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Question

Determine the constants (denoted by capital letters) so that the equation is an identity. Do the problem in two ways.

(1 − x) / ((3x + 2)(4x − 1)) = (A) / (3x + 2)+ (B) / (4x − 1)

(a) Use the equating-the-coefficients theorem, as in Example 1.
A =
B =

(b) Use the convenient-values method that was explained after Example 1.
A =
B =

 

Answer 1

The equation is .

• a). Equating the coefficients theorem :

Least common denominator of (3x + 2) and (4x - 1) is (3x + 2)(4x - 1).

Cancel common terms.

Equating the coefficients of x : 4A + 3B = - 1------->  (1)

Equating the constant terms : - A + 2B = 1   --------> (2)

Solve the above system of equations :

Simplify equation 2 :

- A + 2B = 1

A = 2B - 1.

Substitute A = 2B - 1 in equation (1).

4(2B - 1) + 3B = - 1

8B - 4 + 3B = - 1

11B = - 1 + 4

11B = 3

B = 3/11.

Substitute B = 3/11 in equation (1).

4A + 3(3/11) = - 1

4A + (9/11) = - 1

4A = - 1 - 9/11

A = (- 11 - 9)/(11 * 4) = - 5/11.

Therefore, the values of constants of the given equation are A = - 5/11 and B = 3/11.

Answer 2

• b). Convenient - values method :

The equation is .

Least common denominator of (3x + 2) and (4x - 1) is (3x + 2)(4x - 1).

Cancel common terms.

If we set x = 1,

⇒ (1 - 1) = A(4 - 1) + B(3 + 2)

3A + 5B = 0 ----------------> (1)

If we set x = 1/4,

⇒ (1 - 1/4) = A((4*1/4) - 1) + B((3*1/4) + 2)

3/4 = A(1 - 1) + B(3/4 + 2)

3/4 = B[(3 + 8)/4 ]

3/4 = (11/4)B

3 = 11B

⇒ B = 3/11.

Substitute B = 3/11 in equation (1).

3A + 5(3/11) = 0

3A + (15/11) = 0

3A = - 15/11

⇒ A = - 5/11.

Therefore, the values of constants of the given equation are A = - 5/11 and B = 3/11.