Solve 20^2-ab-2b^2=39-3^b+3

$$20 ^ { 2 } - a b - 2 b ^ { 2 } = 39 - 3 ^ { b + 3 }$$

Solve for a

$$a=\frac{361+27\times 3^{b}-2b^{2}}{b}$$

$$b\neq 0$$

Steps for Solving Linear Equation

Calculate 20 to the power of 2 and get 400.

$$400-ab-2b^{2}=39-3^{b+3}$$

Subtract 400 from both sides.

$$-ab-2b^{2}=39-3^{b+3}-400$$

Add 2b^{2} to both sides.

$$-ab=39-3^{b+3}-400+2b^{2}$$

Subtract 400 from 39 to get -361.

$$-ab=-361-3^{b+3}+2b^{2}$$

The equation is in standard form.

$$\left(-b\right)a=2b^{2}-3^{b+3}-361$$

Divide both sides by -b.

$$\frac{\left(-b\right)a}{-b}=\frac{2b^{2}-27\times 3^{b}-361}{-b}$$

Dividing by -b undoes the multiplication by -b.

$$a=\frac{2b^{2}-27\times 3^{b}-361}{-b}$$

Divide -361-27\times 3^{b}+2b^{2} by -b.

$$a=\frac{27\times 3^{b}+361}{b}-2b$$