Solve {c}{(1-2/x+1)divx^2-2x+1/x+1=y}{x=sqrt{2}+1}

$$\left. \begin{array} { c } { ( 1 - \frac { 2 } { x + 1 } ) \div \frac { x ^ { 2 } - 2 x + 1 } { x + 1 } = y } \\ { x = \sqrt { 2 } + 1 } \end{array} \right.$$

Solve for x, y

$$x=\sqrt{2}+1\approx 2.414213562$$

$$y=\frac{\sqrt{2}}{2}\approx 0.707106781$$

Solution Steps

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

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

Add 1 and 1 to get 2.

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

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

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

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

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

Square \sqrt{2}. Square 2.

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

Subtract 4 from 2 to get -2.

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

Cancel out -2 and -2.

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

The opposite of -\left(\sqrt{2}-2\right) is \sqrt{2}-2.

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

Subtract 2 from 1 to get -1.

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

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

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

The square of \sqrt{2} is 2.

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

Add 2 and 1 to get 3.

$$\frac{-1+\sqrt{2}}{\frac{3+2\sqrt{2}-2\left(\sqrt{2}+1\right)+1}{\sqrt{2}+1+1}}=y$$

Add 3 and 1 to get 4.

$$\frac{-1+\sqrt{2}}{\frac{4+2\sqrt{2}-2\left(\sqrt{2}+1\right)}{\sqrt{2}+1+1}}=y$$

Add 1 and 1 to get 2.

$$\frac{-1+\sqrt{2}}{\frac{4+2\sqrt{2}-2\left(\sqrt{2}+1\right)}{\sqrt{2}+2}}=y$$

Rationalize the denominator of \frac{4+2\sqrt{2}-2\left(\sqrt{2}+1\right)}{\sqrt{2}+2} by multiplying numerator and denominator by \sqrt{2}-2.

$$\frac{-1+\sqrt{2}}{\frac{\left(4+2\sqrt{2}-2\left(\sqrt{2}+1\right)\right)\left(\sqrt{2}-2\right)}{\left(\sqrt{2}+2\right)\left(\sqrt{2}-2\right)}}=y$$

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

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

Square \sqrt{2}. Square 2.

$$\frac{-1+\sqrt{2}}{\frac{\left(4+2\sqrt{2}-2\left(\sqrt{2}+1\right)\right)\left(\sqrt{2}-2\right)}{2-4}}=y$$

Subtract 4 from 2 to get -2.

$$\frac{-1+\sqrt{2}}{\frac{\left(4+2\sqrt{2}-2\left(\sqrt{2}+1\right)\right)\left(\sqrt{2}-2\right)}{-2}}=y$$

Divide -1+\sqrt{2} by \frac{\left(4+2\sqrt{2}-2\left(\sqrt{2}+1\right)\right)\left(\sqrt{2}-2\right)}{-2} by multiplying -1+\sqrt{2} by the reciprocal of \frac{\left(4+2\sqrt{2}-2\left(\sqrt{2}+1\right)\right)\left(\sqrt{2}-2\right)}{-2}.

$$\frac{\left(-1+\sqrt{2}\right)\left(-2\right)}{\left(4+2\sqrt{2}-2\left(\sqrt{2}+1\right)\right)\left(\sqrt{2}-2\right)}=y$$

Use the distributive property to multiply -1+\sqrt{2} by -2.

$$\frac{2-2\sqrt{2}}{\left(4+2\sqrt{2}-2\left(\sqrt{2}+1\right)\right)\left(\sqrt{2}-2\right)}=y$$

Use the distributive property to multiply -2 by \sqrt{2}+1.

$$\frac{2-2\sqrt{2}}{\left(4+2\sqrt{2}-2\sqrt{2}-2\right)\left(\sqrt{2}-2\right)}=y$$

Combine 2\sqrt{2} and -2\sqrt{2} to get 0.

$$\frac{2-2\sqrt{2}}{\left(4-2\right)\left(\sqrt{2}-2\right)}=y$$

Subtract 2 from 4 to get 2.

$$\frac{2-2\sqrt{2}}{2\left(\sqrt{2}-2\right)}=y$$

Use the distributive property to multiply 2 by \sqrt{2}-2.

$$\frac{2-2\sqrt{2}}{2\sqrt{2}-4}=y$$

Rationalize the denominator of \frac{2-2\sqrt{2}}{2\sqrt{2}-4} by multiplying numerator and denominator by 2\sqrt{2}+4.

$$\frac{\left(2-2\sqrt{2}\right)\left(2\sqrt{2}+4\right)}{\left(2\sqrt{2}-4\right)\left(2\sqrt{2}+4\right)}=y$$

Consider \left(2\sqrt{2}-4\right)\left(2\sqrt{2}+4\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{\left(2-2\sqrt{2}\right)\left(2\sqrt{2}+4\right)}{\left(2\sqrt{2}\right)^{2}-4^{2}}=y$$

Expand \left(2\sqrt{2}\right)^{2}.

$$\frac{\left(2-2\sqrt{2}\right)\left(2\sqrt{2}+4\right)}{2^{2}\left(\sqrt{2}\right)^{2}-4^{2}}=y$$

Calculate 2 to the power of 2 and get 4.

$$\frac{\left(2-2\sqrt{2}\right)\left(2\sqrt{2}+4\right)}{4\left(\sqrt{2}\right)^{2}-4^{2}}=y$$

The square of \sqrt{2} is 2.

$$\frac{\left(2-2\sqrt{2}\right)\left(2\sqrt{2}+4\right)}{4\times 2-4^{2}}=y$$

Multiply 4 and 2 to get 8.

$$\frac{\left(2-2\sqrt{2}\right)\left(2\sqrt{2}+4\right)}{8-4^{2}}=y$$

Calculate 4 to the power of 2 and get 16.

$$\frac{\left(2-2\sqrt{2}\right)\left(2\sqrt{2}+4\right)}{8-16}=y$$

Subtract 16 from 8 to get -8.

$$\frac{\left(2-2\sqrt{2}\right)\left(2\sqrt{2}+4\right)}{-8}=y$$

Swap sides so that all variable terms are on the left hand side.

$$y=\frac{\left(2-2\sqrt{2}\right)\left(2\sqrt{2}+4\right)}{-8}$$

Use the distributive property to multiply 2-2\sqrt{2} by 2\sqrt{2}+4 and combine like terms.

$$y=\frac{-4\sqrt{2}+8-4\left(\sqrt{2}\right)^{2}}{-8}$$

The square of \sqrt{2} is 2.

$$y=\frac{-4\sqrt{2}+8-4\times 2}{-8}$$

Multiply -4 and 2 to get -8.

$$y=\frac{-4\sqrt{2}+8-8}{-8}$$

Subtract 8 from 8 to get 0.

$$y=\frac{-4\sqrt{2}}{-8}$$

Divide -4\sqrt{2} by -8 to get \frac{1}{2}\sqrt{2}.

$$y=\frac{1}{2}\sqrt{2}$$

The system is now solved.

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