Problem 92 - Introductory Algebra for College Students

Question

Solve for \(s: \quad P=2 s+b\) (Section 2.4, Example 3)

Step-by-Step Solution

Answer

The solution to the equation for s is \(s = (P - b) / 2\).

Step 1: Subtract \(b\) from both sides

The goal in this step is to undo the addition of \(b\) to \(2s\). Do that by subtracting \(b\) from both sides of the equation. This leads to the new equation: \(P - b = 2s\).

Step 2: Divide by 2

Now, to get \(s\) alone, divide both sides of the equation by 2. This undoes the multiplication in the term \(2s\). This leads to the new equation: \(s = (P - b) / 2\).

Step 3: Check the solution

To check if this expression for \(s\) is correct, you can substitute \(s = (P - b) / 2\) into the original equation and it should still hold true.

Key Concepts

Solving for a Variable

Solving for a variable is a fundamental skill in algebra. It involves isolating the variable you're interested in on one side of the equation. Essentially, you're trying to "free" the variable from any numbers or terms attached to it.
For instance, in the equation \(P = 2s + b\), we want to solve for \(s\). This means we need \(s\) to be by itself on one side of the equation. Here’s a simplified guide:

• Identify which variable you need to solve for.
• Consider what operations are currently applied to that variable.
• Use inverse operations to "undo" those actions, step by step.

Once you successfully isolate the variable, you’ve solved the equation. It's crucial to ensure your final answer is correct by plugging it back into the original equation to see if it holds true.

Equation Manipulation

Equation manipulation refers to the methods used to rearrange an equation to isolate a variable. The goal is to perform steps that simplify the equation without altering its original balance.
One effective way to manipulate an equation like \(P = 2s + b\) is by performing inverse operations:

• If a term is added to the variable, subtract it from both sides.
• If a term is multiplied by the variable, divide it on both sides.

In our case, we subtract \(b\) to remove it from the right-hand side, simplifying to \(P - b = 2s\). Next, we divide everything by 2 to isolate \(s\), resulting in \(s = (P - b)/2\).
Remember, all steps must maintain equality. That’s why every operation done to one side should be done to the other side as well.

Introductory Algebra

Introductory algebra serves as the cornerstone for more advanced mathematics. It revolves around understanding variables, constants, equations, and their relationships. In this stage, learning revolves around simpler equations and basic manipulation techniques to solve them.
Algebra terms:

• Variables: Symbols that represent numbers (like \(s\) in our problem).
• Constants: Numbers on their own (like \(b\)).
• Operations: Actions performed (addition, subtraction, multiplication, division).

In introductory algebra, you focus on solving basic linear equations, such as one-variable equations, using techniques like combining like terms and using inverse operations.
Success in introductory algebra helps a smooth transition to more challenging topics like quadratic equations and polynomials in higher-level math.