Question

Solve the equation
$\dfrac{x-1}{x-2}+\dfrac{x+1}{x+2}=$$2\dfrac{4}{5}$

Answer 1

Hint: Take L.C.M of denominators of the fractions given in the left hand side of the provided equation. Use the algebraic identity of ${{a}^{2}}-{{b}^{2}}=(a-b)(a+b)$, if required. Further simplify the expression and hence, find two values of ‘x’.

Complete step-by-step answer:

We have
$\dfrac{x-1}{x-2}+\dfrac{x+1}{x+2}=2\dfrac{4}{5}$ …….. (i)
As, we need to find the value of x from the above, so let’s simplify the relation by taking L.C.M. of denominators in the left hand side of the equation and convert the mixed fraction of the right hand side in equation (i) to a simple fraction.
So, we get

$\dfrac{(x-1)(x+2)+(x+1)(x-2)}{(x-2)(x+2)}=\dfrac{14}{5}......(ii)$
As, we know the algebraic identity of ${{a}^{2}}-{{b}^{2}}=(a-b)(a+b)......(iii)$
So, we can use this identity to solve the denominator of the fraction given in left hand side of the equation (ii) as
$\dfrac{(x-1)(x+2)+(x+1)(x-2)}{{{x}^{2}}-{{(2)}^{2}}}=\dfrac{14}{5}$
Now, simplifying the numerator of left hand side, we get
$\dfrac{{{x}^{2}}-x+2x-2+{{x}^{2}}-2x+x-2}{{{x}^{2}}-4}=\dfrac{14}{5}$
$\dfrac{2{{x}^{2}}-4}{{{x}^{2}}-4}=\dfrac{14}{5}$
On cross multiplying the above equation, we get above equation as
$\begin{align}
  & 5(2{{x}^{2}}-4)=14({{x}^{2}}-4) \\
 & 10{{x}^{2}}-20=14{{x}^{2}}-56 \\
 & 56-20=4{{x}^{2}} \\
 & 4{{x}^{2}}=36 \\
 & \\
\end{align}$
On dividing the whole equation by 4, we get
$\begin{align}
  & \dfrac{4{{x}^{2}}}{4}=\dfrac{36}{4}=9 \\
 & {{x}^{2}}=9 \\
\end{align}$

On taking the square root on both sides of the equation, we get
$x=\pm 3$
Hence, values of x solving the given expression in the problem are -3, 3.

Note: One may go wrong with the last step i.e.
${{x}^{2}}=9$
And hence we can write value of x as
$x=3$ only.
So, it can be a general mistake by students. We can ignore ‘-3’ from the expression ${{x}^{2}}-9$, which is wrong. As ${{x}^{2}}=9$ or ${{x}^{2}}-9=0$ is a quadratic, so it should have two roots. So, take care of this step for future reference. Algebraic Calculation is the key point of the equation.