Chapter SH, Problem 8.18EP

To determine

To find: the solution of the equation and verify it.

Answer to Problem 8.18EP

At $\text{x=}-16$ is solution of the equation and its verified.

Explanation of Solution

Given:

  $\frac{\text{14}}{\text{x-5}}\text{-}\frac{\text{7}}{\text{x+5}}\text{=}\frac{\text{-7}}{{\text{x}}^{\text{2}}\text{-25}}$ .

Concept used:

Here the denominator is taken common by taking lowest common multiple.

The solution can be verified by putting the value of $\text{x}$ back to the equation and when $\text{L}\text{.HS=R}\text{.HS}$ The equation is verified.

Calculation:

  $\frac{\text{14}}{\text{x-5}}\text{-}\frac{\text{7}}{\text{x+5}}\text{=}\frac{\text{-7}}{{\text{x}}^{\text{2}}\text{-25}}$ ...... $\left(\text{i}\right)$

  $\Rightarrow \frac{\text{14}\left(\text{x+5}\right)-\text{7}\left(\text{x-5}\right)}{\left(\text{x-5}\right)\left(\text{x+5}\right)}\text{=}\frac{\text{-7}}{\left(\text{x-5}\right)\left(\text{x+5}\right)}$ .

By cross multiplying each other:

  $\Rightarrow \text{14}\left(\text{x+5}\right)-\text{7}\left(\text{x-5}\right)\text{=-7}$ .

  $\Rightarrow \text{14x+70}-\text{7x+35=-7}$ .

  $\Rightarrow \text{x=}\frac{\text{-112}}{7}$ .

  $\Rightarrow \text{x=}-16$ .

By putting the $\text{x=}-16$ in equation $\left(\text{i}\right)$ :

  $\frac{\text{14}}{\left(-16\right)\text{-5}}\text{-}\frac{\text{7}}{\left(-16\right)\text{+5}}\text{=}\frac{\text{-7}}{{\left(-16\right)}^{\text{2}}\text{-25}}$ .

  $\frac{\text{14}}{-21}+\frac{\text{7}}{11}\text{=}\frac{\text{-7}}{\text{231}}$ .

  $-\frac{\text{14}\times \text{11+7}\times \text{21}}{21\times 11}\text{=}\frac{\text{-7}}{\text{231}}$ .

  $\frac{\text{-}\left(\text{14}\times \text{11}\right)\text{+7}\times \text{21}}{21\times 11}\text{=}\frac{\text{-7}}{\text{231}}$ .

  $\frac{\text{-}7}{231}\text{=}\frac{\text{-7}}{\text{231}}$ .

Hence, at $\text{x=}-16$ is solution of the equation and its verified.