Question

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

Explanation:

Step1: Cross - multiply the equation

Given the equation \(\frac{1}{g - 3}=\frac{-g}{10g}\), cross - multiplying (which is valid when \(g eq3\) and \(g eq0\) to avoid division by zero) gives us \(1\times10g=-g\times(g - 3)\).

Step2: Simplify both sides

Simplify the left - hand side to get \(10g\), and the right - hand side: \(-g\times(g - 3)=-g^{2}+3g\). So the equation becomes \(10g=-g^{2}+3g\).

Step3: Rearrange into standard quadratic form

Bring all terms to one side to form a quadratic equation. Add \(g^{2}\) to both sides and subtract \(3g\) from both sides: \(g^{2}+10g - 3g = 0\), which simplifies to \(g^{2}+7g = 0\).

Step4: Factor the quadratic equation

Factor out the common factor \(g\) from the left - hand side: \(g(g + 7)=0\).

Step5: Solve for \(g\)

Using the zero - product property (if \(ab = 0\), then either \(a = 0\) or \(b = 0\)), we have two cases:
- Case 1: \(g=0\). But if we substitute \(g = 0\) back into the original equation \(\frac{1}{g - 3}=\frac{-g}{10g}\), the right - hand side \(\frac{-g}{10g}\) becomes undefined (since we have \(\frac{0}{0}\)), so \(g = 0\) is an extraneous solution.
- Case 2: \(g+7 = 0\), which gives us \(g=-7\). We also need to check if \(g=-7\) is valid. Substitute \(g=-7\) into the original equation:
Left - hand side: \(\frac{1}{-7 - 3}=\frac{1}{-10}=-\frac{1}{10}\)
Right - hand side: \(\frac{-(-7)}{10\times(-7)}=\frac{7}{-70}=-\frac{1}{10}\)
So \(g=-7\) is a valid solution. Wait, we made a mistake in step 2. Let's go back to step 2.

Step1 (Revised): Cross - multiply the equation

Given \(\frac{1}{g - 3}=\frac{-g}{10g}\), first simplify \(\frac{-g}{10g}\) (for \(g eq0\)) to \(\frac{-1}{10}\). So the equation becomes \(\frac{1}{g - 3}=-\frac{1}{10}\)

Step2 (Revised): Cross - multiply again

Cross - multiplying gives \(10=-(g - 3)\)

Step3 (Revised): Solve for \(g\)

Expand the right - hand side: \(10=-g + 3\). Add \(g\) to both sides: \(g+10 = 3\). Subtract 10 from both sides: \(g=3 - 10=-7\). Also, we need to check the case when \(g = 0\) (from the original denominator \(10g\)), but as we saw before, \(g = 0\) makes the right - hand side undefined. And when \(g = 3\), the left - hand side is undefined. So the only valid solution is \(g=-7\)? Wait, no, let's start over.

Original equation: \(\frac{1}{g - 3}=\frac{-g}{10g}\)

First, note that \(g eq0\) (because of \(10g\) in the denominator) and \(g eq3\) (because of \(g - 3\) in the denominator).

Simplify \(\frac{-g}{10g}\) (for \(g eq0\)): \(\frac{-g}{10g}=-\frac{1}{10}\)

So the equation is \(\frac{1}{g - 3}=-\frac{1}{10}\)

Cross - multiply: \(10=-(g - 3)\)

\(10=-g + 3\)

Add \(g\) to both sides: \(g + 10=3\)

Subtract 10: \(g=3 - 10=-7\)

Wait, but when we first cross - multiplied without simplifying \(\frac{-g}{10g}\), we had:

\(10g=-g(g - 3)\)

\(10g=-g^{2}+3g\)

\(g^{2}+7g = 0\)

\(g(g + 7)=0\)

\(g = 0\) or \(g=-7\). But \(g = 0\) is not allowed (since \(\frac{-g}{10g}\) is undefined when \(g = 0\)), so the only solution is \(g=-7\)? But the problem says there may be 1 or 2 solutions. Wait, maybe I made a mistake in simplifying \(\frac{-g}{10g}\). Let's check again.

\(\frac{-g}{10g}\): the \(g\) in the numerator and denominator cancels out when \(g eq0\), so \(\frac{-g}{10g}=-\frac{1}{10}\) (for \(g eq0\)). So the equation is \(\frac{1}{g - 3}=-\frac{1}{10}\)

Cross - multiply: \(10=-(g - 3)\)

\(10=-g + 3\)

\(g=3 - 10=-7\)

So the solution is \(g=-7\). But where did the other solution come from? When we cross - multiplied without simplifying, we included \(g = 0\) as a potential solution, but \(g = 0\) is extraneous. So the only solution is \(g=-7\). Wait, maybe the original equation was written incorrectly? Or maybe I misread it. Let me check again.

Original equation: \(\frac{1}{g - 3}=\frac{-g}{10g}\)

Yes, that's what's written. So after simplifying \(\frac{-g}{10g}\) to \(-\frac{1}{10}\) (for \(g eq0\)), we solve \(\frac{1}{g - 3}=-\frac{1}{10}\), which gives \(g=-7\). And \(g = 0\) is extraneous.

Answer:

\(g=-7\) (and the other "solution" \(g = 0\) is extraneous, so the only valid solution is \(g=-7\))