Question

Solve for X: $\dfrac{X-1}{X+2}+\dfrac{X-3}{X-4}=\dfrac{10}{3}$

Answer 1

Hint: This type of question can be solved by first finding the Least Common Multiple (LCM) of both the terms on the left-hand side of the equation. We then need to multiply the obtained terms and expand. Next, we cross multiply with the fraction on the right-hand side of the equation. We then simplify the equation to a quadratic equation and then solve for x.

Complete step by step solution:
Given the equation $\dfrac{X-1}{X+2}+\dfrac{X-3}{X-4}=\dfrac{10}{3},$ we need to find the value of X. In order to do so, we simplify the left-hand side of the equation first. For this, we find the least common multiple (LCM) of both the denominators on the left-hand side of the equation. Therefore, the LCM of $X+2$ and $X-4$ is $\left( X+2 \right).\left( X-4 \right).$ We now use this in the equation as follows,
$\Rightarrow \dfrac{\left( X-1 \right).\left( X-4 \right)+\left( X-3 \right).\left( X+2 \right)}{\left( X+2 \right).\left( X-4 \right)}=\dfrac{10}{3}$
Multiplying the terms together in the numerator and denominator,
$\Rightarrow \dfrac{{{X}^{2}}-X-4X+4+{{X}^{2}}-3X+2X-6}{{{X}^{2}}-4X+2X-8}=\dfrac{10}{3}$
Adding up the terms in the numerator and denominator,
$\Rightarrow \dfrac{2{{X}^{2}}-6X-2}{{{X}^{2}}-2X-8}=\dfrac{10}{3}$
Cross multiplying both sides of the equation,
$\Rightarrow 3\left( 2{{X}^{2}}-6X-2 \right)=10\left( {{X}^{2}}-2X-8 \right)$
Multiplying the term outside with the term in the brackets,
$\Rightarrow 6{{X}^{2}}-18X-6=10{{X}^{2}}-20X-80$
Taking all the terms to the right-hand side of the equation,
$\Rightarrow 10{{X}^{2}}-20X-80-6{{X}^{2}}+18X+6=0$
Adding the terms with same powers of X,
$\Rightarrow 4{{X}^{2}}-2X-74=0$
We now have the quadratic equation in the form of $a{{X}^{2}}+bX+c=0.$ To find the solution to this, we use the formula,
$X=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
From our equation we know that a=4, b=-2 and c=-74.
Substituting these values,
$\Rightarrow X=\dfrac{-\left( -2 \right)\pm \sqrt{{{\left( -2 \right)}^{2}}-4.4.-74}}{2.4}$
Simplifying the term inside the square root,
$\Rightarrow X=\dfrac{+2\pm \sqrt{4+1184}}{8}$
$\Rightarrow X=\dfrac{+2\pm \sqrt{1188}}{8}$
Substituting the value of $\sqrt{1188},$
$\Rightarrow X=\dfrac{+2\pm 34.4674}{8}$
Adding the terms in the numerator and dividing by 8, we get one root of the equation.
$\Rightarrow X=\dfrac{+36.4674}{8}$
$\Rightarrow X=4.5584$
Subtracting the terms in the numerator and dividing by 8, we get one more root of the equation.
$\Rightarrow X=\dfrac{-32.4674}{8}$
$\Rightarrow X=-4.0584$
Hence the values of X for the given equation are 4.5584 and -4.0584.

Note: To solve this question, students need to understand how to simplify algebraic equations in fraction form. Most importantly, they should know how to apply the equation in order to simplify for the roots. Sometimes the term in the square root may be negative. In this case, we need to consider the term i which is nothing but a complex number whose value is equal to $\sqrt{-1}.$