4/9y2 - 4//3y+1=5/1-3y

To solve the equation \( \frac{4}{9}y^2 - \frac{4}{3}y + 1 = \frac{5}{1 - 3y} \), we need to clear the fractions and combine terms. Let's first eliminate the fraction on the right side by multiplying both sides of the equation by \( 1 - 3y \): \[ (1 - 3y) \left( \frac{4}{9}y^2 - \frac{4}{3}y + 1 \right) = 5 \] Now, distribute \( (1 - 3y) \) on the left side: \[ \frac{4}{9}y^2(1 - 3y) - \frac{4}{3}y(1 - 3y) + 1(1 - 3y) = 5 \] Distributing each term: 1. \( \frac{4}{9}y^2(1 - 3y) = \frac{4}{9}y^2 - \frac{12}{9}y^3 = \frac{4}{9}y^2 - \frac{4}{3}y^3 \) 2. \( -\frac{4}{3}y(1 - 3y) = -\frac{4}{3}y + 4y^2 \) 3. \( 1(1 - 3y) = 1 - 3y \) Combining these, we get: \[ \frac{4}{9}y^2 - \frac{4}{3}y^3 - \frac{4}{3}y + 4y^2 + 1 - 3y = 5 \] Now, convert all terms to a common denominator of 9 to make it easier to combine like terms: \[ -\frac{4}{3}y^3 + \left(\frac{4}{9}y^2 + \frac{36}{9}y^2\right) + 1 - \left(\frac{27}{9}y + \frac{12}{9}y\right) = 5 \] This simplifies to: \[ -\frac{4}{3}y^3 + \frac{40}{9}y^2 - \frac{39}{9}y + 1 = 5 \] Now, subtract 5 from both sides to set the equation to zero: \[ -\frac{4}{3}y^3 + \frac{40}{9}y^2 - \frac{39}{9}y + 1 - 5 = 0 \] This simplifies to: \[ -\frac{4}{3}y^3 + \frac{40}{9}y^2 - \frac{39}{9}y - 4 = 0 \] To eliminate the denominators, multiply through by 9: \[ -12y^3 + 40y^2 - 39y - 36 = 0 \] Now you can proceed to solve the cubic equation \( -12y^3 + 40y^2 - 39y - 36 = 0 \) either by using synthetic division, factoring (if possible), or numerical methods. This is the form of the equation you would need to solve to find the values of \( y \). Do you want to proceed with a specific method to solve this cubic equation?