Problem 75 - Prealgebra

Question

The problems below review some of the material on solving equations. Reviewing these problems will help you with the next section. Solve. $$\frac{1}{7} a=-7$$

Step-by-Step Solution

Answer

The solution is \( a = -49 \).

Step 1: Understand the Equation

The given equation is \( \frac{1}{7} a = -7 \). Here, the variable \( a \) is being multiplied by \( \frac{1}{7} \). Our goal is to solve for \( a \).

Step 2: Clear the Fraction

To eliminate the fraction, multiply both sides of the equation by 7. This gives:\[ 7 \times \left( \frac{1}{7} a \right) = 7 \times (-7) \]This simplifies to \( a = -49 \).

Step 3: Verify the Solution

Substitute \( a = -49 \) back into the original equation to check our work:\[ \frac{1}{7} \times (-49) = -7 \]Calculating the left side gives \( -7 \), which matches the right side of the equation. This confirms our solution is correct.

Key Concepts

Prealgebra

Prealgebra is the foundational mathematics that prepares you for the study of algebra. It includes basic arithmetic like addition, subtraction, multiplication, and division, as well as introductory concepts in algebra. Understanding prealgebra is essential because it lays the groundwork for more complex mathematical concepts. In this context, our exercise is an example of solving a simple equation, which is a key skill in prealgebra. When tackling prealgebra problems, the first step is to understand the elements involved. For instance, in the equation \( \frac{1}{7} a = -7 \), recognizing that \( \frac{1}{7} \) is a fraction and \( a \) is a variable is important. Being comfortable with variables and how they interact with numbers is crucial in prealgebra. This exercise helps you learn to solve for unknowns by performing inverse operations, a fundamental concept that will be used in algebra and beyond.

Fraction Multiplication

Fraction multiplication is a critical operation in mathematics, especially when dealing with equations involving fractions. To multiply a fraction by a number, or another fraction, you multiply the numerators together to get the new numerator, and the denominators together to get the new denominator.Let's apply this with our example equation \( \frac{1}{7} a = -7 \). Here, \( \frac{1}{7} \) is being multiplied by \( a \). To solve for \( a \), we aim to "clear" the fraction. This means we need to eliminate the fraction so that \( a \) stands alone. We do this by multiplying both sides of the equation by 7, which is the reciprocal of \( \frac{1}{7} \).

• Multiplying \( 7 \times \left( \frac{1}{7} a \right) \) simplifies to \( a \) because \( 7 \times \frac{1}{7} = 1 \).
• Doing the same to the other side, \( 7 \times (-7) \), gives us \( -49 \).

The fraction disappears, making it easier to see that the solution is \( a = -49 \). This step is crucial because it demonstrates how multiplication can simplify fractions within equations.

Equation Verification

Equation verification is the process of checking your solution to ensure that it satisfies the original equation. After finding a solution, it’s important to verify it by plugging the value back into the original equation. This confirms your work and helps prevent mistakes.In our example, after solving for \( a \) and finding \( a = -49 \), we substituted this value back into the original equation: \( \frac{1}{7} a = -7 \). When we substitute \( a \) with \(-49\), the equation becomes:\[ \frac{1}{7} \times (-49) = -7 \]This simplifies to \(-7\), which matches the right side of the equation, \(-7\). Since both sides of the equation are equal, this confirms that our solution \( a = -49 \) is indeed correct. Verification is an essential part of solving equations. It provides confidence in your answer and ensures that no calculation errors were made along the way. Each time you solve an equation, take a moment to verify your solution to reinforce this critical problem-solving habit.