Question

Solve the proportion. Check for extraneous solutions. $$\frac{x}{3}=\frac{2}{5}$$

Short answer

The solution to the proportion is \(x = \frac{6}{5}\), and upon checking, it was found that there are no extraneous solutions.

Step-by-step solution

Cross-multiply

In this step, cross-multiply the given proportion. Multiply \(x\) by \(5\) and \(3\) by \(2\), which gives the equation \(5x = 6\).

Solve for x

Now, isolate \(x\) by dividing both sides of the equation by 5. This gives \(x = \frac{6}{5}\).

Check for extraneous solutions

Plug the obtained value of \(x\) back into the original equation to see if it's valid. After substitution, \(\frac{6/5}{3} = \frac{2}{5}\) simplifies to \( \frac{2}{5} = \frac{2}{5}\). The left side is equal to the right side, confirming that our solution is correct. Therefore, there are no extraneous solutions.

Key concepts

Cross-Multiplication

When confronted with a proportion problem, cross-multiplication is one of the key strategies to solve it efficiently. This method involves multiplying the numerator of one fraction by the denominator of the other. Let's delve into the procedure with an example:
Given the proportion \( \frac{x}{3} = \frac{2}{5} \) you cross-multiply by multiplying \( x \) by \( 5 \) and \( 3 \) by \( 2 \) which results in the equation \( 5x = 6 \). This process creates an equation that is often easier to solve because it removes the fractions and simplifies the relationship between the quantities involved.

Why Cross-Multiply?

• It simplifies the proportion into a linear equation.
• Makes it possible to isolate the variable and solve for it.
• Eliminates the need for finding a common denominator.

Isolating Variables

After cross-multiplication, you'll usually end up with a simple equation that requires isolating the variable to find its value. Isolating the variable means getting the variable on one side of the equation and everything else on the other. Following our initial proportion example, we have \( 5x = 6 \).
To isolate \( x \) and solve the equation, divide both sides by \( 5 \) which yields \( x = \frac{6}{5} \). This technique is crucial because:

Benefits of Isolating Variables

• It clearly shows the solution for the variable.
• Reduces the risk of making algebraic errors.
• Prepares a clear path for further verification of the solution.

Remember, consistency in performing the same operation on both sides of the equation is key to maintaining its balance and integrity.

Checking for Extraneous Solutions

After finding the value of the variable, it's important to validate the solution to ensure it's not extraneous. An extraneous solution is a number that emerges from the process of solving the problem but isn't a valid solution to the original equation. To check for extraneous solutions, we substitute the value of the variable back into the original equation.
For example, by plugging \( \frac{6}{5} \) into our original proportion \( \frac{x}{3} = \frac{2}{5} \) and simplifying, we find that both sides of the proportion remain equal.

Why Check for Extraneous Solutions?

• It confirms the validity of the solution within the initial problem context.
• Helps identify and eliminate any non-solutions that may have surfaced during algebraic manipulations.

Always perform this step to ensure the soundness of your answer.

Proportion Problems

Proportion problems deal with determining how two ratios or fractions relate to each other, usually in the form of \( \frac{a}{b} = \frac{c}{d} \). These problems often appear in real-life scenarios, such as in recipes, scale models, and map readings. The goal is to find a missing value that keeps both ratios equivalent.

Approach to Solving Proportions

• Identify which quantity is represented by the variable.
• Use cross-multiplication to eliminate fractions and create an equation.
• Isolate the variable to solve the equation.
• Check the solution to ensure it fits the original proportion without causing contradictions.

Mastery of solving proportions is not only important for schoolwork but for many practical applications beyond the classroom.