'over' has no effect (or distributive effect) on c. Moreover 7 over x -3 are two terms : the first is 7 over x connetced by - sign to the 2nd term 3.
In a similar manner 4 over x^2-9 = 4/x^2-9.
Therefore, the given expression , (7 over x-3) - (4 over x^2 - 9) is equal to:
= (7x-4+6x^2)/x^2 or
But if you intend to mean (7 over x-3) - (4 over x^2 - 9) to be 7/(x-3) - 4/(x^2-9), then you have to write your expression like:
7 over (x-3) - 4 over (x^2-9) which means same as [7 over (x-3)] - [4 over (x-9)], the rectangular brackets is only redundant but not necessary and won't spoil the inteded meaning also.
The denominators, x-3 and x^2-9 has the LCM x^2-9 and could be made common denominator.Also x^2-9 = (x-3)(x+3).
So, 7 over (x-3) - 4 over (x^2-9) = 7(x+3)/[(x-3)(x+3)] - 4/(x^2-9).
(7 over x-3) - (4 over x^2 - 9) has to be "translated" into:
[7 / (x-3)] - [4 /( x^2 - 9)]
The second step is to notice that you have the advantage that
x^2 - 9=(x-3)*(x+3)
[7 / (x-3)] - [4 /( x^2 - 9)]=[7 / (x-3)] -[4 /(x-3)*(x+3)]
Now, in order to subtract the second ratio from the first one, all we have to do is to find out the common denominator, which is obviously, (x-3)*(x+3). It is easy to see that the first ratio has to be multiplied with (x+3).
[7(x+3)-4 / (x-3)*(x+3) = (7x+21-4)/(x-3)*(x+3)
7 over x-3) - (4 over x^2 - 9)
`7/(x-3) - (4/(x^2-9))`
``The first fraction can be written as 7(x-3) / (x^2 - 9)
Now that the denominators are same, we can subtract as normal.
( 7(x-3) - 4 ) / (x^2 -9)
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