Solve r/y-2=3y+1/3y-4

$$\frac { r } { y - 2 } = \frac { 3 y + 1 } { 3 y - 4 }$$

Solve for r

$$r=-\frac{\left(2-y\right)\left(3y+1\right)}{3y-4}$$

$$y\neq 2\text{ and }y\neq \frac{4}{3}$$

Steps for Solving Linear Equation

Multiply both sides of the equation by \left(y-2\right)\left(3y-4\right), the least common multiple of y-2,3y-4.

$$\left(3y-4\right)r=\left(y-2\right)\left(3y+1\right)$$

Use the distributive property to multiply 3y-4 by r.

$$3yr-4r=\left(y-2\right)\left(3y+1\right)$$

Use the distributive property to multiply y-2 by 3y+1 and combine like terms.

$$3yr-4r=3y^{2}-5y-2$$

Combine all terms containing r.

$$\left(3y-4\right)r=3y^{2}-5y-2$$

Divide both sides by 3y-4.

$$\frac{\left(3y-4\right)r}{3y-4}=\frac{\left(y-2\right)\left(3y+1\right)}{3y-4}$$

Dividing by 3y-4 undoes the multiplication by 3y-4.

$$r=\frac{\left(y-2\right)\left(3y+1\right)}{3y-4}$$

Solve for y

$$\left\{\begin{matrix}\\y=-\frac{\sqrt{9r^{2}-18r+49}}{6}+\frac{r}{2}+\frac{5}{6}\text{, }&\text{unconditionally}\\y=\frac{\sqrt{9r^{2}-18r+49}}{6}+\frac{r}{2}+\frac{5}{6}\text{, }&r\neq 0\end{matrix}\right.$$