Problem 11 - Elementary Algebra

Question

For Problems \(1-12\), solve each of the equations. These equations are the types you will be using in Problems 13-40. $$ 24\left(t-\frac{2}{3}\right)=18 t+8 $$

Step-by-Step Solution

Answer

The solution is \( t = 4 \).

Step 1: Distribute 24 on the Left Side

Start by distributing the number 24 to both terms inside the parentheses on the left side of the equation. This gives us:\[ 24t - 24 \times \frac{2}{3} = 18t + 8 \] Simplify \( 24 \times \frac{2}{3} \) to get \( 16 \). The equation now reads:\[ 24t - 16 = 18t + 8 \]

Step 2: Move All Terms Involving t to One Side

Subtract \( 18t \) from both sides of the equation to bring all terms involving \( t \) to the left:\[ 24t - 18t - 16 = 8 \] Simplify to:\[ 6t - 16 = 8 \]

Step 3: Move Constant Terms to the Right Side

Add 16 to both sides to isolate the variable term on the left:\[ 6t - 16 + 16 = 8 + 16 \]Simplify to:\[ 6t = 24 \]

Step 4: Solve for t

Divide both sides by 6 to solve for \( t \):\[ \frac{6t}{6} = \frac{24}{6} \]This simplifies to:\[ t = 4 \]

Key Concepts

Distributive Property

The distributive property is a fundamental concept in algebra that helps simplify expressions when a number is multiplied by a sum or difference within parentheses. In our exercise, we have the equation \( 24(t - \frac{2}{3}) = 18t + 8 \). To apply the distributive property, we multiply the number 24 by each term inside the parentheses: \( 24t \) and \( 24 \times -\frac{2}{3} \).

• This results in \( 24t - 16 \), as 24 multiplied by \(-\frac{2}{3}\) equals -16.
• The key here is to ensure each term inside the parentheses is appropriately multiplied by the factor outside.

Once distributed, the equation becomes more manageable and simpler to manipulate for further steps.

Isolating Variables

Isolating the variable is crucial for solving equations. It involves rearranging the equation so the variable exists on one side by itself. Initially, we had the transformed equation \( 24t - 16 = 18t + 8 \). To isolate the variable \( t \), we need to get all terms involving \( t \) on one side of the equation.

• We do this by subtracting \( 18t \) from both sides, simplifying to \( 6t - 16 = 8 \).
• This step helps move all terms with the variable to one side, clearing the way to solve for \( t \).

Properly isolating the variable facilitates solving, ensuring you can manipulate the equation as needed.

Simplifying Equations

Simplifying equations involves combining like terms and reducing the equation step-by-step to solve for the unknown variable. After moving terms involving \( t \) to one side, the equation \( 6t - 16 = 8 \) needs further simplification.

• Adding 16 to both sides removes the constant term from the left, vital for isolating \( t \).
• You're left with \( 6t = 24 \), a much simpler equation.

By continually reducing the number of operations and terms, you make the equation easier to handle, ensuring clarity at every step, and preventing errors during calculations.

Basic Algebra

Basic algebra serves as the foundation for solving equations. It includes simple operations like addition, subtraction, multiplication, and division, which we use to manipulate equations and find the value of the unknown variable. In the final step of our solution, we have the equation \( 6t = 24 \).

• We divide both sides by 6 to discover \( t \). This is essentially performing the reverse operation to simplify the expression.
• Doing so, we find \( t = 4 \).

Understanding basic algebraic operations is key to solving linear equations effectively, allowing you to break down and tackle more complex problems with confidence and precision.