Solve operatorname{tai}x=frac{sqrt{2}-1}{sqrt{2}+1}

$$\operatorname { tai } x = \frac { \sqrt { 2 } - 1 } { \sqrt { 2 } + 1 }$$

Solve for a

$$a=\frac{\left(2\sqrt{2}-3\right)i}{tx}$$

$$x\neq 0\text{ and }t\neq 0$$

Steps for Solving Linear Equation

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

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

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}.

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

Square \sqrt{2}. Square 1.

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

Subtract 1 from 2 to get 1.

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

Anything divided by one gives itself.

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

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

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

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

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

The square of \sqrt{2} is 2.

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

Add 2 and 1 to get 3.

$$taix=3-2\sqrt{2}$$

The equation is in standard form.

$$itxa=3-2\sqrt{2}$$

Divide both sides by itx.

$$\frac{itxa}{itx}=\frac{3-2\sqrt{2}}{itx}$$

Dividing by itx undoes the multiplication by itx.

$$a=\frac{3-2\sqrt{2}}{itx}$$

Divide 3-2\sqrt{2} by itx.

$$a=-\frac{\left(3-2\sqrt{2}\right)i}{tx}$$

Solve for t

$$t=\frac{\left(2\sqrt{2}-3\right)i}{ax}$$

$$x\neq 0\text{ and }a\neq 0$$

Steps for Solving Linear Equation

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

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

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}.

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

Square \sqrt{2}. Square 1.

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

Subtract 1 from 2 to get 1.

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

Anything divided by one gives itself.

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

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

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

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

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

The square of \sqrt{2} is 2.

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

Add 2 and 1 to get 3.

$$taix=3-2\sqrt{2}$$

The equation is in standard form.

$$iaxt=3-2\sqrt{2}$$

Divide both sides by iax.

$$\frac{iaxt}{iax}=\frac{3-2\sqrt{2}}{iax}$$

Dividing by iax undoes the multiplication by iax.

$$t=\frac{3-2\sqrt{2}}{iax}$$

Divide 3-2\sqrt{2} by iax.

$$t=-\frac{\left(3-2\sqrt{2}\right)i}{ax}$$