Question

select the correct answer.
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 the equation

To solve the rational equation \(\frac{4}{3x + 1}=\frac{x}{2x+10}\), we cross - multiply. Cross - multiplying gives us \(4(2x + 10)=x(3x + 1)\).

Step2: Expand both sides

Expand the left - hand side: \(4\times2x+4\times10 = 8x + 40\).
Expand 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 \(3x^{2}-7x - 40 = 0\) using the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\), where \(a = 3\), \(b=-7\), and \(c=-40\).
First, calculate the discriminant \(\Delta=b^{2}-4ac=(-7)^{2}-4\times3\times(-40)=49 + 480 = 529\).
Then, \(x=\frac{7\pm\sqrt{529}}{2\times3}=\frac{7\pm23}{6}\).
We have two solutions:
\(x_1=\frac{7 + 23}{6}=\frac{30}{6}=5\) and \(x_2=\frac{7-23}{6}=\frac{-16}{6}=-\frac{8}{3}\).

Step5: Check for extraneous solutions

We need to check if these solutions make the denominators of the original rational equation zero.
For the denominator \(3x + 1\):
- When \(x = 5\), \(3x+1=3\times5 + 1=16 eq0\).
- When \(x=-\frac{8}{3}\), \(3x + 1=3\times(-\frac{8}{3})+1=-8 + 1=-7 eq0\).
For the denominator \(2x + 10\):
- When \(x = 5\), \(2x+10=2\times5 + 10=20 eq0\).
- When \(x=-\frac{8}{3}\), \(2x + 10=2\times(-\frac{8}{3})+10=-\frac{16}{3}+\frac{30}{3}=\frac{14}{3} eq0\).

Both solutions are valid. Wait, but let's re - check the quadratic solution. Wait, the quadratic equation was \(3x^{2}-7x - 40 = 0\). Wait, maybe I made a mistake in the sign when rearranging. Let's go back to the cross - multiplication step.

Original equation: \(\frac{4}{3x + 1}=\frac{x}{2x + 10}\)

Cross - multiply: \(4(2x + 10)=x(3x + 1)\)

\(8x+40 = 3x^{2}+x\)

\(3x^{2}+x-8x - 40=0\)

\(3x^{2}-7x - 40 = 0\)

Wait, discriminant calculation: \((-7)^2-4*3*(-40)=49 + 480 = 529\), which is correct. \(\sqrt{529}=23\), correct.

\(x=\frac{7\pm23}{6}\), so \(x=\frac{30}{6}=5\) and \(x=\frac{-16}{6}=-\frac{8}{3}\), correct.

Checking denominators:

For \(3x + 1\):

- \(x = 5\): \(3*5+1 = 16 eq0\)
- \(x=-\frac{8}{3}\): \(3*(-\frac{8}{3})+1=-8 + 1=-7 eq0\)

For \(2x + 10\):

- \(x = 5\): \(2*5+10 = 20 eq0\)
- \(x=-\frac{8}{3}\): \(2*(-\frac{8}{3})+10=-\frac{16}{3}+\frac{30}{3}=\frac{14}{3} eq0\)

Wait, 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 equation. Let's re - solve the original equation.

\(\frac{4}{3x + 1}=\frac{x}{2x + 10}\)

Cross - multiply: \(4(2x + 10)=x(3x + 1)\)

\(8x+40=3x^{2}+x\)

\(3x^{2}-7x - 40 = 0\)

Factor the quadratic: \(3x^{2}-15x+8x - 40 = 0\)

\(3x(x - 5)+8(x - 5)=0\)

\((3x + 8)(x - 5)=0\)

Ah! Here is the mistake. I factored it wrong earlier. So the correct factoring is \((3x + 8)(x - 5)=0\), so \(3x+8 = 0\) or \(x - 5=0\), so \(x = 5\) or \(x=-\frac{8}{3}\) (wait, \(3x+8 = 0\) gives \(x=-\frac{8}{3}\), which is the same as before). Wait, but when we factor \(3x^{2}-7x - 40\), let's use the formula correctly. \(a = 3\), \(b=-7\), \(c=-40\).

\(x=\frac{7\pm\sqrt{49+480}}{6}=\frac{7\pm23}{6}\), so \(x = 5\) or \(x=-\frac{8}{3}\). Both are valid as denominators are non - zero. But the options:

Wait, maybe the original equation was written wrong? Wait, the original equation is \(\frac{4}{3x + 1}=\frac{x}{2x + 10}\). Wait, let's check the discriminant again. \((-7)^2-4*3*(-40)=49 + 480 = 529\), which is \(23^2\), correct. So the two solutions are valid. But the options have C as "two valid solutions and no extraneous solutions". But wait, maybe I made a mistake in the problem. Wait, no, let's check again.

Wait, the quadratic equation \(3x^{2}-7x - 40 = 0\) has two real roots, and both roots do not make the denominators zero, so there are two valid solutions and no extraneous solutions. But let's check the answer options. Option C says "The equation has two valid solutions and no extraneous solutions." But wait, maybe I made a mistake in the cross - multiplication. Wait, no, cross - multiplying a rational equation \(\frac{a}{b}=\frac{c}{d}\) (where \(b eq0\) and \(d eq0\)) gives \(ad = bc\). So that step is correct.

Wait, maybe the problem was supposed to have a different equation. But according to the given equation, the two solutions are valid. So the answer should be C? But wait, let's check again.

Wait, when \(x = 5\):

Left - hand side: \(\frac{4}{3*5+1}=\frac{4}{16}=\frac{1}{4}\)

Right - hand side: \(\frac{5}{2*5 + 10}=\frac{5}{20}=\frac{1}{4}\). Correct.

When \(x=-\frac{8}{3}\):

Left - hand side: \(\frac{4}{3*(-\frac{8}{3})+1}=\frac{4}{-8 + 1}=\frac{4}{-7}=-\frac{4}{7}\)

Right - hand side: \(\frac{-\frac{8}{3}}{2*(-\frac{8}{3})+10}=\frac{-\frac{8}{3}}{-\frac{16}{3}+\frac{30}{3}}=\frac{-\frac{8}{3}}{\frac{14}{3}}=-\frac{8}{14}=-\frac{4}{7}\). Correct.

So both solutions are valid. So the equation has two valid solutions and no extraneous solutions. So the answer is C. Wait, but the options are A, B, C, D. So the correct option is C.

Answer:

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