Problem 73 - Introductory Algebra for College Students

Question

Add or subtract as indicated. Simplify the result, if possible. $$\frac{y+3}{5 y^{2}}-\frac{y-5}{15 y}$$

Step-by-Step Solution

Answer

The simplified result of the given expression is \(\frac{2(y + 7)}{15y}\).

Step 1: Identify the common denominator

In this expression, the two denominators are \(5y^{2}\) and \(15y\). A common denominator for these two fractions is their least common multiple (LCM). A multiple of \(5y^{2}\) can be written as \(5y^2 × m\), and a multiple of \(15y\) can be written as \(15y × n\). The LCM is the smallest value for which \(m\) and \(n\) are whole numbers. In this case, the LCM (and thus the common denominator) is \(15y^{2}\).

Step 2: Rewrite the fractions

We need to rewrite each fraction so they both have the denominator \(15y^{2}\). The first fraction becomes \(\frac{(y+3) \times 3y}{15y^{2}}\) and the second fraction becomes \(\frac{(y-5) × y}{15y^{2}}\). This results in the new expression: \(\frac{3y(y+3)}{15y^{2}} - \frac{y(y-5)}{15y^{2}}\).

Step 3: Combine like terms

Now that the fractions have the same denominator, they can be subtracted. This results in: \(\frac{3y(y+3) - y(y-5)}{15y^{2}}\).

Step 4: Distribute and simplify

Next, you distribute the \(3y\) and \(-y\) to terms inside the brackets to get: \(\frac{3y^2+9y - y^2 +5y}{15y^{2}}\). This simplifies to \(\frac{2y^{2} + 14y}{15y^{2}}\). This simplifies further by dividing through by \(y\) (given that \(y \neq 0\)) to get: \(\frac{2y + 14}{15y}\).

Step 5: Simplify further

The numerator has common factors that can be factored out which gives \(\frac{2(y + 7)}{15y}\). This is the simplest form of the original expression given.