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

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

Solve for x, y, z

$$z=\sqrt{2}\approx 1.414213562$$

Solution Steps

Consider the first equation. Rationalize the denominator of \frac{1}{\sqrt{2}+1} by multiplying numerator and denominator by \sqrt{2}-1.

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

Consider \left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

Square \sqrt{2}. Square 1.

$$x=\frac{\sqrt{2}-1}{2-1}$$

Subtract 1 from 2 to get 1.

$$x=\frac{\sqrt{2}-1}{1}$$

Anything divided by one gives itself.

$$x=\sqrt{2}-1$$

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

$$y=\sqrt{2}-1+1$$

Add -1 and 1 to get 0.

$$y=\sqrt{2}$$

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

$$z=\sqrt{2}$$

The system is now solved.

$$x=\sqrt{2}-1 y=\sqrt{2} z=\sqrt{2}$$