Solve frac{b_{1}h_{1}}{2}+1.5h+1.5h+frac{bh_{2}}{2}=31.35

$$\frac { b _ { 1 } h _ { 1 } } { 2 } + 1.5 h + 1.5 h + \frac { b h _ { 2 } } { 2 } = 31.35$$

Solve for b

$$\left\{\begin{matrix}b=-\frac{10b_{1}h_{1}+60h-627}{10h_{2}}\text{, }&h_{2}\neq 0\\b\in \mathrm{R}\text{, }&h=-\frac{b_{1}h_{1}}{6}+\frac{209}{20}\text{ and }h_{2}=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Multiply both sides of the equation by 2.

$$b_{1}h_{1}+3h+3h+bh_{2}=62.7$$

Combine 3h and 3h to get 6h.

$$b_{1}h_{1}+6h+bh_{2}=62.7$$

Subtract b_{1}h_{1} from both sides.

$$6h+bh_{2}=62.7-b_{1}h_{1}$$

Subtract 6h from both sides.

$$bh_{2}=62.7-b_{1}h_{1}-6h$$

The equation is in standard form.

$$h_{2}b=62.7-6h-b_{1}h_{1}$$

Divide both sides by h_{2}.

$$\frac{h_{2}b}{h_{2}}=\frac{62.7-6h-b_{1}h_{1}}{h_{2}}$$

Dividing by h_{2} undoes the multiplication by h_{2}.

$$b=\frac{62.7-6h-b_{1}h_{1}}{h_{2}}$$

Solve for b_1

$$\left\{\begin{matrix}b_{1}=-\frac{10bh_{2}+60h-627}{10h_{1}}\text{, }&h_{1}\neq 0\\b_{1}\in \mathrm{R}\text{, }&h=-\frac{bh_{2}}{6}+\frac{209}{20}\text{ and }h_{1}=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Multiply both sides of the equation by 2.

$$b_{1}h_{1}+3h+3h+bh_{2}=62.7$$

Combine 3h and 3h to get 6h.

$$b_{1}h_{1}+6h+bh_{2}=62.7$$

Subtract 6h from both sides.

$$b_{1}h_{1}+bh_{2}=62.7-6h$$

Subtract bh_{2} from both sides.

$$b_{1}h_{1}=62.7-6h-bh_{2}$$

The equation is in standard form.

$$h_{1}b_{1}=62.7-6h-bh_{2}$$

Divide both sides by h_{1}.

$$\frac{h_{1}b_{1}}{h_{1}}=\frac{62.7-6h-bh_{2}}{h_{1}}$$

Dividing by h_{1} undoes the multiplication by h_{1}.

$$b_{1}=\frac{62.7-6h-bh_{2}}{h_{1}}$$