Solve {l}{x=sqrt{2}-1}{y={(2x+1)}{(2x-1)}}{z=-{(x+1)}{(3x-2)}}{text{Solvefor}a,btext{where}}{a=y}{b=z}

$$\left. \begin{array} { l } { x = \sqrt{2} - 1 }\\ { y = {(2 x + 1)} {(2 x - 1)} }\\ { z = -{(x + 1)} {(3 x - 2)} }\\ { \text{Solve for } a,b \text{ where} } \\ { a = y }\\ { b = z } \end{array} \right.$$

Solve for x, y, z, a, b

$$a=11-8\sqrt{2}\approx -0.313708499$$

$$b=5\sqrt{2}-6\approx 1.071067812$$

Solution Steps

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

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

Consider \left(2\left(\sqrt{2}-1\right)+1\right)\left(2\left(\sqrt{2}-1\right)-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}. Square 1.

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

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

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

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

$$y=4\left(\sqrt{2}\right)^{2}-8\sqrt{2}+4-1$$

The square of \sqrt{2} is 2.

$$y=4\times 2-8\sqrt{2}+4-1$$

Multiply 4 and 2 to get 8.

$$y=8-8\sqrt{2}+4-1$$

Add 8 and 4 to get 12.

$$y=12-8\sqrt{2}-1$$

Subtract 1 from 12 to get 11.

$$y=11-8\sqrt{2}$$

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

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

Add -1 and 1 to get 0.

$$z=\left(-\sqrt{2}\right)\left(3\left(\sqrt{2}-1\right)-2\right)$$

Use the distributive property to multiply 3 by \sqrt{2}-1.

$$z=\left(-\sqrt{2}\right)\left(3\sqrt{2}-3-2\right)$$

Subtract 2 from -3 to get -5.

$$z=\left(-\sqrt{2}\right)\left(3\sqrt{2}-5\right)$$

Use the distributive property to multiply -\sqrt{2} by 3\sqrt{2}-5.

$$z=3\left(-\sqrt{2}\right)\sqrt{2}-5\left(-\sqrt{2}\right)$$

Multiply -5 and -1 to get 5.

$$z=3\left(-\sqrt{2}\right)\sqrt{2}+5\sqrt{2}$$

Multiply \sqrt{2} and \sqrt{2} to get 2.

$$z=3\left(-1\right)\times 2+5\sqrt{2}$$

Multiply 3 and -1 to get -3.

$$z=-3\times 2+5\sqrt{2}$$

Multiply -3 and 2 to get -6.

$$z=-6+5\sqrt{2}$$

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

$$a=11-8\sqrt{2}$$

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

$$b=-6+5\sqrt{2}$$

The system is now solved.

$$x=\sqrt{2}-1 y=11-8\sqrt{2} z=-6+5\sqrt{2} a=11-8\sqrt{2} b=-6+5\sqrt{2}$$