Question

Solve the equation and check your answer.
StartFraction 3 x Over 16 x plus 12 EndFraction equals StartFraction 1 Over x EndFraction

Answer 1

To solve the equation

\[
\frac{3x}{16x + 12} = \frac{1}{x},
\]

we can cross-multiply to eliminate the fractions. This gives us:

\[
3x \cdot x = 1 \cdot (16x + 12).
\]

Simplifying both sides yields:

\[
3x^2 = 16x + 12.
\]

Next, we can rearrange this into a standard quadratic equation form:

\[
3x^2 - 16x - 12 = 0.
\]

Now, we can solve this quadratic equation using the quadratic formula:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},
\]

where \( a = 3 \), \( b = -16 \), and \( c = -12 \).

First, we calculate the discriminant:

\[
b^2 - 4ac = (-16)^2 - 4 \cdot 3 \cdot (-12) = 256 + 144 = 400.
\]

Now we can plug the values into the quadratic formula:

\[
x = \frac{-(-16) \pm \sqrt{400}}{2 \cdot 3} = \frac{16 \pm 20}{6}.
\]

This gives us two potential solutions:

1. \( x = \frac{16 + 20}{6} = \frac{36}{6} = 6 \)
2. \( x = \frac{16 - 20}{6} = \frac{-4}{6} = -\frac{2}{3} \)

The solutions are \( x = 6 \) and \( x = -\frac{2}{3} \).

**Check the Solutions:**

1. Check \( x = 6 \):

\[
\frac{3(6)}{16(6) + 12} = \frac{18}{96 + 12} = \frac{18}{108} = \frac{1}{6},
\]
and
\[
\frac{1}{6} = \frac{1}{6} \quad \text{(True)}.
\]

2. Check \( x = -\frac{2}{3} \):

\[
\frac{3(-\frac{2}{3})}{16(-\frac{2}{3}) + 12} = \frac{-2}{-\frac{32}{3} + 12} = \frac{-2}{-\frac{32}{3} + \frac{36}{3}} = \frac{-2}{\frac{4}{3}} = -2 \cdot \frac{3}{4} = -\frac{3}{2},
\]
and
\[
\frac{1}{-\frac{2}{3}} = -\frac{3}{2} \quad \text{(True)}.
\]

Thus, both solutions are valid.

The final solutions are:

\[
\boxed{6} \quad \text{and} \quad \boxed{-\frac{2}{3}}.
\]