Solve (x+1/x+2)^2=x+2/x+4

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

Solve for x

$$x = -\frac{4}{3} = -1\frac{1}{3} \approx -1.333333333$$

Steps for Solving Linear Equation

Variable x cannot be equal to -4 since division by zero is not defined. Multiply both sides of the equation by x+4.

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

To raise \frac{x+1}{x+2} to a power, raise both numerator and denominator to the power and then divide.

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

Express \left(x+4\right)\times \frac{\left(x+1\right)^{2}}{\left(x+2\right)^{2}} as a single fraction.

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

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+1\right)^{2}.

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

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+2\right)^{2}.

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

Subtract x from both sides.

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

Use the distributive property to multiply x+4 by x^{2}+2x+1 and combine like terms.

$$\frac{x^{3}+6x^{2}+9x+4}{x^{2}+4x+4}-x=2$$

Factor x^{2}+4x+4.

$$\frac{x^{3}+6x^{2}+9x+4}{\left(x+2\right)^{2}}-x=2$$

To add or subtract expressions, expand them to make their denominators the same. Multiply x times \frac{\left(x+2\right)^{2}}{\left(x+2\right)^{2}}.

$$\frac{x^{3}+6x^{2}+9x+4}{\left(x+2\right)^{2}}-\frac{x\left(x+2\right)^{2}}{\left(x+2\right)^{2}}=2$$

Since \frac{x^{3}+6x^{2}+9x+4}{\left(x+2\right)^{2}} and \frac{x\left(x+2\right)^{2}}{\left(x+2\right)^{2}} have the same denominator, subtract them by subtracting their numerators.

$$\frac{x^{3}+6x^{2}+9x+4-x\left(x+2\right)^{2}}{\left(x+2\right)^{2}}=2$$

Do the multiplications in x^{3}+6x^{2}+9x+4-x\left(x+2\right)^{2}.

$$\frac{x^{3}+6x^{2}+9x+4-x^{3}-4x^{2}-4x}{\left(x+2\right)^{2}}=2$$

Combine like terms in x^{3}+6x^{2}+9x+4-x^{3}-4x^{2}-4x.

$$\frac{2x^{2}+5x+4}{\left(x+2\right)^{2}}=2$$

Variable x cannot be equal to -2 since division by zero is not defined. Multiply both sides of the equation by \left(x+2\right)^{2}.

$$2x^{2}+5x+4=2\left(x+2\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+2\right)^{2}.

$$2x^{2}+5x+4=2\left(x^{2}+4x+4\right)$$

Use the distributive property to multiply 2 by x^{2}+4x+4.

$$2x^{2}+5x+4=2x^{2}+8x+8$$

Subtract 2x^{2} from both sides.

$$2x^{2}+5x+4-2x^{2}=8x+8$$

Combine 2x^{2} and -2x^{2} to get 0.

$$5x+4=8x+8$$

Subtract 8x from both sides.

$$5x+4-8x=8$$

Combine 5x and -8x to get -3x.

$$-3x+4=8$$

Subtract 4 from both sides.

$$-3x=8-4$$

Subtract 4 from 8 to get 4.

$$-3x=4$$

Divide both sides by -3.

$$x=\frac{4}{-3}$$

Fraction \frac{4}{-3} can be rewritten as -\frac{4}{3} by extracting the negative sign.

$$x=-\frac{4}{3}$$