Question

solve for y.
\\(\frac{2}{y - 7} = \frac{-y}{2y - 14}\\)
there may be 1 or 2 solutions.
y = \boxed{} or y = \boxed{}

Explanation:

Step1: Simplify the denominator

Notice that \(2y - 14 = 2(y - 7)\). So the equation \(\frac{2}{y - 7}=\frac{-y}{2y - 14}\) can be rewritten as \(\frac{2}{y - 7}=\frac{-y}{2(y - 7)}\).

Step2: Multiply both sides by \(2(y - 7)\) (note \(y

eq7\))
Multiply each side by \(2(y - 7)\) to eliminate the denominators: \(2\times2=-y\).

Step3: Solve for \(y\)

Simplify the left side: \(4 = -y\), so \(y=-4\). We also need to check if \(y = 7\) is a solution, but plugging \(y = 7\) into the original equation makes the denominators zero, so it's not valid. So the only solution is \(y=-4\) (wait, but let's re - check the steps. Wait, when we multiplied both sides by \(2(y - 7)\), we have \(2\times2=-y\)? Wait, no: \(\frac{2}{y - 7}\times2(y - 7)=2\times2 = 4\), and \(\frac{-y}{2(y - 7)}\times2(y - 7)=-y\). So \(4=-y\), so \(y = - 4\). And since \(y=-4\) does not make the denominators zero (\(y-7=-4 - 7=-11
eq0\), \(2y-14=2\times(-4)-14=-8 - 14=-22
eq0\)), it is a valid solution. Wait, but the problem said there may be 1 or 2 solutions. Wait, maybe I made a mistake in the first step. Let's start over.

Original equation: \(\frac{2}{y - 7}=\frac{-y}{2y - 14}\)

Factor the denominator on the right: \(2y-14 = 2(y - 7)\), so the equation is \(\frac{2}{y - 7}=\frac{-y}{2(y - 7)}\)

Multiply both sides by \(2(y - 7)\) (assuming \(y
eq7\)):

\(2\times2=-y\)

\(4=-y\)

\(y = - 4\)

Wait, but let's check the original equation with \(y=-4\):

Left side: \(\frac{2}{-4 - 7}=\frac{2}{-11}=-\frac{2}{11}\)

Right side: \(\frac{-(-4)}{2\times(-4)-14}=\frac{4}{-8 - 14}=\frac{4}{-22}=-\frac{2}{11}\)

So it works. And \(y = 7\) is extraneous. So the solution is \(y=-4\). Wait, but the problem has two boxes, maybe I made a mistake. Wait, let's do cross - multiplication instead.

Cross - multiply: \(2(2y - 14)=-y(y - 7)\)

Expand both sides: \(4y-28=-y^{2}+7y\)

Bring all terms to one side: \(y^{2}-3y - 28 = 0\)

Factor the quadratic: \(y^{2}-3y - 28=(y - 7)(y + 4)=0\)

So \(y - 7 = 0\) or \(y+4 = 0\), so \(y = 7\) or \(y=-4\). But \(y = 7\) makes the denominators of the original equation zero (since \(y - 7 = 0\) and \(2y-14 = 0\) when \(y = 7\)), so \(y = 7\) is an extraneous solution. So the only valid solution is \(y=-4\). Wait, but when we did cross - multiplication, we got a quadratic equation. Let's check the cross - multiplication step again.

Cross - multiplying \(\frac{2}{y - 7}=\frac{-y}{2y - 14}\) gives \(2(2y - 14)=-y(y - 7)\)

Left side: \(4y-28\)

Right side: \(-y^{2}+7y\)

Bring all terms to left: \(4y-28 + y^{2}-7y=0\)

\(y^{2}-3y - 28 = 0\)

Factor: We need two numbers that multiply to \(-28\) and add to \(-3\). The numbers are \(-7\) and \(4\). So \((y - 7)(y + 4)=0\)

So \(y = 7\) or \(y=-4\). But \(y = 7\) is not in the domain (since it makes the denominators zero), so the only solution is \(y=-4\). Wait, but the problem has two boxes. Maybe the problem had a typo, or maybe I messed up. Wait, let's check the original equation again. If we consider that when we multiply both sides by \(2(y - 7)\), we have to note that \(y
eq7\), but when we solve the quadratic, we get \(y = 7\) and \(y=-4\), but \(y = 7\) is extraneous. So the solution is \(y=-4\) (only one solution, but the problem said 1 or 2). Maybe the initial equation was different? Wait, the original equation is \(\frac{2}{y - 7}=\frac{-y}{2y - 14}\). So after cross - multiplying, we get the quadratic, but one solution is extraneous. So the valid solution is \(y=-4\).

Answer:

\(y=-4\) (and the other "solution" \(y = 7\) is extraneous, so the only solution is \(y=-4\))