Question

Solve for $x:$
$\dfrac{{x + 1}}{{x - 1}} + \dfrac{{x - 2}}{{x + 2}} = 4 - \dfrac{{2x + 3}}{{x - 2}};x \ne 1, - 2,2.$

Answer 1

Hint: According to given in the question we have to determine the value of $x$ for the given expression $\dfrac{{x + 1}}{{x - 1}} + \dfrac{{x - 2}}{{x + 2}} = 4 - \dfrac{{2x + 3}}{{x - 2}};x \ne 1, - 2,2.$ So, first of all we have to apply the cross multiplication for the given expression.
Now, we have to multiply each of the terms of the expression obtained after applying cross multiplication.
Now, we have to take the terms obtained in the right hand side of the expression to the left hand side then we have to eliminate the terms which can be eliminated by adding and subtracting the terms of the expression just obtained.
Now, we will obtain the quadratic expression and we have to solve the expression to obtain the roots which can be done by finding the factors of the multiplication of the constant term and the coefficient of ${x^2}$ then we can easily obtain the required value of x.

Complete step-by-step solution:
Step 1: First of all we have to multiply each of the terms of the expression obtained after applying cross multiplication.
$ \Rightarrow \dfrac{{(x + 1)(x + 2) + (x - 1)(x - 2)}}{{(x - 1)(x + 2)}} = \dfrac{{4(x - 2) - 2x + 3}}{{x - 2}}$………………….(1)
Step 2: Now, we have to simplify the expression (1) as obtained in the solution step 1 by multiplying each term of the expression. Hence,
$ \Rightarrow \dfrac{{({x^2} + 2x + x + 2) + ({x^2} - x - 2x + 2)}}{{({x^2} + 2x - x - 2)}} = \dfrac{{(4x - 8) - 2x + 3}}{{x - 2}}$
$ \Rightarrow \dfrac{{(2{x^2} + 4)}}{{({x^2} + x - 2)}} = \dfrac{{(2x - 11)}}{{x - 2}}$
$
   \Rightarrow (2{x^2} + 4)(x - 2) = (2x - 11)({x^2} + x - 2) \\
   \Rightarrow 2{x^3} - 4{x^2} + 4x - 8 = 2{x^3} + 2{x^2} - 4x - 11{x^2} - 11x + 22 \\
   \Rightarrow 2{x^3} - 4{x^2} + 4x - 8 = 2{x^3} - 9{x^2} - 15x + 22
 $
Step 3: Now, we simplify the expression obtained in the solution step 2
$ \Rightarrow 5{x^2} + 19x - 30 = 0$
Now, we simplify this quadratic equation by factor the coefficient of $x$ in the term of coefficient of ${x^2}$ and C.
$
   \Rightarrow 5{x^2} + (25 - 6)x - 30 = 0 \\
   \Rightarrow 5{x^2} + 25x - 6x - 30 = 0 \\
   \Rightarrow 5x(x + 5) - 6(x + 5) = 0 \\
   \Rightarrow (x + 5)(5x - 6) = 0 \\
   \Rightarrow x = - 5,\dfrac{6}{5}
 $

Hence, the value of $x$ is $ - 5,\dfrac{6}{5}$ for the given expression $\dfrac{{x + 1}}{{x - 1}} + \dfrac{{x - 2}}{{x + 2}} = 4 - \dfrac{{2x + 3}}{{x - 2}};x \ne 1, - 2,2.$

Note: On solving the quadratic equation which is in the form of $a{x^2} + bx + c = 0$. We always get two possible roots of $x$
To obtain the value of $x$ we have to necessarily use a cross multiplication rule then we have to simply the obtained expression.