Question

solve for y.
\\(\frac{2}{y - 3} = \frac{y}{3y - 10}\\)
there may be 1 or 2 solutions.
y = \boxed{} or y = \boxed{}

Explanation:

Step1: Cross - multiply the equation

Given the equation \(\frac{2}{y - 3}=\frac{y}{3y-10}\), cross - multiplying (if \(y
eq3\) and \(y
eq\frac{10}{3}\) to avoid division by zero) gives \(2(3y - 10)=y(y - 3)\).

Step2: Expand both sides

Expand the left - hand side: \(2\times3y-2\times10 = 6y-20\).
Expand the right - hand side: \(y\times y-y\times3=y^{2}-3y\).
So the equation becomes \(6y - 20=y^{2}-3y\).

Step3: Rearrange into standard quadratic form

Bring all terms to one side to get a quadratic equation: \(y^{2}-3y-6y + 20 = 0\), which simplifies to \(y^{2}-9y + 20=0\).

Step4: Factor the quadratic equation

We need to find two numbers that multiply to \(20\) and add up to \(-9\). The numbers are \(-4\) and \(-5\).
So, \(y^{2}-9y + 20=(y - 4)(y - 5)=0\).

Step5: Solve for y

Set each factor equal to zero:

• If \(y - 4=0\), then \(y = 4\).
• If \(y - 5=0\), then \(y = 5\).

We need to check if these solutions are valid (i.e., they do not make the original denominators zero).
For \(y = 4\): \(y-3=4 - 3=1
eq0\) and \(3y - 10=12 - 10 = 2
eq0\).
For \(y = 5\): \(y - 3=5 - 3 = 2
eq0\) and \(3y-10=15 - 10 = 5
eq0\).

Answer:

\(y = 4\) or \(y = 5\)