Problem 76 - Prealgebra

Question

The problems below review some of the material on solving equations. Reviewing these problems will help you with the next section. Solve. $$\frac{1}{5} a=-3$$

Step-by-Step Solution

Answer

The value of \( a \) is \(-15\).

Step 1: Understand the Equation

The given equation is \( \frac{1}{5}a = -3 \). This is a simple linear equation where \( a \) is multiplied by \( \frac{1}{5} \), meaning one-fifth of \( a \) is equal to \(-3\). Our goal is to solve for \( a \).

Step 2: Isolate the Variable

To isolate \( a \), we need to eliminate the fraction by multiplying both sides of the equation by 5. This gives us:\[ 5 \times \frac{1}{5}a = 5 \times (-3) \]

Step 3: Simplify Both Sides

On the left side, \( 5 \times \frac{1}{5} \) simplifies to 1, leaving us with \( a \). On the right side, \( 5 \times (-3) \) equals \(-15\). So we now have:\[ a = -15 \]

Step 4: Verify the Solution

Substitute \( a = -15 \) back into the original equation to verify:\[ \frac{1}{5} (-15) = -3 \]Calculate \( \frac{1}{5} \times -15 \) which equals \(-3\). Therefore, the original equation holds true.

Key Concepts

Isolate the Variable

Isolating the variable is a key step in solving linear equations. In the equation \( \frac{1}{5}a = -3 \), the goal is to solve for \( a \). This means we need \( a \) by itself on one side of the equation. The variable is currently multiplied by \( \frac{1}{5} \). To isolate \( a \), we do the opposite of multiplying by \( \frac{1}{5} \), which is multiplying by 5.
By performing this operation on both sides:\[ 5 \times \frac{1}{5}a = 5 \times (-3) \]
The 5 on the left cancels with the fraction \( \frac{1}{5} \), leaving \( a \) alone. This is a fundamental concept because isolating the variable turns a complex equation into a simple statement: what the variable equals. This step is crucial for solving equations because it reveals the value of the unknown directly.

Verify the Solution

After finding a solution, verifying helps ensure it's correct. This step involves plugging the solution back into the original equation to see if it holds true. Here, we found \( a = -15 \) for \( \frac{1}{5}a = -3 \).
Substitute \( -15 \) in place of \( a \) in the equation:\[ \frac{1}{5} (-15) = -3 \]
Calculating gives \( -3 = -3 \), confirming the left side equals the right side.
Verification is important because it checks your work and ensures no mistakes were made. This can also build confidence in your problem-solving skills, knowing that the solution is accurate.

Simplifying Equations

Simplifying equations involves reducing expressions to their simplest form. It makes solving equations much easier and more manageable. In the solution \( a = -15 \), simplification was key. We first simplified the left side by canceling \( 5 \times \frac{1}{5} \), resulting in 1.
Similarly, practicing simplification can involve:

• Combining like terms
• Addressing parentheses by distributing coefficients
• Cleaning up fractions and decimals

These practices streamline equations and help in effectively solving them. More stable and reliable solution paths are established, minimizing potential errors down the line. Simplifying early in the process paves the way for easier and clearer steps throughout any equation handling.