Problem 6 - 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. $$ 800(0.11) t=1600 $$

Step-by-Step Solution

Answer

The solution is \( t = 18.18 \).

Step 1: Identify the Equation

The given equation is \(800(0.11)t = 1600\). We need to solve for the variable \(t\).

Step 2: Simplify the Equation

First, simplify the left side of the equation. Calculate \(800 \times 0.11\):\[ 800 \times 0.11 = 88 \]So, the equation becomes \(88t = 1600\).

Step 3: Solve for the Variable

Now, isolate \(t\) by dividing both sides of the equation by 88:\[ t = \frac{1600}{88} \]

Step 4: Perform the Division

Calculate the division on the right side:\[ t = \frac{1600}{88} = 18.18\, (rounded\, to\, two\, decimal\, places)\]

Key Concepts

Algebra

Algebra is the branch of mathematics that deals with symbols and the rules for manipulating these symbols. In the equation given, the symbol used is \( t \), which represents a variable. A variable is a symbol used to represent an unknown number or value. In algebra, we often deal with equations, which are mathematical statements indicating that two expressions are equal. The key objective in algebra is to solve equations by finding the value of the variable that makes the equation true.

When tackling algebraic problems, we use 'operators' such as addition, subtraction, multiplication, and division to manipulate the symbols. Here, the main goal was to find the value of \( t \), an unknown quantity in the equation \( 88t = 1600 \). This involves using certain algebraic techniques that maintain the balance of the equation.

Simplifying Equations

Simplifying equations is a crucial step in solving algebraic problems. It involves reducing the equation to its simplest form to make solving easier.

In our example, the original equation was \( 800(0.11)t = 1600 \). Simplification starts by calculating any constants or products. Here, we simplified \( 800 \times 0.11 \), resulting in \( 88 \).

• Identify constants and numbers that can be calculated.
• Carry out operations like multiplication or addition.
• Rewrite the equation in a simpler form.

Upon simplifying, the equation becomes \( 88t = 1600 \). The process of simplifying helps make the problem more manageable and reduces the chance of errors in subsequent steps.

Variable Isolation

Variable isolation is the process of manipulating an equation to have the variable by itself on one side.

In our problem, the aim was to isolate \( t \) in the equation \( 88t = 1600 \). Isolation is accomplished through inverse operations. In this case, since \( t \) was multiplied by 88, we did the inverse, which is division, to isolate \( t \).

• Determine the operation being used on the variable.
• Apply the opposite operation to both sides of the equation.
• Simplify until the variable is alone on one side.

Thus, by dividing both sides by 88, we found that \( t = \frac{1600}{88} \), which simplifies to 18.18. This step is essential in ensuring that we correctly solve for the variable in any algebraic equation.