Chapter 1, Problem 1RE

Determine the restrictions on x for the equation

$\frac{3}{x^2-4}+\frac{4}{2x-7}=\frac{2}{3}$

To determine

To calculate: The restriction of provide equation $\frac{3}{x^2-4}+\frac{4}{2x-7}=\frac{2}{3}$ on $x$.

Answer to Problem 1RE

Solution:

The restriction of the provided equation $\frac{3}{x^2-4}+\frac{4}{2x-7}=\frac{2}{3}$ are $x\ne 2,x\ne -2,x\ne \frac{7}{2}$.

Explanation of Solution

Given information:

The provided equation is $\frac{3}{x^2-4}+\frac{4}{2x-7}=\frac{2}{3}$.

Formula used:

If $a,b$ are two any real number, then

$a^2-b^2=\left(a-b\right)\left(a+b\right)$

Calculation:

Consider the provided equation, $\frac{3}{x^2-4}+\frac{4}{2x-7}=\frac{2}{3}$

Factor the denominators of the equation $\frac{3}{x^2-4}+\frac{4}{2x-7}=\frac{2}{3}$

$\frac{3}{\left(x+2\right)\left(x-2\right)}+\frac{4}{2x-7}=\frac{2}{3}$

To find restricted for the above equation,

Take each denominator equal to zero,

First, take $\left(x+2\right)$ equal to zero.

$\begin{array}{c}x+2=0\\ x=-2\end{array}$

Now, take $\left(x-2\right)$ equal to zero

$\begin{array}{c}x-2=0\\ x=2\end{array}$

Now, take $2x-7$ equal to zero

$\begin{array}{c}2x-7=0\\ 2x=7\\ x=\frac{7}{2}\end{array}$

Hence, the restriction value for the equation are $x\ne 2,x\ne -2,x\ne \frac{7}{2}$.