Partial Fraction Decomposition Calculator

Your input: perform the partial fraction decomposition of $$$\frac{7}{\left(x + 2\right) \left(2 x - 3\right)}$$$

The form of the partial fraction decomposition is

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

Write the right-hand side as a single fraction:

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

The denominators are equal, so we require the equality of the numerators:

$$7=\left(x + 2\right) A + \left(2 x - 3\right) B$$

Expand the right-hand side:

$$7=x A + 2 x B + 2 A - 3 B$$

Collect up the like terms:

$$7=x \left(A + 2 B\right) + 2 A - 3 B$$

The coefficients near the like terms should be equal, so the following system is obtained:

$$\begin{cases} A + 2 B = 0\\2 A - 3 B = 7 \end{cases}$$

Solving it (for steps, see system of equations calculator), we get that $$$A=2$$$, $$$B=-1$$$

Therefore,

$$\frac{7}{\left(x + 2\right) \left(2 x - 3\right)}=\frac{2}{2 x - 3}+\frac{-1}{x + 2}$$

Answer: $$$\frac{7}{\left(x + 2\right) \left(2 x - 3\right)}=\frac{2}{2 x - 3}+\frac{-1}{x + 2}$$$