Question

Solve the proportion. Check for extraneous solutions. $$\frac{2}{x}=\frac{x-1}{6}$$

Short answer

The solutions to the proportion are \(x = 4\) and \(x = -3\). Both solutions have been checked and are valid, so there are no extraneous solutions.

Step-by-step solution

Simplify the proportion

In the given proportion \(\frac{2}{x}=\frac{x-1}{6}\), start by performing cross multiplication to get rid of the fractions. With cross multiplication, you multiply the numerator of the first fraction by the denominator of the second fraction and then set it equal to multiplying the denominator of the first fraction by the numerator of the second fraction to get \(2*6 = x*(x-1)\), simplifying this will gives \(12 = x^2 - x\)

Rearrange the equation

Rewrite the equation into standard form by subtracting 12 from both sides of the equation. This gives \(x^2 - x - 12 = 0\) which is now a quadratic equation.

Factor the quadratic equation

Factor the quadratic equation \(x^2 - x - 12 = 0\) to its factors. The factors should be such that multiply to -12 and add up to -1. From inspection, the factors are \(x - 4\) and \(x + 3\). So, we should get \((x - 4)(x + 3) = 0\)

Solve for x

Now, set each factor equal to zero and solve. So, \(x - 4 = 0\) gives \(x = 4\) and \(x + 3 = 0\) gives \(x = -3\)

Check for extraneous solutions

By substituting these values back into the original proportion, if we get a true statement then the solution is valid, if not it's extraneous. For \(x = 4\), we have \(\frac{2}{4} = \frac{4 - 1}{6}\) which simplifies to \(\frac{1}{2} = \frac{1}{2}\). This is true, so \(x = 4\) is a valid solution. For \(x = -3\), we have \(\frac{2}{-3} = \frac{-3 - 1}{6}\) which simplifies to \(-\frac{2}{3} = -\frac{2}{3}\). This is true, so \(x = -3\) is also a valid solution.