Problem 10

Question

Solve the equations and inequalities for the following problems. $$ \text { Solve } 2 p-6 q+1=-2 \text { for } p $$

Step-by-Step Solution

Verified
Answer
Question: Solve the given linear equation for the variable p: \(2p - 6q + 1 = -2\) Answer: \(p = \frac{6q - 3}{2}\)
1Step 1: Identify the given equation
The equation we are given is: $$ 2p - 6q + 1 = -2 $$
2Step 2: Move all terms containing "q" and constants to the other side
To isolate p, move the \(-6q\) term and the constant term to the right side of the equation. This can be done by performing the following steps: 1. Add \(6q\) to both sides of the equation 2. Subtract 1 from both sides of the equation Doing so, we get: $$ 2p = 6q - 2 - 1 $$
3Step 3: Simplify the right side of the equation
Combine the constants \(-2\) and \(-1\) on the right side to get: $$ 2p = 6q - 3 $$
4Step 4: Solve for "p"
Now, to solve for p, we'll need to get rid of the coefficient in front of it, which is 2. Divide both sides of the equation by 2 to isolate p: $$ p = \frac{6q - 3}{2} $$
5Step 5: Final Answer
After isolating p, our final answer is as the equation: $$ p = \frac{6q - 3}{2} $$

Key Concepts

Understanding AlgebraThe Art of Isolating VariablesUnpacking Linear Equations
Understanding Algebra
Algebra is a fascinating branch of mathematics that revolves around working with symbols and letters that represent numbers. This makes it incredibly versatile when dealing with unknown quantities. In this context, variables like \( p \) and \( q \) stand in for specific values. Mastering algebra involves becoming comfortable with how these symbols interact in equations.

Algebra is often likened to a toolbox, whereby each technique or rule is a tool used to solve problems. Key tools include understanding how to manipulate variables, perform operations, and apply rules like the distributive property or the rules for balancing equations.
  • Variables act as placeholders for numerical values.
  • Operations include addition, subtraction, multiplication, and division.
By accurately using these tools, you will be able to translate real-world situations into mathematical language, allowing engineers, scientists, and economists to solve complex problems. In our problem, we utilized algebraic techniques to isolate \( p \), which helps us see what \( p \) equals in terms of \( q \).
The Art of Isolating Variables
Isolating variables is a critical skill in algebra, especially when solving equations. It involves rearranging an equation to get one specific variable by itself on one side of the equation. The ultimate goal is to express the variable in terms of other variables or constants. In our problem, we aimed to solve for \( p \).

The process of isolating a variable often involves several steps:
  • Move terms involving the variable to one side.
  • Move other terms to the opposite side.
  • Simplify as needed to get the variable alone.
In our specific example, we had \( 2p - 6q + 1 = -2 \). We started by moving all terms involving \( q \) and constant numbers to the right side, eventually leading to \( 2p = 6q - 3 \). By dividing all terms by 2, we could finally isolate \( p \) as \( p = \frac{6q - 3}{2} \). Understanding the steps and rationale behind each move helps clarify the process of solving equations.
Unpacking Linear Equations
Linear equations are a staple in algebra. These equations form straight lines when graphed and typically follow the format \( ax + b = c \). They are characterized by variables that are only to the first power, meaning there's no squaring or other higher-degree power involved.

In our problem, the linear equation was \( 2p - 6q + 1 = -2 \). Recognizing this helps set the path for solving it, as linear equations have consistent solutions that involve straightforward methodologies:
  • Start by simplifying and rearranging terms if necessary.
  • Make sure each variable term is on one side and constants on the other.
  • Isolate the variable you're solving for.
This type of equation is simple yet powerful, showing how even basic mathematical rules can solve complex problems. Linear equations are omnipresent in various fields, providing insights and solutions across a multitude of scenarios.