`\frac{2}{x+3}-\frac{1}{x}=\frac{-4}{x^{2}+3x}`

Question

`\frac{2}{x+3}-\frac{1}{x}=\frac{-4}{x\left(x+3\right)}`

`\frac{2}{x+3}\cdot\frac{x}{x}-\frac{1}{x}\cdot\frac{\left(x+3\right)}{\left(x+3\right)}=\frac{-4}{x\left(x+3\right)}`

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

...keep going! What is the solution?

Answer 1

Multiplying both sides by \( x(x+3) \) to clear the denominators:

\( 2x - (x+3) = -4 \)

Simplify the left side:

\( 2x - x - 3 = -4 \)

\( x - 3 = -4 \)

Add 3 to both sides:

\( x = -1 \)

Therefore, the solution to the equation is \( x = -1 \).