Problem 3 - Elementary Algebra

Question

For Problems 1-12, solve each equation. You will be using these types of equations in Problems \(13-41\). $$ 0.2(20)+x=0.3(20+x) $$

Step-by-Step Solution

Answer

The solution to the equation is \( x = 2.857 \).

Step 1: Simplify both sides

The first step involves simplifying the given equation. Start by distributing the numbers outside the parentheses to the terms inside them.For the left side: \[ 0.2(20) = 4 \]For the right side:\[ 0.3(20 + x) = 0.3 \times 20 + 0.3 \times x = 6 + 0.3x \] Now, the equation looks like this:\[ 4 + x = 6 + 0.3x \]

Step 2: Rearrange terms

Move all terms involving the variable \(x\) to one side and the constant terms to the other side.Subtract \(0.3x\) from both sides to isolate the \(x\) terms:\[ x - 0.3x = 6 - 4 \]This simplifies to:\[ 0.7x = 2 \]

Step 3: Solve for x

The equation now involves isolating \(x\) by dividing both sides by \(0.7\):\[ x = \frac{2}{0.7} \]Upon performing the division, you will find:\[ x = 2.857 \]

Key Concepts

Solving Equations

At its core, solving an equation means finding the value of the variable that makes the equation true. For example, if you have an equation like \( 4 + x = 6 + 0.3x \), your main goal is to find the value of \( x \) that will make both sides of the equation equal.

To effectively solve equations, you often need to use different mathematical operations. These operations include addition, subtraction, multiplication, and division. The aim is to "undo" the operations that are affecting the variable. This allows you to isolate the variable and solve for its value. In practice, you will apply these operations across both sides of the equation to maintain equality.

• Perform operations in reverse order of mathematical operations.
  
• Apply the same operation to both sides to keep the equation balanced.
  
• Your final answer is the solution for the variable.
  

By mastering the steps and operations used in solving equations, you can solve more complicated equations with ease.

Distribution Property

The distribution property is a fundamental concept in algebra that aids in solving equations. It allows you to eliminate parentheses by distributing a factor across the terms inside the parentheses. For example, consider the expression \( 0.3(20 + x) \). By applying the distribution property, you multiply \(0.3\) with each term inside the parentheses separately:

• Multiply \( 0.3 \) by \( 20 \) to get \( 6 \).
• Multiply \( 0.3 \) by \( x \) to get \( 0.3x \).

The distributed expression becomes \( 6 + 0.3x \).
Using the distribution property simplifies equations and makes it easier to move forward with solving them. Practice applying this property with different numbers and variables to gain confidence in distributing terms in algebraic equations.

Simplification of Equations

Simplifying equations involves combining like terms and removing unnecessary parts to make the equation easier to solve. Once you've applied the distribution property, you'll often find yourself with multiple terms that can be liked together or simplified further.

For instance, in the equation \( 4 + x = 6 + 0.3x \), notice that different constant terms and variable terms are present. The goal of simplification is to consolidate these components:

• Combine constants on each side if possible.
• Combine like terms involving the variable.
  

After simplifying using these techniques, the equation becomes more manageable, making further steps such as rearranging terms straightforward.

Isolation of Variables

Isolating the variable is about getting the variable you're solving for alone on one side of the equation. This helps you see exactly what the value of the variable is.

In our example, after distributing and simplifying, we had the equation \( 0.7x = 2 \). To isolate \( x \), we simply needed to divide both sides by \( 0.7 \).

• Perform division across both sides to keep equation balanced.
• The solution is the value of the isolated variable.

Once \( x \) is alone, it's easy to determine its value as \( x = 2.857 \).
Mastering the concept of isolation will greatly assist in managing complex equations.