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1/a+3-4/a-3=2a/a^2-9
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Well it's like : \[(1/(a+3)) - 4/(a-3) = \] we multiply 1/(a+3) by (a-3), and -4/(a-3) by (a+3), as follow : \[= (1\times (a-3)/(a+3)\times(a-3)) - (4\times (a+3)/(a-3)\times(a+3))\] So (a+3)*(a-3) = a^2 - 9 then; \[= (a-3)/(a^{2} - 9) - (4a+12)/(a^{2} - 9)\] = \[ = (a - 3 - 4a - 12)/(a^{2} - 9)\] \[ = (- 3a - 15)/(a^{2} - 9)\]
For future reference, this is better suited to the Math group. Closing this question as off topic.
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