Solve left(3b-2bright)left(1-mright)^2=5

$$\left( 3b-2b \right) { \left(1-m \right) }^{ 2 } = 5$$

Solve for b

$$b=\frac{5}{\left(m-1\right)^{2}}$$

$$m\neq 1$$

Steps for Solving Linear Equation

Combine 3b and -2b to get b.

$$b\left(1-m\right)^{2}=5$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(1-m\right)^{2}.

$$b\left(1-2m+m^{2}\right)=5$$

Use the distributive property to multiply b by 1-2m+m^{2}.

$$b-2bm+bm^{2}=5$$

Combine all terms containing b.

$$\left(1-2m+m^{2}\right)b=5$$

The equation is in standard form.

$$\left(m^{2}-2m+1\right)b=5$$

Divide both sides by m^{2}-2m+1.

$$\frac{\left(m^{2}-2m+1\right)b}{m^{2}-2m+1}=\frac{5}{m^{2}-2m+1}$$

Dividing by m^{2}-2m+1 undoes the multiplication by m^{2}-2m+1.

$$b=\frac{5}{m^{2}-2m+1}$$

Divide 5 by m^{2}-2m+1.

$$b=\frac{5}{\left(m-1\right)^{2}}$$

Solve for m

$$m=1-\sqrt{\frac{5}{b}}$$

$$m=1+\sqrt{\frac{5}{b}}\text{, }b>0$$