Question

Question:
If \[\dfrac{{1 - 11x}}{{\left( {x + 1} \right)\left( {x + 6} \right)\left( {x - 4} \right)}} = \dfrac{A}{{x + 1}} + \dfrac{B}{{x + 6}} + \dfrac{C}{{x - 4}}\], then B is equal to
A) \[ - \dfrac{{12}}{{25}}\]
B) \[\dfrac{{67}}{{50}}\]
C) \[ - \dfrac{{43}}{{50}}\]
D) \[ - \dfrac{{67}}{{50}}\]

Answer 1

Hint:
In the given question, we have been given a relation between a fraction (which has denominators which is a product of three polynomials) and three terms with constants in their numerators and with each term containing one of the three polynomials in their denominator. We have to calculate the value of one of the constants by solving this system of equations. For doing that, we are going to multiply the denominator of the left side of the equation to the other side, which is going to cancel out each of the denominators of the three terms. Then we are going to put in such a value of the argument that all but the needed terms become zero. And then we are going to solve for the unknown and find our answer.

Complete Step by Step Solution:
The given expression is \[\dfrac{{1 - 11x}}{{\left( {x + 1} \right)\left( {x + 6} \right)\left( {x - 4} \right)}} = \dfrac{A}{{x + 1}} + \dfrac{B}{{x + 6}} + \dfrac{C}{{x - 4}}\].
We are going to cross-multiply \[\left( {x + 1} \right)\left( {x + 6} \right)\left( {x - 4} \right)\] to the other side.
\[1 - 11x = A\left( {x + 6} \right)\left( {x - 4} \right) + B\left( {x - 4} \right)\left( {x + 1} \right) + C\left( {x + 1} \right)\left( {x + 6} \right)\]
Now, we are going to put in \[x = - 6\] so that the A and C terms become zero,
\[1 - 11\left( { - 6} \right) = A \times 0 + B\left( { - 10} \right) \times \left( { - 5} \right) + C \times 0\]
Now, solving for B,
\[50B = 67\]
Thus, \[B = \dfrac{{67}}{{50}}\]

Hence, the correct option is B.

Note:
In the given question we had to calculate the value of a variable. We were given an expression in which we had been given the use of this variable whose value was to be found. We first removed the denominators from all the terms as the denominator on one side of the equality was same as the LCM of the three denominators on the other side of equality. Then we got the three variables used in the equation being multiplied with some part of denominators’ LCM. Then we just put in the value of the argument which caused the other two non-necessary variables to become zero, which caused us to have just the one variable we wanted and then we just solved and got the answer.