Solve 1/2ak=1/2(a+3)(h-3)

$$\frac { 1 } { 2 } a k = \frac { 1 } { 2 } ( a + 3 ) ( h - 3 )$$

Solve for a

$$\left\{\begin{matrix}a=-\frac{3\left(3-h\right)}{3+k-h}\text{, }&k\neq h-3\\a\in \mathrm{R}\text{, }&h=3\text{ and }k=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Cancel out \frac{1}{2} on both sides.

$$ak=\left(a+3\right)\left(h-3\right)$$

Use the distributive property to multiply a+3 by h-3.

$$ak=ah-3a+3h-9$$

Subtract ah from both sides.

$$ak-ah=-3a+3h-9$$

Add 3a to both sides.

$$ak-ah+3a=3h-9$$

Combine all terms containing a.

$$\left(k-h+3\right)a=3h-9$$

The equation is in standard form.

$$\left(3+k-h\right)a=3h-9$$

Divide both sides by k+3-h.

$$\frac{\left(3+k-h\right)a}{3+k-h}=\frac{3h-9}{3+k-h}$$

Dividing by k+3-h undoes the multiplication by k+3-h.

$$a=\frac{3h-9}{3+k-h}$$

Divide -9+3h by k+3-h.

$$a=\frac{3\left(h-3\right)}{3+k-h}$$

Solve for h

$$\left\{\begin{matrix}h=\frac{ak+3a+9}{a+3}\text{, }&a\neq -3\\h\in \mathrm{R}\text{, }&k=0\text{ and }a=-3\end{matrix}\right.$$

Steps for Solving Linear Equation

Cancel out \frac{1}{2} on both sides.

$$ak=\left(a+3\right)\left(h-3\right)$$

Use the distributive property to multiply a+3 by h-3.

$$ak=ah-3a+3h-9$$

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

$$ah-3a+3h-9=ak$$

Add 3a to both sides.

$$ah+3h-9=ak+3a$$

Add 9 to both sides.

$$ah+3h=ak+3a+9$$

Combine all terms containing h.

$$\left(a+3\right)h=ak+3a+9$$

Divide both sides by a+3.

$$\frac{\left(a+3\right)h}{a+3}=\frac{ak+3a+9}{a+3}$$

Dividing by a+3 undoes the multiplication by a+3.

$$h=\frac{ak+3a+9}{a+3}$$