Find numbers A and B such that 1/(x-2)(x+3)=A/x-2+B/x+3

Question

Please answer this question using clear steps and/or explanation

Answer 1

We use the partial fractions method.  We write the right side of the equation using the common denomniator.  Since the denominator on the left side is (x - 2)(x + 3), we need to have the denomimator on the right side to be the same as the denominator on the left side.

 

 

1 / (x - 2)(x + 3) = [A(x + 3) + B(x - 2)] / (x - 2)(x + 3)

 

 

Next, we equate numerators.

 

1 = A(x + 3) + B(x - 2)

 

1 = Ax + 3A + Bx - 2B

 

1 = (A + B)x + 3A - 2B

 

 

Next, we equate coefficients according to their terms.  Notice that the coefficient of x on the left side is 0.

 

x:  A + B = 0

 

const:  3A - 2B = 1

 

 

We have a system of equations.

 

A + B = 0               eq1

 

3A - 2B = 1             eq2

 

 

We can substitute eq1 into eq2, so that we have a single equation with only one variable.   We can get eq2 in terms of A.  From eq1,

 

B = -A

 

 

3A - 2(-A) = 1

 

Solve for A from this equation.

 

3A + 2A = 1

 

5A = 1

 

A = 1/5

 

 

Next, substitute this value of A into eq1.  This makes

 

B = -1/5

 

 

Now to check our values of A and B, we plug them into the right side of the equation.

 

[(1 / 5) / (x - 2)] + [(-1 / 5) / (x + 3)] =

 

[1 / (5x - 10)] - [1 / (5x + 15)]

 

 

Factor each term's denominator.

 

[1 / (5(x - 2))] - [ 1 / (5(x + 3))]

 

 

Using a common denominator,

 

[(x + 3) - (x - 2)] / [5(x - 2)(x + 3)]

 

 

Simplifying, we get

 

5 / [5(x - 2)(x + 3)] =

 

1 / (x - 2)(x + 3)

 

 

This equals the expression of the left side.