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Multiply both sides of this equation by the LCD, which is (x+3)(x-1).
The result will be:
5x + 7 = A(x-1) + B(x+3)
You can substitute any two values of x into this equation, and obtain two equations and two unknowns. However, the easiest way to do it is to let x=1 and let x= -3, since these particular values of x will "zero out" half of the problem and give immediate results:
If x= 1
5x + 7 = A(x-1) + B(x+3)
5 + 7 = A(0) + B(1+3)
12=4B, so B= 3
If x = -3,
5x + 7 = A(x-1) + B(x+3)
-15+7 = A(-3-1) + B(0)
-8 = -4A, so A = 2
Therefore the partial fractions would be:
R^2 at SCC
