Problem 11 - Algebra 1

Question

Use the subtraction rule to rewrite the subtraction expression as an equivalent addition expression. Then evaluate the expression. $$ \frac{2}{3}-\left(-\frac{1}{6}\right)-\frac{1}{3} $$

Step-by-Step Solution

Answer

The result of evaluating the expression is \(\frac{7}{6}\).

Step 1: Transform subtraction into addition

We can transform the exercise into an addition operation by changing the subtraction of a number into addition of its opposite. The given expression \(\frac{2}{3}-\left(-\frac{1}{6}\right)-\frac{1}{3}\) can be converted into an equivalent addition expression where the sign of the number after subtraction becomes positive and vice versa. After the transformation, the expression becomes \(\frac{2}{3}+\frac{1}{6}+\frac{1}{3}\).

Step 2: Simplify the expression

Add the fractions using the common denominator which is 6 in this case. Simplify \(\frac{2}{3}+\frac{1}{6}+\frac{1}{3}\) to \(\frac{4}{6}+\frac{1}{6}+\frac{2}{6}\).

Step 3: Evaluate the simplified expression

Now, add these fractions which will give \(\frac{7}{6}\).

Key Concepts

Transforming Subtraction to Addition

Subtraction in algebra can often be a source of difficulty when trying to simplify expressions. However, a useful trick that helps clarify these processes is transforming subtraction into addition. This method involves changing a subtraction operation into the addition of the opposite value.

For example, the subtraction expression \( \frac{2}{3} - \left(-\frac{1}{6}\right) - \frac{1}{3} \) can be confusing due to the presence of a negative fraction. By transforming subtraction to addition, we take the negative sign of the subtrahend and change it to a positive, effectively turning the subtrahend into its additive inverse. The given expression then becomes \( \frac{2}{3} + \frac{1}{6} + \left(-\frac{1}{3}\right) \) by converting the second term's negative into a positive and keeping the unchanged minus before the last fraction as the addition of a negative number.

This transformation simplifies the visualization of the problem, allowing for easier manipulation of the terms and avoidance of common mistakes associated with keeping track of multiple negative signs in subtraction.

Adding Fractions

When adding fractions, it's crucial to find a common denominator, which is a shared multiple of the denominators of the fractions involved. Once the common denominator is determined, the numerators are adjusted accordingly to ensure that the fractions are equivalent to their original value.

Continuing with our expression \( \frac{2}{3} + \frac{1}{6} + \left(-\frac{1}{3}\right) \), we can see that the common denominator is 6. Therefore, we convert \( \frac{2}{3} \) to \( \frac{4}{6} \) and \( \frac{1}{3} \) to \( \frac{2}{6} \). This process doesn't change the value of the fractions; it merely provides a standardized way to combine them. We then have \( \frac{4}{6} + \frac{1}{6} + \left(-\frac{2}{6}\right) \).

Now, the numerators can be easily added or subtracted since they are all over the same denominator. This is a traditional method that helps students avoid errors and makes calculations more straightforward.

Simplifying Expressions

Simplifying algebraic expressions is a fundamental skill in algebra which allows us to express complex ideas more clearly and solve equations efficiently. Once we have successfully transformed subtraction into addition and determined the common denominator for adding fractions, we can then proceed to simplify the expression.

As in our example, after finding the common denominator and combining the numerators, we have \( \frac{4}{6} + \frac{1}{6} - \frac{2}{6} \). Adding these gives us \( \frac{4 + 1 - 2}{6} \) which simplifies to \( \frac{3}{6} \). To further simplify, we can reduce this fraction by dividing both numerator and denominator by the greatest common divisor, which is 3 in this case, resulting in \( \frac{1}{2} \).

The final simplified expression is easier to read, use, and understand. This approach to simplifying does not only apply to fractions but to all algebraic expressions, making it a versatile and powerful tool in a student's mathematical toolkit.