Problem 30 - Prealgebra

Question

Using the addition property of equality first, solve each of the following equations. $$-\frac{1}{5} a+3=7$$

Step-by-Step Solution

Answer

The solution to the equation is \(a = -20\).

Step 1: Understand the Equation

The given equation is \(-\frac{1}{5}a + 3 = 7\). Our task is to find the value of \(a\) by first eliminating the constant term on the left side using the addition property of equality.

Step 2: Isolate the Variable Term

To isolate the term involving \(a\), subtract 3 from both sides of the equation, which removes the constant on the left side: \[-\frac{1}{5}a + 3 - 3 = 7 - 3\] This simplifies to: \[-\frac{1}{5}a = 4\]

Step 3: Solve for \(a\)

To solve for \(a\), we need to remove the coefficient \(-\frac{1}{5}\) by performing the opposite operation. Multiply both sides by \(-5\) to cancel out \(-\frac{1}{5}\): \[-5 \times \left(-\frac{1}{5}a\right) = 4 \times (-5)\] This simplifies to: \[a = -20\]

Step 4: Verify the Solution

Substitute \(a = -20\) back into the original equation to check the solution: \[-\frac{1}{5}(-20) + 3 = 7\] Calculate the left-hand side: \[4 + 3 = 7\] Since both sides equal 7, the solution \(a = -20\) is verified.

Key Concepts

Solving Linear Equations

When we talk about solving linear equations, we mean finding the value of the variable that makes the equation true. A linear equation is an equation of the first degree, where the variable, usually represented by a letter like \(a\), appears with an exponent of 1.
Essentially, solving linear equations involves a few basic steps:

• First, simplify both sides of the equation if necessary, combining like terms and clearing out any parenthesis or fractions.
• Then, use properties of equality to transform the equation in such a way that the variable is isolated on one side. We'll delve into these properties soon.
• Finally, solve for the variable and verify your solution by substituting it back into the original equation to ensure both sides equal.

Solving linear equations is like peeling an onion; you remove one layer at a time until you reach the core, represented by the isolated variable. It is an essential skill in algebra and lays the groundwork for more advanced math topics.

Prealgebra Concepts

Prealgebra provides the basic tools necessary to solve equations and understand algebra. It's the foundation that introduces variables, operation properties, and the balance of equations. Here’s a quick rundown of some critical prealgebra concepts that relate to solving equations:

• Variables: These are symbols, often letters, that stand in for unknown numbers. In equations like \(-\frac{1}{5}a + 3 = 7\), \(a\) is our variable.
• Constants: These are numbers without variables attached. The number 3 in our equation is a constant.
• Coefficients: These numbers multiply a variable. Here, \(-\frac{1}{5}\) is the coefficient of \(a\).
• Equality: This denotes that two expressions are equivalent and is represented by the \(=\) sign.

Understanding these concepts helps students navigate equations and algebraic expressions efficiently. It allows for manipulation of equations to find unknown values, which is the cornerstone of algebra.

Isolate Variable in Equation

Isolating the variable in an equation means reworking the equation to get the variable on one side and everything else on the other. This often involves several operations and is a critical step in solving equations.
In our exercise,

• We started with the equation \(-\frac{1}{5}a + 3 = 7\).
• We used subtraction to remove the 3 from the left side, yielding \(-\frac{1}{5}a = 4\).
• Next, we need to address the coefficient \(-\frac{1}{5}\), which is attached to \(a\).

To isolate \(a\), we did the opposite of division by \(-5\) (flipping operations). Thus, multiplying both sides by \(-5\) freed \(a\) from the coefficient, resulting in \(a = -20\).
Through these steps, we use equality properties to keep the balance of the equation, ensuring that what we do to one side, we do to the other. This ensures the equation remains valid, and we achieve the solution accurately.