Solve (2(x + 4))/(x + 3) - (x ^ 2 + 3x - 24)/(x ^ 2 - x - 12) | Equatix - Math Solver

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Example

3(r+2s)=2t-4

a⋅(b-2)=3b

3(r+2s)=2t-4

a⋅(b-2)=3b

Question

$$\frac{2(x+4)}{x+3}-\frac{x^{2}+3x-24}{x^{2}-x-12}$$

Evaluate

$\frac{x^{2}-3x-8}{\left(x-4\right)\left(x+3\right)}$

Short Solution Steps

Factor $x^{2}-x-12$.

$$\frac{2\left(x+4\right)}{x+3}-\frac{x^{2}+3x-24}{\left(x-4\right)\left(x+3\right)}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of $x+3$ and $\left(x-4\right)\left(x+3\right)$ is $\left(x-4\right)\left(x+3\right)$. Multiply $\frac{2\left(x+4\right)}{x+3}$ times $\frac{x-4}{x-4}$.

$$\frac{2\left(x+4\right)\left(x-4\right)}{\left(x-4\right)\left(x+3\right)}-\frac{x^{2}+3x-24}{\left(x-4\right)\left(x+3\right)}$$

Since $\frac{2\left(x+4\right)\left(x-4\right)}{\left(x-4\right)\left(x+3\right)}$ and $\frac{x^{2}+3x-24}{\left(x-4\right)\left(x+3\right)}$ have the same denominator, subtract them by subtracting their numerators.

$$\frac{2\left(x+4\right)\left(x-4\right)-\left(x^{2}+3x-24\right)}{\left(x-4\right)\left(x+3\right)}$$

Do the multiplications in $2\left(x+4\right)\left(x-4\right)-\left(x^{2}+3x-24\right)$.

$$\frac{2x^{2}-8x+8x-32-x^{2}-3x+24}{\left(x-4\right)\left(x+3\right)}$$

Combine like terms in $2x^{2}-8x+8x-32-x^{2}-3x+24$.

$$\frac{x^{2}-3x-8}{\left(x-4\right)\left(x+3\right)}$$

Expand $\left(x-4\right)\left(x+3\right)$.

$$\frac{x^{2}-3x-8}{x^{2}-x-12}$$

Expand

$\frac{x^{2}-3x-8}{\left(x-4\right)\left(x+3\right)}$

Short Solution Steps

Factor $x^{2}-x-12$.

$$\frac{2\left(x+4\right)}{x+3}-\frac{x^{2}+3x-24}{\left(x-4\right)\left(x+3\right)}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of $x+3$ and $\left(x-4\right)\left(x+3\right)$ is $\left(x-4\right)\left(x+3\right)$. Multiply $\frac{2\left(x+4\right)}{x+3}$ times $\frac{x-4}{x-4}$.

$$\frac{2\left(x+4\right)\left(x-4\right)}{\left(x-4\right)\left(x+3\right)}-\frac{x^{2}+3x-24}{\left(x-4\right)\left(x+3\right)}$$

Since $\frac{2\left(x+4\right)\left(x-4\right)}{\left(x-4\right)\left(x+3\right)}$ and $\frac{x^{2}+3x-24}{\left(x-4\right)\left(x+3\right)}$ have the same denominator, subtract them by subtracting their numerators.

$$\frac{2\left(x+4\right)\left(x-4\right)-\left(x^{2}+3x-24\right)}{\left(x-4\right)\left(x+3\right)}$$

Do the multiplications in $2\left(x+4\right)\left(x-4\right)-\left(x^{2}+3x-24\right)$.

$$\frac{2x^{2}-8x+8x-32-x^{2}-3x+24}{\left(x-4\right)\left(x+3\right)}$$

Combine like terms in $2x^{2}-8x+8x-32-x^{2}-3x+24$.

$$\frac{x^{2}-3x-8}{\left(x-4\right)\left(x+3\right)}$$

Expand $\left(x-4\right)\left(x+3\right)$.

$$\frac{x^{2}-3x-8}{x^{2}-x-12}$$