Problem 8 - Elementary Algebra

Question

For Problems 1-12, solve each equation. You will be using these types of equations in Problems \(13-41\). $$ 0.2(15)-0.2 x+x=0.4(15) $$

Step-by-Step Solution

Answer

The solution to the equation is \( x = 3.75 \).

Step 1: Simplify Both Sides of the Equation

First, distribute the numbers outside the parentheses on both sides of the equation. On the left side, compute: \[ 0.2(15) = 3 \] This transforms the original equation to: \[ 3 - 0.2x + x = 0.4(15) \]On the right side of the equation, calculate: \[ 0.4(15) = 6 \]This transforms the equation to: \[ 3 - 0.2x + x = 6 \]

Step 2: Combine Like Terms on the Left

Combine the terms involving 'x' on the left side of the equation. The equation now is:\[ -0.2x + x = 0.8x \]So, simplify the left side as:\[ 3 + 0.8x = 6 \]

Step 3: Isolate the Variable Term

Subtract 3 from both sides of the equation to get the term with the variable 'x' by itself:\[ 0.8x = 6 - 3 \]This simplifies to:\[ 0.8x = 3 \]

Step 4: Solve for 'x'

Divide both sides of the equation by 0.8 to solve for 'x'.\[ x = \frac{3}{0.8} \]Calculate \( \frac{3}{0.8} \) which simplifies to 3.75.Therefore, \( x = 3.75 \).

Key Concepts

Distributive Property

The distributive property is a fundamental algebra concept used to simplify expressions and equations. It allows us to remove parentheses by distributing a number or variable outside the parentheses to each term within the parentheses.
This property is symbolically represented as: \[ a(b + c) = ab + ac \] In the exercise given, the distributive property was applied to both sides of the equation.

• On the left side, \( 0.2 \) was distributed to \( 15 \), resulting in \( 0.2 \times 15 = 3 \).
• On the right side, \( 0.4 \) was distributed to \( 15 \), leading to \( 0.4 \times 15 = 6 \).

This step is crucial as it transforms an equation that includes parentheses into a simpler arithmetic expression, making it easier to work with.

Combining Like Terms

Combining like terms is a key process in simplifying algebraic expressions, making calculations more straightforward. "Like terms" are terms that have identical variable parts, meaning their variables and the powers of those variables are the same.
In the given equation after distribution, we have terms involving 'x', \( -0.2x + x \).

• To simplify, combine these to form \( 0.8x \).

This represents a significant step in algebra as it reduces the complexity and helps in achieving a cleaner expression. It aids in making the equation more manageable by grouping similar terms together, which is essential before attempting to isolate the variable.

Isolating Variables

Isolating the variable is a critical skill in solving equations. It involves rearranging the equation so that the variable is alone on one side of the equation. In our example, after combining like terms, we had:
\[ 3 + 0.8x = 6 \]
To isolate \( x \), we need to move all constant terms to the opposite side.

• Subtract 3 from both sides, resulting in \( 0.8x = 3 \).

This is an important procedure because once the variable is isolated, we can directly solve for it. Step-by-step manipulation and understanding of these adjustments are important to maintain the equation’s balance and ensure a correct solution.

Solving Linear Equations

Solving linear equations involves finding the value of the variable that makes the equation true. After isolating \( x \), we had the equation:
\[ 0.8x = 3 \] To find \( x \), divide both sides by the coefficient of \( x \), which is 0.8.

• This process results in \( x = \frac{3}{0.8} \).
• Carry out the division to get \( x = 3.75 \).

This final step confirms the solution to the equation. Solving linear equations often requires a sequence of operations, such as adding, subtracting, multiplying, or dividing, all aimed at simplifying the equation until the solution for the variable is clear and precise.