Problem 85 - Beginning and Intermediate Algebra

Question

Simplify each expression. \(-\frac{4}{3}(y-12)-\frac{1}{6} y\)

Step-by-Step Solution

Answer

\(-\frac{3}{2}y + 16\)

Step 1 - Distribute the first term

Distribute \(-\frac{4}{3}\) across the terms inside the parenthesis: \(-\frac{4}{3}(y-12) = -\frac{4}{3}y + \frac{48}{3}\). This simplifies to \(-\frac{4}{3}y + 16\).

Step 2 - Write the expression with the distributed terms

Now, replace the original parenthesis with the distributed terms: \(-\frac{4}{3}y + 16 - \frac{1}{6}y\).

Step 3 - Find a common denominator for the terms with \(y\)

Convert \(-\frac{4}{3}y\) and \(-\frac{1}{6}y\) to have the same denominator. The least common multiple of 3 and 6 is 6. So, \(-\frac{4}{3}y = -\frac{8}{6}y\). Now, the expression is \(-\frac{8}{6}y + 16 - \frac{1}{6}y\).

Step 4 - Combine the \(y\) terms

Combine \(-\frac{8}{6}y\) and \(-\frac{1}{6}y\): \(-\frac{8}{6}y - \frac{1}{6}y = -\frac{9}{6}y\). Simplify \(-\frac{9}{6}y\) to \(-\frac{3}{2}y\). The expression is now \(-\frac{3}{2}y + 16\).

Key Concepts

Distributive Property

Understanding the Distributive Property is key to simplifying algebraic expressions. This property allows you to distribute a multiplier across terms inside parentheses. In the given problem, we need to distribute \(-\frac{4}{3}\) across the terms inside the parenthesis, yielding: \(-\frac{4}{3}(y-12) = -\frac{4}{3}y + \frac{48}{3}\). This simplifies further to \(-\frac{4}{3}y + 16\).

By distributing, we removed the parentheses. This is a vital first step because it breaks down the original expression into simpler parts that we can further manipulate.

Combining Like Terms

Combining like terms means to combine terms that have the same variable and exponent. In our exercise, after distribution, we have \(-\frac{4}{3}y + 16 - \frac{1}{6}y\).

To combine \-\frac{4}{3}y\ and \-\frac{1}{6}y\, we first need to give them a common denominator. The common denominator for 3 and 6 is 6. Thus, \-\frac{4}{3}y\ is converted to \-\frac{8}{6}y\.

Now, combining \-\frac{8}{6}y\ and \-\frac{1}{6}y\ results in \-\frac{9}{6}y\. This simplification process is critical because it reduces the expression to a more manageable form: \(-\frac{3}{2}y + 16\).

Common Denominator

Finding a common denominator is necessary when you need to add or subtract fractions with different denominators. In this exercise, after distributing, we had to combine \(-\frac{4}{3}y\) and \(-\frac{1}{6}y\). The least common multiple (LCM) of 3 and 6 is 6.

Converting \(-\frac{4}{3}y\) to a denominator of 6 gives us \(-\frac{8}{6}y\). This makes it straightforward to subtract fractions: \(-\frac{8}{6}y - \frac{1}{6}y = -\frac{9}{6}y\), which simplifies to \(-\frac{3}{2}y\).

Using common denominators allows us to easily combine fractional terms, resulting in a simplified and final expression: \(-\frac{3}{2}y + 16\). This concept is widely applicable in solving much more complex algebra problems.