Chapter 11.5, Problem 2LC

To determine

To solve the equation $\frac{5}{x^2}=\frac{6}{x}-1$ and check the solution.

Answer to Problem 2LC

  $x=1,5$

Explanation of Solution

Given:

Given equation

  $\frac{5}{x^2}=\frac{6}{x}-1$

Concept Used:

• Least Common Denominator (LCD) is the smallest number that is divisible by the denominator of all the fractions that are being added or subtracted.
• Solving an equation containing rational expressions
1. Multiply both sides of equations by LCD(least common denominator) of all denominators.
2. Remove any grouping symbols and solve the resulting equation for the variable asked in equation

Calculation:

In order to solve the equation $\frac{5}{x^2}=\frac{6}{x}-1$ , observe that the LCD of all the denominators is $x^2$ . So multiply both sides by this LCD to get rid of the denominators and simplify further as shown below

  $\begin{array}{c}\frac{5}{x^2}=\frac{6}{x}-1\\ \frac{5}{x^2}\cdot x^2=\frac{6}{x}\cdot x^2-1\cdot x^2\\ 5=6x-x^2\\ x^2-6x+5=0\\ x^2-5x-x+5=0\\ x\left(x-5\right)-1\left(x-5\right)=0\\ \left(x-5\right)\left(x-1\right)=0\\ x-5=0\text{or}x-1=0\\ x=5\text{or}x=1\end{array}$

Thus, the solutions of the given equationare $x=1,5$ .

Now to check the solutions, substitute $x=1$ in the given equation and check whether left hand side value is equal to the right hand side value or no. So, it gives

  $\begin{array}{c}\frac{5}{1^2}=\frac{6}{1}-1\\ 5=6-1\\ 5=5\text{True Statement}\end{array}$

Again substitute $x=5$ in the given equation and check whether left hand side value is equal to the right hand side value or no. So, it gives

  $\begin{array}{c}\frac{5}{5^2}=\frac{6}{5}-1\\ \frac{1}{5}=\frac{6-5}{5}\\ \frac{1}{5}=\frac{1}{5}\text{True Statement}\end{array}$

Thus, the solution is correct.