Problem 14 - Algebra 2 and Trigonometry

Question

In \(3-20,\) solve each equation and check. $$ \frac{3}{a}+\frac{1}{2}=\frac{11}{5 a} $$

Step-by-Step Solution

Answer

The solution is \(a = \frac{-8}{5}\).

Step 1: Combine Terms with a Common Denominator

To eliminate the fractions, find a common denominator for all the fractions in the equation. The denominators are \(a\), \(2\), and \(5a\). The least common denominator (LCD) is \(10a\). Multiply every term by the LCD to get:\[10a \times \left( \frac{3}{a} + \frac{1}{2} \right) = 10a \times \frac{11}{5a}\] which simplifies to:\[30 + 5a = 22\]

Step 2: Isolate the Variable

Simplify the equation by solving for \(a\):Subtract 30 from both sides to isolate terms involving \(a\):\[5a = 22 - 30\]which simplifies to:\[5a = -8\]

Step 3: Solve for the Variable

Continue by solving for \(a\) by dividing both sides by 5:\[a = \frac{-8}{5}\]

Step 4: Check the Solution

Substitute \(a = \frac{-8}{5}\) back into the original equation to verify the solution is correct:Substitute into the left side:\[\frac{3}{\left(\frac{-8}{5}\right)} + \frac{1}{2} = -\frac{15}{8} + \frac{1}{2}\]Combine these fractions by converting \(\frac{1}{2}\) to \(\frac{4}{8}\):\[-\frac{15}{8} + \frac{4}{8} = -\frac{11}{8}\]Substitute into the right side:\[\frac{11}{5 \times \left(\frac{-8}{5}\right)} = -\frac{11}{8}\]Both sides equal \(-\frac{11}{8}\), confirming that the solution is correct.

Key Concepts

Solving Equations

Solving equations is the process of finding the value of the unknown variable that makes the equation true. In algebra, equations can often contain fractions or multiple terms which can make them appear complex. The goal is to manipulate the equation so that the variable stands alone on one side of the equation.

• First, identify all the terms that involve the variable.
• Next, use basic algebraic operations such as addition, subtraction, multiplication, or division to isolate the variable.

Once the variables are isolated, you can solve for the variable by performing the inverse operation. This step-by-step action is vital for confirming the exact point where each variable balances the equation.

Least Common Denominator

When dealing with equations involving fractions, finding the least common denominator (LCD) is a key step. This technique helps clear out fractions, making equations simpler to handle.

• The LCD is the smallest multiple that is divisible by all the denominators in the equation.
• For instance, in the exercise, the denominators are \(a\), \(2\), and \(5a\), so the LCD is \(10a\).

Once you determine the LCD, you can multiply each term in the equation by it, thereby eliminating the fractions and simplifying the equation considerably. This method sets up a clear path to solve the problem without the complexity of fractional calculations.

Fractions

Fractions can make equations look daunting, but they are approachable with the right methods. Fractions are numbers expressed as one whole number divided by another. In algebra, we often need to deal with fractions by:

• Adding, subtracting, multiplying, or dividing them just like whole numbers.
• Finding a common denominator to perform addition or subtraction easily.

Understanding how to manipulate fractions through equivalent fractions, finding reciprocals, or common denominators can greatly enhance your problem-solving skills. Remember that converting fractions using the least common denominator simplifies the process of solving equations involving multiple fractions.

Checking Solutions

Once you solve an equation, it is essential to verify your solution. This ensures you did not make any mistakes during your calculations. Here's how you can check your solutions effectively:

• Substitute the value you obtained for the variable back into the original equation.
• Simplify both sides of the equation step by step.
• If both sides are equal, your solution is correct.

For example, with \(a = -\frac{8}{5}\), substituting back into the original equation and simplifying, confirms that both sides equal -\(\frac{11}{8}\). This reassures that the solution was found accurately and confirms the coordination between the original condition and your answer.