Solve {l}{z=x^2y(4-x-y)}{x=sqrt{5}-1}{y=frac{5-sqrt{5}}{2}}

$$\left\{ \begin{array} { l } { z = x ^ { 2 } y ( 4 - x - y ) } \\ { x = \sqrt { 5 } - 1 } \\ { y = \frac { 5 - \sqrt { 5 } } { 2 } } \end{array} \right.$$

Solve for z, x, y

$$x=\sqrt{5}-1\approx 1.236067977$$

$$y = \frac{5 - \sqrt{5}}{2} \approx 1.381966011$$

$$z=70-30\sqrt{5}\approx 2.917960675$$

Solution Steps

Consider the third equation. Divide each term of 5-\sqrt{5} by 2 to get \frac{5}{2}-\frac{1}{2}\sqrt{5}.

$$y=\frac{5}{2}-\frac{1}{2}\sqrt{5}$$

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

$$z=\left(\sqrt{5}-1\right)^{2}\left(\frac{5}{2}-\frac{1}{2}\sqrt{5}\right)\left(4-\left(\sqrt{5}-1\right)-\left(\frac{5}{2}-\frac{1}{2}\sqrt{5}\right)\right)$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{5}-1\right)^{2}.

$$z=\left(\left(\sqrt{5}\right)^{2}-2\sqrt{5}+1\right)\left(\frac{5}{2}-\frac{1}{2}\sqrt{5}\right)\left(4-\left(\sqrt{5}-1\right)-\left(\frac{5}{2}-\frac{1}{2}\sqrt{5}\right)\right)$$

The square of \sqrt{5} is 5.

$$z=\left(5-2\sqrt{5}+1\right)\left(\frac{5}{2}-\frac{1}{2}\sqrt{5}\right)\left(4-\left(\sqrt{5}-1\right)-\left(\frac{5}{2}-\frac{1}{2}\sqrt{5}\right)\right)$$

Add 5 and 1 to get 6.

$$z=\left(6-2\sqrt{5}\right)\left(\frac{5}{2}-\frac{1}{2}\sqrt{5}\right)\left(4-\left(\sqrt{5}-1\right)-\left(\frac{5}{2}-\frac{1}{2}\sqrt{5}\right)\right)$$

To find the opposite of \sqrt{5}-1, find the opposite of each term.

$$z=\left(6-2\sqrt{5}\right)\left(\frac{5}{2}-\frac{1}{2}\sqrt{5}\right)\left(4-\sqrt{5}+1-\left(\frac{5}{2}-\frac{1}{2}\sqrt{5}\right)\right)$$

Add 4 and 1 to get 5.

$$z=\left(6-2\sqrt{5}\right)\left(\frac{5}{2}-\frac{1}{2}\sqrt{5}\right)\left(5-\sqrt{5}-\left(\frac{5}{2}-\frac{1}{2}\sqrt{5}\right)\right)$$

To find the opposite of \frac{5}{2}-\frac{1}{2}\sqrt{5}, find the opposite of each term.

$$z=\left(6-2\sqrt{5}\right)\left(\frac{5}{2}-\frac{1}{2}\sqrt{5}\right)\left(5-\sqrt{5}-\frac{5}{2}+\frac{1}{2}\sqrt{5}\right)$$

Subtract \frac{5}{2} from 5 to get \frac{5}{2}.

$$z=\left(6-2\sqrt{5}\right)\left(\frac{5}{2}-\frac{1}{2}\sqrt{5}\right)\left(\frac{5}{2}-\sqrt{5}+\frac{1}{2}\sqrt{5}\right)$$

Combine -\sqrt{5} and \frac{1}{2}\sqrt{5} to get -\frac{1}{2}\sqrt{5}.

$$z=\left(6-2\sqrt{5}\right)\left(\frac{5}{2}-\frac{1}{2}\sqrt{5}\right)\left(\frac{5}{2}-\frac{1}{2}\sqrt{5}\right)$$

Multiply \frac{5}{2}-\frac{1}{2}\sqrt{5} and \frac{5}{2}-\frac{1}{2}\sqrt{5} to get \left(\frac{5}{2}-\frac{1}{2}\sqrt{5}\right)^{2}.

$$z=\left(6-2\sqrt{5}\right)\left(\frac{5}{2}-\frac{1}{2}\sqrt{5}\right)^{2}$$

Use the distributive property to multiply 6-2\sqrt{5} by \left(\frac{5}{2}-\frac{1}{2}\sqrt{5}\right)^{2}.

$$z=6\left(\frac{5}{2}-\frac{1}{2}\sqrt{5}\right)^{2}-2\sqrt{5}\left(\frac{5}{2}-\frac{1}{2}\sqrt{5}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\frac{5}{2}-\frac{1}{2}\sqrt{5}\right)^{2}.

$$z=6\left(\frac{25}{4}-\frac{5}{2}\sqrt{5}+\frac{1}{4}\left(\sqrt{5}\right)^{2}\right)-2\sqrt{5}\left(\frac{5}{2}-\frac{1}{2}\sqrt{5}\right)^{2}$$

The square of \sqrt{5} is 5.

$$z=6\left(\frac{25}{4}-\frac{5}{2}\sqrt{5}+\frac{1}{4}\times 5\right)-2\sqrt{5}\left(\frac{5}{2}-\frac{1}{2}\sqrt{5}\right)^{2}$$

Multiply \frac{1}{4} and 5 to get \frac{5}{4}.

$$z=6\left(\frac{25}{4}-\frac{5}{2}\sqrt{5}+\frac{5}{4}\right)-2\sqrt{5}\left(\frac{5}{2}-\frac{1}{2}\sqrt{5}\right)^{2}$$

Add \frac{25}{4} and \frac{5}{4} to get \frac{15}{2}.

$$z=6\left(\frac{15}{2}-\frac{5}{2}\sqrt{5}\right)-2\sqrt{5}\left(\frac{5}{2}-\frac{1}{2}\sqrt{5}\right)^{2}$$

Use the distributive property to multiply 6 by \frac{15}{2}-\frac{5}{2}\sqrt{5}.

$$z=45-15\sqrt{5}-2\sqrt{5}\left(\frac{5}{2}-\frac{1}{2}\sqrt{5}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\frac{5}{2}-\frac{1}{2}\sqrt{5}\right)^{2}.

$$z=45-15\sqrt{5}-2\sqrt{5}\left(\frac{25}{4}-\frac{5}{2}\sqrt{5}+\frac{1}{4}\left(\sqrt{5}\right)^{2}\right)$$

The square of \sqrt{5} is 5.

$$z=45-15\sqrt{5}-2\sqrt{5}\left(\frac{25}{4}-\frac{5}{2}\sqrt{5}+\frac{1}{4}\times 5\right)$$

Multiply \frac{1}{4} and 5 to get \frac{5}{4}.

$$z=45-15\sqrt{5}-2\sqrt{5}\left(\frac{25}{4}-\frac{5}{2}\sqrt{5}+\frac{5}{4}\right)$$

Add \frac{25}{4} and \frac{5}{4} to get \frac{15}{2}.

$$z=45-15\sqrt{5}-2\sqrt{5}\left(\frac{15}{2}-\frac{5}{2}\sqrt{5}\right)$$

Use the distributive property to multiply -2\sqrt{5} by \frac{15}{2}-\frac{5}{2}\sqrt{5}.

$$z=45-15\sqrt{5}-15\sqrt{5}+5\left(\sqrt{5}\right)^{2}$$

The square of \sqrt{5} is 5.

$$z=45-15\sqrt{5}-15\sqrt{5}+5\times 5$$

Multiply 5 and 5 to get 25.

$$z=45-15\sqrt{5}-15\sqrt{5}+25$$

Combine -15\sqrt{5} and -15\sqrt{5} to get -30\sqrt{5}.

$$z=45-30\sqrt{5}+25$$

Add 45 and 25 to get 70.

$$z=70-30\sqrt{5}$$

The system is now solved.

$$z=70-30\sqrt{5} x=\sqrt{5}-1 y=\frac{5}{2}-\frac{1}{2}\sqrt{5}$$