Problem 13 - Elementary Algebra

Question

Solve each of the equations. $$s=42+0.4 s$$

Step-by-Step Solution

Answer

The solution is \(s = 70\).

Step 1: Move terms to one side

The first step is to simplify the equation by moving all terms involving the variable \(s\) to one side of the equation. Start with the equation:\[ s = 42 + 0.4s \]Subtract \(0.4s\) from both sides:\[ s - 0.4s = 42 \]

Step 2: Simplify the equation

Simplify the left side of the equation by combining like terms. The equation becomes:\[ 0.6s = 42 \]

Step 3: Solve for the variable

Now solve for \(s\) by dividing both sides of the equation by 0.6 to isolate \(s\):\[ s = \frac{42}{0.6} \]

Step 4: Calculate the result

Divide 42 by 0.6 to find the value of \(s\):\[ s = 70 \]

Key Concepts

Linear Equations

Linear equations are fundamental components of algebra that involve variables raised to the first power and set equal to a constant or another linear expression. In the exercise we encountered, the equation \( s = 42 + 0.4s \) is a linear equation because both sides of the equation ultimately boil down to linear terms without any variables raised to a power higher than one. Linear equations typically take the form of \( ax + b = c \), where "x" is the variable, and "a", "b", and "c" are constants. In our given example, the equation involves the variable "s" and constant values 42 and 0.4. Understanding linear equations is crucial as they form the basis for more complex mathematical operations. By grasping how they work, you can resolve situations where a variable needs to be solved in a straightforward and systematic manner.

Isolation of Variables

Isolation of variables refers to the process of manipulating an equation to get the variable of interest, typically represented by a letter, alone on one side of the equation. This is essential in solving equations as it allows you to find the important values. How do we achieve this? Let's consider our equation: initially, it was \( s = 42 + 0.4s \). Our goal was to have "s" by itself for easy calculation and solution. We did this by subtracting \( 0.4s \) from both sides, moving all terms involving "s" to one side:

• Original: \( s = 42 + 0.4s \)
• Subtract \( 0.4s \) from both sides: \( s - 0.4s = 42 \)
• This results in: \( 0.6s = 42 \)

What we did here was embrace the balance method! Each step ensures both sides remain equal, and we bring variables to one side for clarity.

Simplification of Expressions

Simplification involves rewriting expressions in their simplest form. It is a crucial step in solving equations because it reduces complex expressions into an easily manageable state. In our case, combining terms that involve the same variable is a simplification step, where we simplified \( s - 0.4s \) to \( 0.6s \).

• The expression \( s = 0.4s + 42 \) simplifies to \( 0.6s = 42 \) after subtracting \( 0.4s \) from both sides.
• We recognized that \( s \) and \( 0.4s \) are similar terms because they both involve the variable "s".

By combining these terms, we reduced the complexity of the equation, making it easier to solve. Simplifying is also a way to catch errors and streamline calculations, so it's a valuable skill, not just for math but for problem-solving in general.

Combining Like Terms

Combining like terms is a part of the simplification process where terms with the same variable or the same powers are added or subtracted from each other. In the context of our exercise, we started with the equation \( s - 0.4s = 42 \), which included terms that were "like" because they each contained the variable "s."

• Initial expression: \( s - 0.4s = 42 \)
• Recognize both "s" terms: \( s = 1 \times s \) and \( 0.4s \)
• Combine them: \( 0.6s \) which came from performing \( 1s - 0.4s \).

By combining these like terms, the equation became much simpler to handle, allowing us to isolate the variable efficiently. It aligns with the goal of getting the variable on one side and all constants or different terms on the other side. This keeps everything neat and avoids unnecessary complications when solving linear equations.