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Cross-multiplication is a nifty tool for solving equations involving fractions. It helps us get rid of those pesky denominators, making the equation easier to work with. The basic idea is to multiply the numerator of one fraction by the denominator of the other fraction, doing this operation across the equation. Thus, if you have an equation of the form \( \frac{a}{b} = \frac{c}{d} \), cross-multiplication gives you \( a \times d = b \times c \).

In our exercise, \( \frac{3}{x} = \frac{x}{12} \), cross-multiplication involves multiplying the 3 by 12 and the \( x \) by \( x \), leading to \( 3 \times 12 = x \times x \) or \( 36 = x^2 \). By eliminating the fractions, you now have a simple quadratic equation to solve. This is a powerful method that streamlines solving such rational equations.

Extraneous solutions can appear when manipulating equations, especially when dealing with squares and roots. An extraneous solution is a solution derived during the problem-solving process that does not satisfy the original equation. This can happen when operations, like squaring both sides of an equation, introduce "false" solutions.

In our case, solving \( 36 = x^2 \) gives us two potential solutions: \( x = 6 \) and \( x = -6 \). While these solutions satisfy the quadratic equation we obtained after cross-multiplication, they must be checked against the original fraction equation. Specific operations, such as squaring, can sometimes yield these fake results, tricking us into thinking they are valid. That's why checking for extraneous solutions is crucial in ensuring the solutions truly work for the initial given conditions.

Checking solutions is a crucial step in verifying that the answers we derive from an algebraic equation truly work for the original problem. After finding potential solutions, we substitute these back into the original equation to see if the left-hand side equals the right-hand side.

For our equation \( \frac{3}{x} = \frac{x}{12} \), we found \( x = 6 \) and \( x = -6 \). When plugging \( x = 6 \) back into the equation, both sides simplify to 0.5, meaning it checks out as a valid solution. Similarly, for \( x = -6 \), both sides simplify to -0.5, confirming this solution also holds.

• Substitute the value back into the original equation
• Simplify both sides to check for equality
• Confirm if the solution is valid

By checking, you ensure that your approach is correct and that you're not left with any solutions that don't fit the original problem.