Solve (f+1)(3-2r)+(2r+1)(2r-1)+(2r+2)(1-3)=0

$$( f + 1 ) ( 3 - 2 r ) + ( 2 r + 1 ) ( 2 r - 1 ) + ( 2 r + 2 ) ( 1 - 3 ) = 0$$

Solve for f

$$f=-\frac{2\left(2r^{2}-3r-1\right)}{3-2r}$$

$$r\neq \frac{3}{2}$$

Steps for Solving Linear Equation

Use the distributive property to multiply f+1 by 3-2r.

$$3f-2fr+3-2r+\left(2r+1\right)\left(2r-1\right)+\left(2r+2\right)\left(1-3\right)=0$$

Consider \left(2r+1\right)\left(2r-1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 1.

$$3f-2fr+3-2r+\left(2r\right)^{2}-1+\left(2r+2\right)\left(1-3\right)=0$$

Expand \left(2r\right)^{2}.

$$3f-2fr+3-2r+2^{2}r^{2}-1+\left(2r+2\right)\left(1-3\right)=0$$

Calculate 2 to the power of 2 and get 4.

$$3f-2fr+3-2r+4r^{2}-1+\left(2r+2\right)\left(1-3\right)=0$$

Subtract 1 from 3 to get 2.

$$3f-2fr+2-2r+4r^{2}+\left(2r+2\right)\left(1-3\right)=0$$

Subtract 3 from 1 to get -2.

$$3f-2fr+2-2r+4r^{2}+\left(2r+2\right)\left(-2\right)=0$$

Use the distributive property to multiply 2r+2 by -2.

$$3f-2fr+2-2r+4r^{2}-4r-4=0$$

Combine -2r and -4r to get -6r.

$$3f-2fr+2-6r+4r^{2}-4=0$$

Subtract 4 from 2 to get -2.

$$3f-2fr-2-6r+4r^{2}=0$$

Add 2 to both sides. Anything plus zero gives itself.

$$3f-2fr-6r+4r^{2}=2$$

Add 6r to both sides.

$$3f-2fr+4r^{2}=2+6r$$

Subtract 4r^{2} from both sides.

$$3f-2fr=2+6r-4r^{2}$$

Combine all terms containing f.

$$\left(3-2r\right)f=2+6r-4r^{2}$$

Divide both sides by 3-2r.

$$\frac{\left(3-2r\right)f}{3-2r}=\frac{2+6r-4r^{2}}{3-2r}$$

Dividing by 3-2r undoes the multiplication by 3-2r.

$$f=\frac{2+6r-4r^{2}}{3-2r}$$

Divide 2+6r-4r^{2} by 3-2r.

$$f=\frac{2\left(1+3r-2r^{2}\right)}{3-2r}$$

Solve for r

$$r=\frac{\sqrt{f^{2}-6f+17}+f+3}{4}$$

$$r=\frac{-\sqrt{f^{2}-6f+17}+f+3}{4}$$