- Identify constraints \(x
eq -1/3\) and \(x
eq -5\) where the denominator becomes zero.
- Cross-multiply the original equation to obtain the quadratic form \(3x^2 - 7x - 40 = 0\).
- Calculate the two solutions \(x = 5\) and \(x = -8/3\) using the quadratic formula.
- Confirm both solutions are valid as they do not violate the initial constraints, leading to the conclusion that the equation has \(\boxed{\text{two valid solutions and no extraneous solutions}}\).
Explanation
- Problem Analysis and Restrictions
The given equation is \(\frac{4}{3x+1} = \frac{x}{2x+10}\). This is a rational equation, which requires us to identify any values of \(x\) that make the denominators zero. If any solutions we find correspond to these values, they are considered extraneous.
First, set the denominators to zero to find the restricted values:
- \(3x + 1 = 0 \implies 3x = -1 \implies x = -1/3\)
- \(2x + 10 = 0 \implies 2x = -10 \implies x = -5\)
Thus, the solutions cannot be \(x = -1/3\) or \(x = -5\).
- Solving the Equation
To solve the equation, we cross-multiply to eliminate the denominators:
\[4(2x + 10) = x(3x + 1)\]
Expanding both sides gives:
\[8x + 40 = 3x^2 + x\]
Now, rearrange all terms into a standard quadratic equation of the form \(ax^2 + bx + c = 0\):
\[3x^2 + x - 8x - 40 = 0\]
\[3x^2 - 7x - 40 = 0\]
- Checking Solutions and Conclusion
We use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) to find the roots, where \(a = 3\), \(b = -7\), and \(c = -40\):
\[x = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(3)(-40)}}{2(3)}\]
\[x = \frac{7 \pm \sqrt{49 + 480}}{6}\]
\[x = \frac{7 \pm \sqrt{529}}{6}\]
Since \(\sqrt{529} = 23\), the roots are:
\[x_1 = \frac{7 + 23}{6} = \frac{30}{6} = 5\]
\[x_2 = \frac{7 - 23}{6} = \frac{-16}{6} = -\frac{8}{3}\]
Both solutions \(5\) and \(-\frac{8}{3}\) are not equal to the restricted values \(-1/3\) or \(-5\). Therefore, both are valid solutions, and there are no extraneous solutions. The correct statement is that the equation has two valid solutions and no extraneous solutions.
Examples
Rational equations are used in physics and engineering to model relationships like electrical resistance in parallel circuits or the time required to complete a task when two people work at different rates. For instance, if two pipes fill a pool at different speeds, the time it takes for them to fill it together can be represented by a rational equation similar to this one. Identifying valid solutions ensures that the calculated time or physical quantity makes sense in the context of the problem and that no division-by-zero errors occur in the model.