Chapter 1, Problem 1RE

To determine

To calculate: The restrictions on $x$ if $\frac{3}{x^2-4}+\frac{4}{2x-7}=\frac{2}{3}$

Answer to Problem 1RE

Solution:

$\underset{\_}{x\ne 2,x\ne -2,x\ne \frac{7}{2}}$

Explanation of Solution

Given:

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

Formula used:

$(x^2-a^2)=(x+a)(x-a)$

Calculation:

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

To determine the restrictions on x, the denominator must not be equal to 0.

Therefore

$x^2-4\ne 0$ and $2x-7\ne 0$.

$x^2\ne 4$ and $x\ne \frac{7}{2}$.

Or

$x\ne 2$, $x\ne -2$, $x\ne \frac{7}{2}$