Question - Solving an Algebraic Fraction Equation

The equation given is \( \frac{4}{2y-1} = 7 - \frac{3}{y} \)

To solve this equation, first find a common denominator for the fractions, which in this case would be \(y(2y-1)\). Then, write each term over the common denominator:

\( \frac{4y}{y(2y-1)} = \frac{7y(2y-1)}{y(2y-1)} - \frac{3(2y-1)}{y(2y-1)} \)

Now simplify the equation by combining terms over the common denominator:

\( \frac{4y}{y(2y-1)} = \frac{14y^2 - 7y - 6y + 3}{y(2y-1)} \)

Since the denominators are the same, we can set the numerators equal to each other:

\( 4y = 14y^2 - 13y + 3 \)

Rearrange the terms to form a quadratic equation:

\( 14y^2 - 13y + 3 - 4y = 0 \)

\( 14y^2 - 17y + 3 = 0 \)

Next, solve the quadratic equation, which might factor or could require the quadratic formula. Unfortunately, this quadratic equation does not factor nicely, so we use the quadratic formula. Since \(a=14\), \(b=-17\), and \(c=3\), we plug these into the quadratic formula:

\( y = \frac{-(-17) \pm \sqrt{(-17)^2 - 4 \cdot 14 \cdot 3}}{2 \cdot 14} \)

\( y = \frac{17 \pm \sqrt{289 - 168}}{28} \)

\( y = \frac{17 \pm \sqrt{121}}{28} \)

\( y = \frac{17 \pm 11}{28} \)

So we have two possible solutions:

\( y = \frac{17 + 11}{28} \) or \( y = \frac{17 - 11}{28} \)

\( y = \frac{28}{28} \) or \( y = \frac{6}{28} \)

\( y = 1 \) or \( y = \frac{3}{14} \)

Therefore, the solutions are \( y = 1 \) or \( y = \frac{3}{14} \).