Solve A_{S}=q/2-frac{2gamma-3sqrt{18}}{2}

$$A _ { S } = \frac { q } { 2 } - \frac { 2 \gamma - 3 \sqrt { 18 } } { 2 }$$

Solve for q

$$q=2A_{S}+2\gamma -9\sqrt{2}$$

Steps for Solving Linear Equation

Factor 18=3^{2}\times 2. Rewrite the square root of the product \sqrt{3^{2}\times 2} as the product of square roots \sqrt{3^{2}}\sqrt{2}. Take the square root of 3^{2}.

$$A_{S}=\frac{q}{2}-\frac{2\gamma -3\times 3\sqrt{2}}{2}$$

Multiply -3 and 3 to get -9.

$$A_{S}=\frac{q}{2}-\frac{2\gamma -9\sqrt{2}}{2}$$

Since \frac{q}{2} and \frac{2\gamma -9\sqrt{2}}{2} have the same denominator, subtract them by subtracting their numerators.

$$A_{S}=\frac{q-\left(2\gamma -9\sqrt{2}\right)}{2}$$

Do the multiplications in q-\left(2\gamma -9\sqrt{2}\right).

$$A_{S}=\frac{q-2\gamma +9\sqrt{2}}{2}$$

Divide each term of q-2\gamma +9\sqrt{2} by 2 to get \frac{1}{2}q-\gamma +\frac{9}{2}\sqrt{2}.

$$A_{S}=\frac{1}{2}q-\gamma +\frac{9}{2}\sqrt{2}$$

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

$$\frac{1}{2}q-\gamma +\frac{9}{2}\sqrt{2}=A_{S}$$

Add \gamma to both sides.

$$\frac{1}{2}q+\frac{9}{2}\sqrt{2}=A_{S}+\gamma$$

Subtract \frac{9}{2}\sqrt{2} from both sides.

$$\frac{1}{2}q=A_{S}+\gamma -\frac{9}{2}\sqrt{2}$$

The equation is in standard form.

$$\frac{1}{2}q=A_{S}+\gamma -\frac{9\sqrt{2}}{2}$$

Multiply both sides by 2.

$$\frac{\frac{1}{2}q}{\frac{1}{2}}=\frac{A_{S}+\gamma -\frac{9\sqrt{2}}{2}}{\frac{1}{2}}$$

Dividing by \frac{1}{2} undoes the multiplication by \frac{1}{2}.

$$q=\frac{A_{S}+\gamma -\frac{9\sqrt{2}}{2}}{\frac{1}{2}}$$

Divide A_{S}+\gamma -\frac{9\sqrt{2}}{2} by \frac{1}{2} by multiplying A_{S}+\gamma -\frac{9\sqrt{2}}{2} by the reciprocal of \frac{1}{2}.

$$q=2A_{S}+2\gamma -9\sqrt{2}$$

Solve for A_S

$$A_{S}=\frac{q-2\gamma +9\sqrt{2}}{2}$$