Question

The value of $ x $ that satisfies the equation $ \dfrac{4}{{x - 3}} + \dfrac{5}{{x - 5}} = \dfrac{9}{{x - 13}} $ is
A. $ 4 $
B. $ 3 $
C. $ 2 $
D. $ 1 $

Answer 1

Hint: This question is related to linear equation concept. An equation for a straight line is known as a linear equation. The term which is involved in a linear equation is either a constant or a single variable or product of a constant. The two variables can never be multiplied. All linear equations have a line graph. Here, in this question we need to solve the equation and check our results. We will use the operations of addition, subtraction, multiplication and division to find the value of $ x $ .

Complete step by step solution:
We will first use the basic mathematical operations i.e., addition, subtraction, multiplication and division to find the value of $ x $ .
Take the LCM of the terms on the left-hand side of the equation, we get;
$
   \Rightarrow \dfrac{{4\left( {x - 5} \right) + 5\left( {x - 3} \right)}}{{\left( {x - 3} \right)\left( {x - 5} \right)}} = \dfrac{9}{{x - 13}} \\
   \Rightarrow \dfrac{{4x - 20 + 5x - 15}}{{\left( {x - 3} \right)\left( {x - 5} \right)}} = \dfrac{9}{{x - 13}} \\
   \Rightarrow \dfrac{{9x - 35}}{{{x^2} - 8x + 15}} = \dfrac{9}{{x - 13}} \\
$
Now, we cross-multiply the above obtained equation i.e., multiply quadratic equation $ {x^2} - 8x + 15 $ to numerator of right-hand side equation i.e., $ 9 $ and multiply $ x - 13 $ to numerator of left-hand side equation i.e., $ 9x - 35 $ .
$
   \Rightarrow \left( {9x - 35} \right)\left( {x - 13} \right) = 9\left( {{x^2} - 8x + 15} \right) \\
   \Rightarrow 9{x^2} - 117x - 35x + 455 = 9{x^2} - 72x + 135 \\
$
Simplify the above equation and we get,
$
   \Rightarrow - 152x + 455 = - 72x + 135 \\
   \Rightarrow 455 - 135 = - 72x + 152x \\
   \Rightarrow 320 = 80x \\
$
Divide the entire equation by $ 80 $ .
$
   \Rightarrow \dfrac{{320}}{{80}} = \dfrac{{80}}{{80}}x \\
   \Rightarrow 4 = x \\
$
Therefore, we get the value of $ x $ as $ 4 $ .

So, the correct answer is Option A.

Note: The given question was an easy one. We can also substitute the value of $ x $ in the given equation to double check our results. Substitute the value of $ x $ as $ 4 $ in the equation.
$
   \Rightarrow \dfrac{4}{{x - 3}} + \dfrac{5}{{x - 5}} = \dfrac{9}{{x - 13}} \\
   \Rightarrow \dfrac{4}{{4 - 3}} + \dfrac{5}{{4 - 5}} = \dfrac{9}{{4 - 13}} \\
   \Rightarrow \dfrac{4}{1} + \dfrac{5}{{ - 1}} = \dfrac{9}{{ - 9}} \\
   \Rightarrow 4 + \left( { - 5} \right) = - 1 \\
   \Rightarrow 4 - 5 = - 1 \\
   \Rightarrow - 1 = - 1 \\
 $
Since, LHS=RHS, we can say that the obtained solution is correct. The only area of mistake in these types of questions is of calculation. Students should perform calculations carefully and avoid making any mistakes.