Question

which statement describes the solutions of this equation?
\\(\frac{4}{3x + 1} = \frac{x}{2x + 10}\\)
\\(\bigcirc\\) a. the equation has one valid solution and no extraneous solutions.
\\(\bigcirc\\) b. the equation has one valid solution and one extraneous solution.
\\(\bigcirc\\) c. the equation has two valid solutions and no extraneous solutions.
\\(\bigcirc\\) d. the equation has no valid solutions and two extraneous solutions.

Explanation:

Step1: Cross - multiply to solve the equation

Given the equation \(\frac{4}{3x + 1}=\frac{x}{2x+10}\), cross - multiplying (which is valid when \(3x + 1 eq0\) and \(2x + 10 eq0\), i.e., \(x eq-\frac{1}{3}\) and \(x eq - 5\)) gives us \(4(2x + 10)=x(3x + 1)\).

Step2: Expand both sides

Expanding the left - hand side: \(4\times2x+4\times10 = 8x + 40\).
Expanding the right - hand side: \(x\times3x+x\times1=3x^{2}+x\).
So the equation becomes \(8x + 40=3x^{2}+x\).

Step3: Rearrange into standard quadratic form

Subtract \(8x\) and \(40\) from both sides to get \(3x^{2}+x-8x - 40 = 0\).
Simplify the like terms: \(3x^{2}-7x - 40 = 0\).

Step4: Solve the quadratic equation

We can solve the quadratic equation \(ax^{2}+bx + c = 0\) (here \(a = 3\), \(b=-7\), \(c = - 40\)) using the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) or by factoring. Let's try factoring.
We need two numbers that multiply to \(3\times(-40)=-120\) and add up to \(-7\). The numbers are \(-15\) and \(8\).
Rewrite the middle term: \(3x^{2}-15x + 8x-40 = 0\).
Factor by grouping: \(3x(x - 5)+8(x - 5)=0\).
Then \((3x + 8)(x - 5)=0\).
Setting each factor equal to zero gives \(3x+8 = 0\) or \(x - 5=0\).
For \(3x+8 = 0\), we have \(x=-\frac{8}{3}\). For \(x - 5=0\), we have \(x = 5\).

Step5: Check for extraneous solutions

Check \(x = 5\):
For the original equation, the denominator \(3x + 1=3\times5+1 = 16 eq0\) and \(2x + 10=2\times5 + 10=20 eq0\). So \(x = 5\) is a valid solution.
Check \(x=-\frac{8}{3}\):
The denominator \(3x + 1=3\times(-\frac{8}{3})+1=-8 + 1=-7 eq0\) and \(2x+10=2\times(-\frac{8}{3})+10=-\frac{16}{3}+\frac{30}{3}=\frac{14}{3} eq0\). So \(x =-\frac{8}{3}\) is also a valid solution? Wait, no, wait. Wait, in step 1, we had the original equation \(\frac{4}{3x + 1}=\frac{x}{2x + 10}\). Wait, maybe I made a mistake in cross - multiplying. Wait, no, cross - multiplying is correct for rational equations when the denominators are non - zero. Wait, but let's re - check the quadratic solution.
Wait, \(3x^{2}-7x - 40 = 0\). Let's use the quadratic formula: \(x=\frac{7\pm\sqrt{(-7)^{2}-4\times3\times(-40)}}{2\times3}=\frac{7\pm\sqrt{49 + 480}}{6}=\frac{7\pm\sqrt{529}}{6}=\frac{7\pm23}{6}\).
So \(x=\frac{7 + 23}{6}=\frac{30}{6}=5\) and \(x=\frac{7-23}{6}=\frac{-16}{6}=-\frac{8}{3}\). Now, check \(x = 5\): denominators \(3x + 1=16\), \(2x + 10 = 20\) (valid). \(x=-\frac{8}{3}\): \(3x+1=3\times(-\frac{8}{3})+1=-8 + 1=-7\), \(2x + 10=2\times(-\frac{8}{3})+10=-\frac{16}{3}+\frac{30}{3}=\frac{14}{3}\) (valid). Wait, but the options don't have an option with two valid solutions? Wait, no, wait, I think I made a mistake in the cross - multiplication step. Wait, the original equation is \(\frac{4}{3x + 1}=\frac{x}{2x + 10}\). Cross - multiplying gives \(4(2x + 10)=x(3x + 1)\), which is \(8x+40 = 3x^{2}+x\), then \(3x^{2}-7x - 40 = 0\). Wait, but let's check the problem again. Wait, maybe the original equation was written incorrectly? Wait, no, the user provided the equation as \(\frac{4}{3x + 1}=\frac{x}{2x + 10}\). Wait, but let's re - check the extraneous solutions. Wait, the domain of the original function: \(3x+1 eq0\Rightarrow x eq-\frac{1}{3}\), \(2x + 10 eq0\Rightarrow x eq - 5\). The solutions we found \(x = 5\) and \(x=-\frac{8}{3}\) are not equal to \(-\frac{1}{3}\) or \(-5\), so both are valid? But the options are A: one valid, no extraneous; B: one valid, one extraneous; C: two valid, no extraneous; D: no valid, two extraneous. Wait, maybe I made a mistake in the quadratic solution. Wait, let's re - solve \(3x^{2}-7x - 40 = 0\). Let's factor again: \(3x^{2}-15x+8x - 40=3x(x - 5)+8(x - 5)=(3x + 8)(x - 5)=0\). So \(x = 5\) and \(x=-\frac{8}{3}\). Both are in the domain (since \(x = 5\) and \(x=-\frac{8}{3}\) are not \(-\frac{1}{3}\) or \(-5\)). Wait, but the options have option C: two valid solutions and no extraneous solutions. But wait, maybe I made a mistake in cross - multiplying. Wait, no, cross - multiplying for \(\frac{a}{b}=\frac{c}{d}\) ( \(b eq0\), \(d eq0\)) gives \(ad = bc\). So that step is correct. So the equation has two valid solutions and no extraneous solutions? But let's check with \(x = 5\): left - hand side \(\frac{4}{3\times5 + 1}=\frac{4}{16}=\frac{1}{4}\), right - hand side \(\frac{5}{2\times5+10}=\frac{5}{20}=\frac{1}{4}\). Correct. For \(x=-\frac{8}{3}\): left - hand side \(\frac{4}{3\times(-\frac{8}{3})+1}=\frac{4}{-8 + 1}=\frac{4}{-7}=-\frac{4}{7}\), right - hand side \(\frac{-\frac{8}{3}}{2\times(-\frac{8}{3})+10}=\frac{-\frac{8}{3}}{-\frac{16}{3}+\frac{30}{3}}=\frac{-\frac{8}{3}}{\frac{14}{3}}=-\frac{8}{14}=-\frac{4}{7}\). Correct. So the equation has two valid solutions and no extraneous solutions. Wait, but the options: option C says "The equation has two valid solutions and no extraneous solutions." But wait, maybe I made a mistake in the problem interpretation. Wait, the original equation: \(\frac{4}{3x + 1}=\frac{x}{2x + 10}\). Let me re - check the quadratic solution. \(3x^{2}-7x - 40 = 0\). Discriminant \(D=b^{2}-4ac=(-7)^{2}-4\times3\times(-40)=49 + 480 = 529=23^{2}\). So roots are \(\frac{7\pm23}{6}\). \(\frac{7 + 23}{6}=\frac{30}{6}=5\), \(\frac{7-23}{6}=\frac{-16}{6}=-\frac{8}{3}\). Both are valid. So the answer should be option C? Wait, but maybe the original problem was written with a different equation? Wait, no, as per the user's question, the equation is \(\frac{4}{3x + 1}=\frac{x}{2x + 10}\). So the correct option is C. Wait, but let's check the options again. Option C: "The equation has two valid solutions and no extraneous solutions." Yes.

Answer:

C. The equation has two valid solutions and no extraneous solutions.