Solve operatorname{arcs}=frac{3+sqrt{2}}{1+2sqrt{2}}

$$\operatorname { arcs } = \frac { 3 + \sqrt { 2 } } { 1 + 2 \sqrt { 2 } }$$

Solve for a

$$a=\frac{5\sqrt{2}+1}{7crs}$$

$$s\neq 0\text{ and }c\neq 0\text{ and }r\neq 0$$

Steps for Solving Linear Equation

Rationalize the denominator of \frac{3+\sqrt{2}}{1+2\sqrt{2}} by multiplying numerator and denominator by 1-2\sqrt{2}.

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

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

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

Calculate 1 to the power of 2 and get 1.

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

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

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

Calculate 2 to the power of 2 and get 4.

$$arcs=\frac{\left(3+\sqrt{2}\right)\left(1-2\sqrt{2}\right)}{1-4\left(\sqrt{2}\right)^{2}}$$

The square of \sqrt{2} is 2.

$$arcs=\frac{\left(3+\sqrt{2}\right)\left(1-2\sqrt{2}\right)}{1-4\times 2}$$

Multiply 4 and 2 to get 8.

$$arcs=\frac{\left(3+\sqrt{2}\right)\left(1-2\sqrt{2}\right)}{1-8}$$

Subtract 8 from 1 to get -7.

$$arcs=\frac{\left(3+\sqrt{2}\right)\left(1-2\sqrt{2}\right)}{-7}$$

Use the distributive property to multiply 3+\sqrt{2} by 1-2\sqrt{2} and combine like terms.

$$arcs=\frac{3-5\sqrt{2}-2\left(\sqrt{2}\right)^{2}}{-7}$$

The square of \sqrt{2} is 2.

$$arcs=\frac{3-5\sqrt{2}-2\times 2}{-7}$$

Multiply -2 and 2 to get -4.

$$arcs=\frac{3-5\sqrt{2}-4}{-7}$$

Subtract 4 from 3 to get -1.

$$arcs=\frac{-1-5\sqrt{2}}{-7}$$

Multiply both numerator and denominator by -1.

$$arcs=\frac{1+5\sqrt{2}}{7}$$

Divide each term of 1+5\sqrt{2} by 7 to get \frac{1}{7}+\frac{5}{7}\sqrt{2}.

$$arcs=\frac{1}{7}+\frac{5}{7}\sqrt{2}$$

The equation is in standard form.

$$crsa=\frac{5\sqrt{2}+1}{7}$$

Divide both sides by rcs.

$$\frac{crsa}{crs}=\frac{5\sqrt{2}+1}{7crs}$$

Dividing by rcs undoes the multiplication by rcs.

$$a=\frac{5\sqrt{2}+1}{7crs}$$

Solve for c

$$c=\frac{5\sqrt{2}+1}{7ars}$$

$$s\neq 0\text{ and }r\neq 0\text{ and }a\neq 0$$

Steps for Solving Linear Equation

Rationalize the denominator of \frac{3+\sqrt{2}}{1+2\sqrt{2}} by multiplying numerator and denominator by 1-2\sqrt{2}.

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

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

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

Calculate 1 to the power of 2 and get 1.

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

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

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

Calculate 2 to the power of 2 and get 4.

$$arcs=\frac{\left(3+\sqrt{2}\right)\left(1-2\sqrt{2}\right)}{1-4\left(\sqrt{2}\right)^{2}}$$

The square of \sqrt{2} is 2.

$$arcs=\frac{\left(3+\sqrt{2}\right)\left(1-2\sqrt{2}\right)}{1-4\times 2}$$

Multiply 4 and 2 to get 8.

$$arcs=\frac{\left(3+\sqrt{2}\right)\left(1-2\sqrt{2}\right)}{1-8}$$

Subtract 8 from 1 to get -7.

$$arcs=\frac{\left(3+\sqrt{2}\right)\left(1-2\sqrt{2}\right)}{-7}$$

Use the distributive property to multiply 3+\sqrt{2} by 1-2\sqrt{2} and combine like terms.

$$arcs=\frac{3-5\sqrt{2}-2\left(\sqrt{2}\right)^{2}}{-7}$$

The square of \sqrt{2} is 2.

$$arcs=\frac{3-5\sqrt{2}-2\times 2}{-7}$$

Multiply -2 and 2 to get -4.

$$arcs=\frac{3-5\sqrt{2}-4}{-7}$$

Subtract 4 from 3 to get -1.

$$arcs=\frac{-1-5\sqrt{2}}{-7}$$

Multiply both numerator and denominator by -1.

$$arcs=\frac{1+5\sqrt{2}}{7}$$

Divide each term of 1+5\sqrt{2} by 7 to get \frac{1}{7}+\frac{5}{7}\sqrt{2}.

$$arcs=\frac{1}{7}+\frac{5}{7}\sqrt{2}$$

The equation is in standard form.

$$arsc=\frac{5\sqrt{2}+1}{7}$$

Divide both sides by ars.

$$\frac{arsc}{ars}=\frac{5\sqrt{2}+1}{7ars}$$

Dividing by ars undoes the multiplication by ars.

$$c=\frac{5\sqrt{2}+1}{7ars}$$