Question

solve the rational equation: $\frac{1}{7x - 4}=\frac{1}{7x^{2}-4x}-\frac{7}{7x^{2}-4x}$
$x = - 6$
$x = - 4$
$x = 6$
$x = 7$

Explanation:

Step1: Factor the denominators

The denominator $7x^{2}-4x=x(7x - 4)$. The given equation is $\frac{1}{7x - 4}=\frac{1}{x(7x - 4)}-\frac{7}{x(7x - 4)}$.

Step2: Find a common - denominator and simplify the right - hand side

Since the right - hand side has a common denominator of $x(7x - 4)$, $\frac{1}{x(7x - 4)}-\frac{7}{x(7x - 4)}=\frac{1 - 7}{x(7x - 4)}=\frac{-6}{x(7x - 4)}$. So the equation becomes $\frac{1}{7x - 4}=\frac{-6}{x(7x - 4)}$.

Step3: Cross - multiply (note $x

eq0$ and $x
eq\frac{4}{7}$)
Cross - multiplying gives $x(7x - 4)\times1=(7x - 4)\times(-6)$. Since $x
eq\frac{4}{7}$ (to avoid division by zero), we can divide both sides of the equation by $(7x - 4)$ (assuming $7x-4
eq0$), getting $x=-6$.

Answer:

$x = - 6$