How to solve 3/(2x + 1) - 2/x

1. Factor the equation (x + 2)(x + 3) = 0. 2. Set each factor equal to zero: x + 2 = 0 and x + 3 = 0. 3. Solve for x: x = -2 and x = -3.

Solve 3/(2x + 1) - 2/x | Equatix

Question

$$\frac{3}{2x+1}-\frac{2}{x}$$

Evaluate

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

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of $2x+1$ and $x$ is $x\left(2x+1\right)$. Multiply $\frac{3}{2x+1}$ times $\frac{x}{x}$. Multiply $\frac{2}{x}$ times $\frac{2x+1}{2x+1}$.

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

Since $\frac{3x}{x\left(2x+1\right)}$ and $\frac{2\left(2x+1\right)}{x\left(2x+1\right)}$ have the same denominator, subtract them by subtracting their numerators.

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

Do the multiplications in $3x-2\left(2x+1\right)$.

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

Combine like terms in $3x-4x-2$.

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

Expand $x\left(2x+1\right)$.

$$\frac{-x-2}{2x^{2}+x}$$

Show Solution

Differentiate w.r.t. x

$\frac{2\left(x^{2}+4x+1\right)}{\left(x\left(2x+1\right)\right)^{2}}$

Steps Using Derivative Rule for Quotient

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{3x}{x\left(2x+1\right)}-\frac{2\left(2x+1\right)}{x\left(2x+1\right)})$$

Since $\frac{3x}{x\left(2x+1\right)}$ and $\frac{2\left(2x+1\right)}{x\left(2x+1\right)}$ have the same denominator, subtract them by subtracting their numerators.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{3x-2\left(2x+1\right)}{x\left(2x+1\right)})$$

Do the multiplications in $3x-2\left(2x+1\right)$.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{3x-4x-2}{x\left(2x+1\right)})$$

Combine like terms in $3x-4x-2$.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{-x-2}{x\left(2x+1\right)})$$

Use the distributive property to multiply $x$ by $2x+1$.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\frac{-x-2}{2x^{2}+x})$$

For any two differentiable functions, the derivative of the quotient of two functions is the denominator times the derivative of the numerator minus the numerator times the derivative of the denominator, all divided by the denominator squared.

$$\frac{\left(2x^{2}+x^{1}\right)\frac{\mathrm{d}}{\mathrm{d}x}(-x^{1}-2)-\left(-x^{1}-2\right)\frac{\mathrm{d}}{\mathrm{d}x}(2x^{2}+x^{1})}{\left(2x^{2}+x^{1}\right)^{2}}$$

The derivative of a polynomial is the sum of the derivatives of its terms. The derivative of a constant term is $0$. The derivative of $ax^{n}$ is $nax^{n-1}$.

$$\frac{\left(2x^{2}+x^{1}\right)\left(-1\right)x^{1-1}-\left(-x^{1}-2\right)\left(2\times 2x^{2-1}+x^{1-1}\right)}{\left(2x^{2}+x^{1}\right)^{2}}$$

Simplify.

$$\frac{\left(2x^{2}+x^{1}\right)\left(-1\right)x^{0}-\left(-x^{1}-2\right)\left(4x^{1}+x^{0}\right)}{\left(2x^{2}+x^{1}\right)^{2}}$$

Multiply $2x^{2}+x^{1}$ times $-x^{0}$.

$$\frac{2x^{2}\left(-1\right)x^{0}+x^{1}\left(-1\right)x^{0}-\left(-x^{1}-2\right)\left(4x^{1}+x^{0}\right)}{\left(2x^{2}+x^{1}\right)^{2}}$$

Multiply $-x^{1}-2$ times $4x^{1}+x^{0}$.

$$\frac{2x^{2}\left(-1\right)x^{0}+x^{1}\left(-1\right)x^{0}-\left(-x^{1}\times 4x^{1}-x^{1}x^{0}-2\times 4x^{1}-2x^{0}\right)}{\left(2x^{2}+x^{1}\right)^{2}}$$

To multiply powers of the same base, add their exponents.

$$\frac{2\left(-1\right)x^{2}-x^{1}-\left(-4x^{1+1}-x^{1}-2\times 4x^{1}-2x^{0}\right)}{\left(2x^{2}+x^{1}\right)^{2}}$$

Simplify.

$$\frac{-2x^{2}-x^{1}-\left(-4x^{2}-x^{1}-8x^{1}-2x^{0}\right)}{\left(2x^{2}+x^{1}\right)^{2}}$$

Combine like terms.

$$\frac{2x^{2}+8x^{1}+2x^{0}}{\left(2x^{2}+x^{1}\right)^{2}}$$

For any term $t$, $t^{1}=t$.

$$\frac{2x^{2}+8x+2x^{0}}{\left(2x^{2}+x\right)^{2}}$$

For any term $t$ except $0$, $t^{0}=1$.

$$\frac{2x^{2}+8x+2\times 1}{\left(2x^{2}+x\right)^{2}}$$

For any term $t$, $t\times 1=t$ and $1t=t$.

$$\frac{2x^{2}+8x+2}{\left(2x^{2}+x\right)^{2}}$$

Show Solution

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Example

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Evaluate

Short Solution Steps

Differentiate w.r.t. x

Steps Using Derivative Rule for Quotient