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- Thread starter Guzman10
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- Jan 26, 2012

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Then once you do that you can let $x$ equal any value you want. Choose $x$ such that $A$ or $B$ cancels out, then you can solve for the other one. Can you make any progress now?

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- Dec 11, 2019

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(x-3)/(x+1) =A + B(x+3)/(x+1)

Now,setting x=-3 or x+3=0,makes the expression on the right containing (x+3) vanish. So,

(-3-3)/(-3+1)=A+0 .Then,

A=3

You could find B by a similar method.

- Jan 29, 2012

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There are several different ways to do this. The most obvious is to add the fractions on the right side: [tex]

\frac{x- 3}{(x+ 1)(x+ 3)}

=[/tex][tex] \frac{A(x+3)}{(x+ 1)(x+ 3)}+ \frac{B(x+ 1)}{(x+1)(x+ 3)}=[/tex][tex] \frac{Ax+ 3A+ Bx+ B}{x^2+ 4x^2+ 3}[/tex]so we must have [tex](A+ B)x+ (3A+ B)= x- 3[/tex]. In order that this be true for all x we must have (A+ B)x= x and 3A+ B= -3. Solve the equations A+ B= 1 and 3A+ B= -3 for A and B.

Multiply both sides of the equation by (x+ 1)(x+ 3): [tex]x- 3= A(x+ 3)+ B(x+ 1)[/tex].

And now we can write [tex]x- 3= (A+ B)x+ 3A+ B[/tex] so we must have the A+ B= 1 and 3A+ B= -3 as before. 3A+ B=-3. A+ B= 1" 2A= -4. A= -2. B= 3

Or, since this is to be true for all x we can simply choose two values for x to get to equations. If we take x= 0 (just because it is an easy number) we have [tex]-3= 3A+ B[/tex] again. If we take x= 1 we have [tex]-2= 4A+ 2B[/tex]. Dividing by 2 gives [tex]2A+ B= -1[/tex]. That last is a new equation but satisfied by the same A and B.

The simplest method is to choose x= -1 and x= -3 because they make the coefficients of one of A and B 0. If x= -1 we have [tex]-1- 3= -4= A(-1+ 3)+ B(-1+ 1)[/tex] so 2A= -4. Taking x= -3 we have [tex]-3- 3= -6= A(-3+ 3)+ B(-3+ 1)[/tex] so -2B= -6.