Solve {l}{1/1+1{1+frac{1/x}}=2}{y=x}{text{Solvefor}ztext{where}}{z=y}

$$\left. \begin{array} { l } { \frac{1}{1 + \frac{1}{1 + \frac{1}{x}}} = 2 }\\ { y = x }\\ { \text{Solve for } z \text{ where} } \\ { z = y } \end{array} \right.$$

Solve for x, y, z

$$z=-\frac{1}{3}\approx -0.333333333$$

Short Solution Steps

Consider the first equation. To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{x}{x}.

$$\frac{1}{1+\frac{1}{\frac{x}{x}+\frac{1}{x}}}=2$$

Since \frac{x}{x} and \frac{1}{x} have the same denominator, add them by adding their numerators.

$$\frac{1}{1+\frac{1}{\frac{x+1}{x}}}=2$$

Variable x cannot be equal to 0 since division by zero is not defined. Divide 1 by \frac{x+1}{x} by multiplying 1 by the reciprocal of \frac{x+1}{x}.

$$\frac{1}{1+\frac{x}{x+1}}=2$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{x+1}{x+1}.

$$\frac{1}{\frac{x+1}{x+1}+\frac{x}{x+1}}=2$$

Since \frac{x+1}{x+1} and \frac{x}{x+1} have the same denominator, add them by adding their numerators.

$$\frac{1}{\frac{x+1+x}{x+1}}=2$$

Combine like terms in x+1+x.

$$\frac{1}{\frac{2x+1}{x+1}}=2$$

Variable x cannot be equal to -1 since division by zero is not defined. Divide 1 by \frac{2x+1}{x+1} by multiplying 1 by the reciprocal of \frac{2x+1}{x+1}.

$$\frac{x+1}{2x+1}=2$$

Variable x cannot be equal to -\frac{1}{2} since division by zero is not defined. Multiply both sides of the equation by 2x+1.

$$x+1=2\left(2x+1\right)$$

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

$$x+1=4x+2$$

Subtract 4x from both sides.

$$x+1-4x=2$$

Combine x and -4x to get -3x.

$$-3x+1=2$$

Subtract 1 from both sides.

$$-3x=2-1$$

Subtract 1 from 2 to get 1.

$$-3x=1$$

Divide both sides by -3.

$$x=-\frac{1}{3}$$

Consider the second equation. Insert the known values of variables into the equation.

$$y=-\frac{1}{3}$$

Consider the third equation. Insert the known values of variables into the equation.

$$z=-\frac{1}{3}$$

The system is now solved.

$$x=-\frac{1}{3} y=-\frac{1}{3} z=-\frac{1}{3}$$