Question

The fraction \[\dfrac{{5x - 11}}{{2{x^2} + x - 6}}\] was obtained by adding the two fractions \[\dfrac{A}{{x + 2}}\] and \[\dfrac{B}{{2x - 3}}\]. The values of A and B must be:
(A) \[A = 5x\]
 \[B = - 11\]
(B) \[A = - 11\]
\[B = 5x\]
(C) \[A = - 1\]
\[B = 3\]
(D) \[A = 3\]
\[B = - 1\]
(E) \[A = 5\]
\[B = - 11\]

Answer 1

Hint: Here, we will be taking the L.C.M on the left-hand side of the equation.

Complete step-by-step answer:
Given, the equation \[\dfrac{A}{{x + 2}} + \dfrac{B}{{2x - 3}} = \dfrac{{5x - 11}}{{2{x^2} + x - 6}}\]
If we take the L.C.M on left hand side of the equation then we get,
\[\dfrac{{A(2x - 3) + B(x + 2)}}{{\left( {x + 2} \right)\left( {2x - 3} \right)}} = \dfrac{{5x - 11}}{{2{x^2} + x - 6}}\]
\[ \Rightarrow \dfrac{{2Ax - 3A + Bx + 2B}}{{2{x^2} + x - 6}} = \dfrac{{5x - 11}}{{2{x^2} + x - 6}}\]
\[ \Rightarrow \dfrac{{(2A + B)x + 2B - 3A}}{{2{x^2} + x - 6}} = \dfrac{{5x - 11}}{{2{x^2} + x - 6}}\]
Since, the denominator is the same on both the sides of the equation therefore we can compare their numerators. On comparing the coefficients of \[x\] on both the sides of the equation, we get
\[2A + B = 5\] …… (1)
Similarly, on comparing the constants on both the sides of the equation, we get
\[ - 3A + 2B = - 11\] …… (2)
If we subtract the equation (1) times 2 and equation (2) then we get,
\[7A = 21\]
\[ \Rightarrow A = 3\]
On putting the value of A in equation (1) we get
\[2 \times 3 + B = 5\]
\[ \Rightarrow 6 + B = 5\]
\[ \Rightarrow B = - 1\]
Therefore, (D) \[A = 3\] is the required solution
                          \[B = - 1\]

Note: In these types of questions always try to compare the coefficients of the terms in the numerators on both side of the equation