1. **State the problem:** Solve the equation $$\frac{y+2}{y-1} - \frac{4-y}{2y} = \frac{7}{3}$$ for $y$. 2. **Identify the common denominator:** The denominators are $y-1$, $2y$, and $3$. The least common denominator (LCD) is $$3 \times 2y \times (y-1) = 6y(y-1).$$ 3. **Multiply both sides by the LCD to clear denominators:** $$6y(y-1) \times \left(\frac{y+2}{y-1} - \frac{4-y}{2y}\right) = 6y(y-1) \times \frac{7}{3}$$ 4. **Simplify each term:** - First term: $$6y(y-1) \times \frac{y+2}{y-1} = 6y(y+2)$$ - Second term: $$6y(y-1) \times \frac{4-y}{2y} = 3(y-1)(4-y)$$ - Right side: $$6y(y-1) \times \frac{7}{3} = 14y(y-1)$$ 5. **Rewrite the equation:** $$6y(y+2) - 3(y-1)(4-y) = 14y(y-1)$$ 6. **Expand terms:** - $$6y(y+2) = 6y^2 + 12y$$ - $$3(y-1)(4-y) = 3[(y)(4) - y^2 - 4 + y] = 3(4y - y^2 -4 + y) = 3(5y - y^2 -4) = 15y - 3y^2 -12$$ - $$14y(y-1) = 14y^2 - 14y$$ 7. **Substitute expansions:** $$6y^2 + 12y - (15y - 3y^2 - 12) = 14y^2 - 14y$$ 8. **Simplify left side:** $$6y^2 + 12y - 15y + 3y^2 + 12 = 14y^2 - 14y$$ $$ (6y^2 + 3y^2) + (12y - 15y) + 12 = 14y^2 - 14y$$ $$9y^2 - 3y + 12 = 14y^2 - 14y$$ 9. **Bring all terms to one side:** $$9y^2 - 3y + 12 - 14y^2 + 14y = 0$$ $$-5y^2 + 11y + 12 = 0$$ 10. **Multiply entire equation by -1 to simplify:** $$5y^2 - 11y - 12 = 0$$ 11. **Use quadratic formula:** $$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=5$, $b=-11$, $c=-12$. 12. **Calculate discriminant:** $$\Delta = (-11)^2 - 4 \times 5 \times (-12) = 121 + 240 = 361$$ 13. **Calculate roots:** $$y = \frac{11 \pm \sqrt{361}}{10} = \frac{11 \pm 19}{10}$$ 14. **Find two solutions:** - $$y_1 = \frac{11 + 19}{10} = \frac{30}{10} = 3$$ - $$y_2 = \frac{11 - 19}{10} = \frac{-8}{10} = -\frac{4}{5}$$ 15. **Check for restrictions:** - Denominators cannot be zero: $y \neq 1$ and $y \neq 0$. - Both $3$ and $-\frac{4}{5}$ are valid. **Final answer:** $$y = 3 \text{ or } y = -\frac{4}{5}$$