# Solve the equation. \frac{3}{2x+5}+\frac{4}{2x-5}=\frac{14x+3}{4x^{2}-25

Solve the equation.
$$\displaystyle{\frac{{{3}}}{{{2}{x}+{5}}}}+{\frac{{{4}}}{{{2}{x}-{5}}}}={\frac{{{14}{x}+{3}}}{{{4}{x}^{{{2}}}-{25}}}}$$

• Questions are typically answered in as fast as 30 minutes

### Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

svartmaleJ
Step 1
Given:
$$\displaystyle{\frac{{{3}}}{{{2}{x}+{5}}}}+{\frac{{{4}}}{{{2}{x}-{5}}}}={\frac{{{14}{x}+{3}}}{{{4}{x}^{{{2}}}-{25}}}}$$
Now we simplify the above equation as,
$$\displaystyle\Rightarrow{\frac{{{3}{\left({2}{x}-{5}\right)}+{4}{\left({2}{x}+{5}\right)}}}{{{4}{x}^{{{2}}}-{25}}}}={\frac{{{14}{x}+{3}}}{{{4}{x}^{{{2}}}-{25}}}}$$
$$\displaystyle\Rightarrow{\frac{{{6}{x}-{15}+{8}{x}+{20}}}{{{4}{x}^{{{2}}}-{25}}}}={\frac{{{14}{x}+{3}}}{{{4}{x}^{{{2}}}-{25}}}}$$
$$\displaystyle\Rightarrow{14}{x}+{5}={14}{x}+{3}$$
$$\displaystyle\Rightarrow{5}={3}$$
Step 2
Which is not possible.
Since,
$$\displaystyle{5}\ne{3}$$
Therefore equation has no solution