Problem 70 - Elementary Algebra

Question

The difference of \(5 x\) and 1 is equal to the sum of \(12 x\) and 1 .

Step-by-Step Solution

Answer

\(x = -\frac{2}{7}\)

Step 1: Write the equation

Based on the problem statement, the equation formed is the difference of \(5x\) and 1 equal to the sum of \(12x\) and 1. This translates to: \(5x - 1 = 12x + 1\).

Step 2: Move all terms involving "x" to one side

Subtract \(12x\) from both sides to get the terms involving \(x\) on one side. This gives: \(5x - 12x = 1 + 1\).

Step 3: Simplify the equation

Simplify both sides of the equation. The left-hand side becomes \(-7x\) and the right-hand side is 2, so the equation is \(-7x = 2\).

Step 4: Solve for "x"

Divide both sides of the equation by \(-7\) to solve for \(x\). This results in \( x = -\frac{2}{7} \).

Key Concepts

Equation Solving

Solving an equation involves finding values for variables that make the equation true. In our example, we started with the equation \(5x - 1 = 12x + 1\). The goal is to find what value of \(x\) satisfies this condition. Just like a scale, each side of the equation needs to be balanced.

• Begin by carefully reading the problem to translate it into an equation.
• Identify operations (addition, subtraction, multiplication, division) needed to isolate the variable.
• Perform inverse operations to move terms from one side of the equation to the other, maintaining balance.

This technique is essential, forming the foundation of algebra for problem-solving across many types of equations.

Linear Equations

Linear equations represent straight lines when graphed. They are essential in modeling relationships where one variable depends linearly on another. Our original equation is a classic linear form, meaning it has degree one.
To spot a linear equation, look for:

• Variables raised only to the first power
• No products of variables
• A constant term that may or may not be present

In standard form, a linear equation looks like \(ax + b = c\). In our case, \(5x - 1 = 12x + 1\) fits this structure. Linear equations are straightforward because they only involve basic arithmetic operations, making them accessible for solving.

Variable Isolation

Isolating the variable means getting the variable by itself on one side of the equation. This process helps clarify the solution for the unknown. Starting from \(5x - 1 = 12x + 1\), we move terms involving \(x\) to one side. Subtracting \(12x\) from both sides results in \(-7x = 2\).

• Opposite operations are used for isolation, like addition to counteract subtraction.
• Remember to perform the same operation on both sides of the equation to keep it balanced.
• When a variable is multiplied by a number, divide both sides by that number.

Finally, dividing by \(-7\) gives \( x = -\frac{2}{7} \). Being able to isolate variables effectively not only solves the current problem but also prepares you for more complex algebraic expressions.