Question

question 2 of 10
select the correct answer.
which statement describes the solutions of this equation?
\\(\frac{x - 5}{2x - 6} = \frac{10}{x^2 - 9}\\)
\\(\bigcirc\\) the equation has one valid solution and no extraneous solutions.
\\(\bigcirc\\) the equation has two valid solutions and no extraneous solutions.
\\(\bigcirc\\) the equation has two valid solutions and one extraneous solution.
\\(\bigcirc\\) the equation has one valid solution and one extraneous solution.

Explanation:

Step1: Factor denominators

Factor \(2x - 6\) as \(2(x - 3)\) and \(x^2 - 9\) as \((x - 3)(x + 3)\). The equation becomes \(\frac{x - 5}{2(x - 3)}=\frac{10}{(x - 3)(x + 3)}\).

Step2: Find domain restrictions

Denominators can't be zero, so \(x eq3\) and \(x eq - 3\).

Step3: Cross - multiply to solve

Cross - multiply: \((x - 5)(x + 3)=20\). Expand the left side: \(x^{2}+3x-5x - 15 = 20\), which simplifies to \(x^{2}-2x-35 = 0\).

Step4: Solve the quadratic equation

Factor \(x^{2}-2x - 35=(x - 7)(x + 5)=0\). So \(x = 7\) or \(x=-5\).

Step5: Check for extraneous solutions

Check \(x = 7\): Substitute into original denominators, \(2(7)-6 = 8 eq0\), \(7^{2}-9 = 40 eq0\). Valid.
Check \(x=-5\): Substitute into original denominators, \(2(-5)-6=-16 eq0\), \((-5)^{2}-9 = 16 eq0\). Wait, wait, earlier mistake. Wait, when we cross - multiplied, we had \((x - 5)(x + 3)=2\times10\)? Wait no, original equation after factoring: \(\frac{x - 5}{2(x - 3)}=\frac{10}{(x - 3)(x + 3)}\). Cross - multiply: \((x - 5)(x + 3)=2\times10\)? No, cross - multiply is \((x - 5)(x + 3)\times1=10\times2\)? Wait, no, correct cross - multiplication: \((x - 5)(x + 3)=2\times10\)? Wait, no, \(\frac{a}{b}=\frac{c}{d}\) implies \(a\times d=b\times c\). So \((x - 5)(x + 3)=2(x - 3)\times10\)? No, wait \(b = 2(x - 3)\), \(d=(x - 3)(x + 3)\), \(a=(x - 5)\), \(c = 10\). So \((x - 5)(x + 3)=2(x - 3)\times10\)? No, no: \((x - 5)\times(x + 3)=2\times10\) when we cancel \((x - 3)\) (but we can only cancel \((x - 3)\) if \(x eq3\)). Wait, I made a mistake in cross - multiplication. Let's correct:

\(\frac{x - 5}{2(x - 3)}=\frac{10}{(x - 3)(x + 3)}\)

Multiply both sides by \(2(x - 3)(x + 3)\) (the least common denominator) to get rid of denominators:

\((x - 5)(x + 3)=20\) (since \(2(x - 3)(x + 3)\times\frac{x - 5}{2(x - 3)}=(x - 5)(x + 3)\) and \(2(x - 3)(x + 3)\times\frac{10}{(x - 3)(x + 3)} = 20\))

Then \(x^{2}+3x-5x - 15=20\)

\(x^{2}-2x-35 = 0\)

Factor: \((x - 7)(x + 5)=0\), so \(x = 7\) or \(x=-5\)

Now check \(x = 3\) is already excluded. Now check \(x = 7\): denominators \(2(7)-6 = 8\), \(7^{2}-9 = 40\), valid.

Check \(x=-5\): denominators \(2(-5)-6=-16\), \((-5)^{2}-9 = 16\), valid. Wait, but that's not matching the options. Wait, no, wait I think I messed up the cross - multiplication. Wait, original equation: \(\frac{x - 5}{2x - 6}=\frac{10}{x^{2}-9}\)

\(2x-6 = 2(x - 3)\), \(x^{2}-9=(x - 3)(x + 3)\)

So cross - multiply: \((x - 5)(x^{2}-9)=10(2x - 6)\)

Expand left side: \((x - 5)(x - 3)(x + 3)=20(x - 3)\)

If \(x eq3\), we can divide both sides by \((x - 3)\):

\((x - 5)(x + 3)=20\)

Which is what I had before. So \(x^{2}-2x-15 - 20=0\), \(x^{2}-2x-35 = 0\), solutions \(x = 7\) and \(x=-5\)

But the options are about one valid and one extraneous. Wait, maybe I made a mistake in the problem. Wait, no, let's re - check. Wait, maybe the original equation was \(\frac{x - 5}{2x - 6}=\frac{10}{x^{2}-9}\)

Wait, let's plug \(x = 3\) into original equation: left denominator \(2(3)-6 = 0\), right denominator \(9 - 9 = 0\), so \(x eq3\).

Wait, maybe the quadratic was solved wrong. Wait, \(x^{2}-2x-35 = 0\), discriminant \(b^{2}-4ac=4 + 140 = 144\), square root of 144 is 12, so \(x=\frac{2\pm12}{2}\), \(x=\frac{14}{2}=7\) or \(x=\frac{-10}{2}=-5\). Both are valid? But the options don't have that. Wait, maybe the original equation was \(\frac{x - 5}{2x - 6}=\frac{10}{x - 9}\)? No, the user wrote \(x^{2}-9\). Wait, maybe I misread the equation. Wait, the equation is \(\frac{x - 5}{2x - 6}=\frac{10}{x^{2}-9}\)

Wait, let's check the options again. The options are about one valid and one extraneous. Wait, maybe I made a mistake in cross - multiplication. Let's do it again:

\(\frac{x - 5}{2(x - 3)}=\frac{10}{(x - 3)(x + 3)}\)

Multiply both sides by \(2(x - 3)(x + 3)\):

\((x - 5)(x + 3)=20\)

\(x^{2}+3x-5x - 15 = 20\)

\(x^{2}-2x-35 = 0\)

\((x - 7)(x + 5)=0\), \(x = 7\) or \(x=-5\)

Now, if we consider that when we multiplied by \(2(x - 3)(x + 3)\), we assumed \(x eq3\) and \(x eq - 3\). Both \(x = 7\) and \(x=-5\) are not 3 or - 3, so both are valid. But the options don't have that. Wait, maybe the original equation was \(\frac{x - 5}{2x - 6}=\frac{10}{x - 9}\)? No, the user's equation is with \(x^{2}-9\). Wait, maybe there's a typo in my calculation. Wait, no, let's check the options again. The options are:

1. One valid, no extraneous.

2. Two valid, no extraneous.

3. Two valid, one extraneous.

4. One valid, one extraneous.

Wait, maybe I made a mistake in the cross - multiplication. Let's try again.

Original equation: \(\frac{x - 5}{2x - 6}=\frac{10}{x^{2}-9}\)

\(2x-6 = 2(x - 3)\), \(x^{2}-9=(x - 3)(x + 3)\)

Cross - multiply: \((x - 5)(x^{2}-9)=10(2x - 6)\)

\((x - 5)(x - 3)(x + 3)=20(x - 3)\)

Now, if \(x = 3\), both sides are 0, but \(x = 3\) makes denominators zero, so \(x = 3\) is extraneous. Now, divide both sides by \((x - 3)\) (for \(x eq3\)):

\((x - 5)(x + 3)=20\)

Which is \(x^{2}-2x-15 - 20=0\), \(x^{2}-2x-35 = 0\), solutions \(x = 7\) and \(x=-5\)

Now, the solutions we found are \(x = 7\) and \(x=-5\), and the extraneous solution is \(x = 3\) (from the original denominator). But the options are about the number of valid and extraneous solutions from the solutions we found (excluding the domain restrictions). Wait, no, when we solve rational equations, we first find the domain, then solve, then check the solutions in the original equation (including domain restrictions).

Wait, the solutions we got from the quadratic are \(x = 7\) and \(x=-5\), both are in the domain (since \(x eq3,-3\)). But that would mean two valid solutions and no extraneous solutions. But that's option 2. But maybe I made a mistake. Wait, let's check the original equation again.

Wait, maybe the equation is \(\frac{x - 5}{2x - 6}=\frac{10}{x^{2}-9}\)

Let's plug \(x = 7\):

Left: \(\frac{7 - 5}{14 - 6}=\frac{2}{8}=\frac{1}{4}\)

Right: \(\frac{10}{49 - 9}=\frac{10}{40}=\frac{1}{4}\). Valid.

Plug \(x=-5\):

Left: \(\frac{-5 - 5}{-10 - 6}=\frac{-10}{-16}=\frac{5}{8}\)

Right: \(\frac{10}{25 - 9}=\frac{10}{16}=\frac{5}{8}\). Valid.

So both solutions are valid, and there are no extraneous solutions from the solutions we found (since \(x = 3\) and \(x=-3\) are not among our solutions). So the equation has two valid solutions and no extraneous solutions.

Answer:

The equation has two valid solutions and no extraneous solutions.