Problem 59 - Intermediate Algebra

Question

Solve each equation. Check each result. See Example 7. $$ \frac{a+1}{3}-\frac{a-1}{5}=\frac{8}{15} $$

Step-by-Step Solution

Answer

The solution is \(a = 0\).

Step 1: Clear the fractions

To eliminate the fractions, find a common denominator for the fractions involved in the equation. The denominators are 3, 5, and 15, and the least common multiple is 15. Multiply every term in the equation by 15 to clear the fractions:\[15 \times \left( \frac{a+1}{3} \right) - 15 \times \left( \frac{a-1}{5} \right) = 15 \times \frac{8}{15}\]This simplifies to:\[5(a+1) - 3(a-1) = 8\]

Step 2: Distribute and simplify

Now distribute the numbers through the parentheses:\[5a + 5 - 3a + 3 = 8\]Combine like terms:\[(5a - 3a) + (5 + 3) = 8\]Which simplifies to:\[2a + 8 = 8\]

Step 3: Solve for 'a'

Subtract 8 from both sides to isolate the term with the variable:\[2a + 8 - 8 = 8 - 8\]Which simplifies to:\[2a = 0\]Next, divide both sides by 2 to solve for \(a\):\[a = 0\]

Step 4: Check the solution

Substitute \(a = 0\) back into the original equation to verify our solution:\[\frac{0+1}{3} - \frac{0-1}{5} = \frac{1}{3} - \frac{-1}{5} = \frac{1}{3} + \frac{1}{5}\]Find a common denominator to simplify this:\[\frac{5}{15} + \frac{3}{15} = \frac{8}{15}\]Since both sides of the equation are equal, the solution \(a = 0\) is verified.

Key Concepts

Fractions in Equations

Handling fractions in linear equations requires some extra care compared to equations with simple whole numbers. Fractions can make calculations feel complicated, but with a few tricks, they become manageable. In the given problem, we have fractions in the equation \( \frac{a+1}{3} - \frac{a-1}{5} = \frac{8}{15} \). The presence of fractions means that each term in the equation is divided by a number, which can make solving the equation directly tricky. To ease the process, it helps to eliminate the fractions by finding a common denominator. This means if you multiply every term by the least common multiple of the denominators, the fractions will disappear, and solving the equation becomes simpler. Remember these steps when dealing with fractions in equations:

• Identify the denominators in the equation.
• Find the least common multiple of these denominators.
• Multiply each term by this least common multiple to clear the fractions.

Doing this helps to transform the equation into a more familiar one, without any fractions, making it easier to solve.

Least Common Multiple

The least common multiple (LCM) is a key concept when solving equations involving fractions. It is the smallest number that is a multiple of all the denominators in the equation. By multiplying each term by the LCM, you can eliminate fractions, simplifying the equation significantly. In the example provided, the denominators are 3, 5, and 15. The LCM of these numbers is 15, as 15 is the smallest number that all these denominators divide into without leaving a remainder. Here's how to find the LCM and use it to solve the equation:

• List the multiples of each denominator until you find the smallest common multiple.
• Multiply every term in the equation by this LCM.
• The fractions are now eliminated, allowing you to solve the equation much like solving a standard linear equation.

Using the LCM empowers you to break down complex equations with fractions into more straightforward, hassle-free problems. It's all about making things easier to handle.

Checking Solutions

Once you've calculated a potential solution to the equation, it's important to ensure it is correct. This step is called "checking solutions," and it helps verify your calculated answer.In the provided example, after solving the equation, you found that \( a = 0 \). To check if this solution is correct, substitute \( a = 0 \) back into the original equation. Doing so will confirm whether both sides of the equation remain equal with this value of \( a \). This is how you check your solution:

• Substitute the solution back into the original equation.
• Simplify the equation to see if both sides match.
• If both sides are equal, your solution is verified as correct.

In the example, substituting \( a = 0 \) gives \( \frac{1}{3} + \frac{1}{5} = \frac{8}{15} \), confirming both sides are equal. Therefore, your solution \( a = 0 \) is indeed correct. Always remember that checking your solutions is an integral part of solving equations to ensure accuracy.