Question

Solve for x: $ \dfrac{{16}}{x} - 1 = \dfrac{{15}}{{x + 1}};x \ne 0, - 1 $

Answer 1

Hint: First we have to start the solution by cross multiplication of RHS and LHS and then we get a quadratic equation. Find its factors using an appropriate method or a direct method. As it is a quadratic equation, we get 2 values for x (roots of the equation). Make sure that the value of x is not equal to 0 or -1.

Complete step-by-step answer:
We are given to find the value of x in $ \dfrac{{16}}{x} - 1 = \dfrac{{15}}{{x + 1}};x \ne 0, - 1 $ .
First we are taking the LCM in the left hand side and multiplying the LCM with its second term ‘1’.
 $ \Rightarrow \dfrac{{16 - x}}{x} = \dfrac{{15}}{{x + 1}} $
Now we are cross multiplying the LHS and RHS
We get
 $ \Rightarrow \left( {16 - x} \right)\left( {x + 1} \right) = x\left( {15} \right) $
As we can see, the LHS has a product of two terms. Multiplying 16 with the second term and x with the second term and subtracting them from one another, we get
 $ \Rightarrow 16\left( {x + 1} \right) - x\left( {x + 1} \right) = 15x $
 $ \Rightarrow 16x + 16 - {x^2} - x = 15x $
Putting all the terms left side, we have
 $ \Rightarrow 16x + 16 - {x^2} - x - 15x = 0 $
 $ \Rightarrow 16x + 16 - {x^2} - 16x = 0 $
Cancelling similar terms with different signs, here it is 16x, we get
 $ \Rightarrow 16 - {x^2} = 0 $
 $ \Rightarrow {x^2} - 16 = 0 $
16 can also be written as 4 times 4 or square of 4.
 $ \Rightarrow {x^2} - {4^2} = 0 $
Considering x as ‘a’ and 4 as ‘b’, the above LHS is in the form $ {a^2} - {b^2} $ which is equal to $ \left( {a + b} \right)\left( {a - b} \right) $
Therefore, the equation becomes
 $ \Rightarrow \left( {x + 4} \right)\left( {x - 4} \right) = 0 $
As the Left hand side is a product, either the first term must be zero or second must be zero or both the terms zero.
So we get $ \Rightarrow \left( {x + 4} \right) = 0 and\left( {x - 4} \right) = 0 $
 $ \therefore x = - 4,x = + 4 $
Therefore, the values of x where $ \dfrac{{16}}{x} - 1 = \dfrac{{15}}{{x + 1}};x \ne 0, - 1 $ is 4, -4.
So, the correct answer is “x=4, x=-4”.

Note: Initially we did not know that the given equation is a quadratic equation, but as we proceeded we knew that it is quadratic. So do not decide the no. of values a variable can have initially itself. And the no. of roots or solutions of an equation depends upon the highest degree of its variable. Here the highest degree is 2, so the equation has 2 solutions.