Problem 11 - Elementary Algebra

Question

In the following exercises, solve using the problem solving strategy for word problems. Remember to write a complete sentence to answer each question. Philip pays \(\$ 1,620\) in rent every month. This amount is \(\$ 120\) more than twice what his brother Paul pays for rent. How much does Paul pay for rent?

Step-by-Step Solution

Answer

Paul pays \( \$750 \) in rent every month.

Step 1: Identify the Unknown

Let’s start by identifying what we need to find. We are asked to find how much Paul pays for rent. Let's call this amount \( x \).

Step 2: Write the Equation

According to the problem, Philip’s rent is \( \$1,620 \), and this amount is \( \$120 \) more than twice Paul's rent. Mathematically, this can be written as an equation: \[ 1620 = 2x + 120 \].

Step 3: Isolate the Variable

To find \( x \), first subtract \( 120 \) from both sides of the equation: \[ 1620 - 120 = 2x \]. This simplifies to \[ 1500 = 2x \].

Step 4: Solve for the Variable

Now, divide both sides of the equation by \( 2 \) to solve for \( x \): \[ x = \frac{1500}{2} \]. This gives us \[ x = 750 \].

Step 5: Write the Complete Sentence

Paul pays \( \$750 \) in rent every month.

Key Concepts

solving equations

Solving equations is a fundamental concept in algebra. It involves finding the value of a variable that makes the equation true. In word problems, this process begins by translating the words into a mathematical equation. For example, in our exercise, we translated 'Philip’s rent is \(1,620, which is \)120 more than twice Paul’s rent' into the equation: ' 1620 = 2x + 120 '. Learning to convert a word problem into a solvable equation is key to mastering algebra and solving equations.

variable isolation

Variable isolation means getting the variable by itself on one side of the equation. This step is necessary to find the value of the unknown quantity. In our example, we started with the equation: |c| |||b|t|d| | ' 1620 = 2x + 120 '. To isolate 'x' , we first subtracted '120' from both sides: ' 1620 - 120 = 2x ' . This simplified the equation to: ' 1500 = 2x ' . Then, we divided both sides by '2' to get: ' x = 750 ' . Hence, ' 750 ' is the value of Paul's rent. Mastering variable isolation is crucial as it forms the basis for solving most algebraic equations.

basic algebraic operations

Basic algebraic operations include addition, subtraction, multiplication, and division. These operations allow us to manipulate equations and solve for unknown variables. In our problem, we used:

• Subtraction: To subtract '$120' from both sides of the equation to simplify it. This gave: ' 1620 - 120 = 2x ' which simplifies to ' 1500 = 2x '
• Division: To divide both sides by '2' to solve for 'x', resulting in: ' x = 750 '.

These steps show how basic operations form the building blocks of solving equations. Understanding and practicing these operations is vital for anyone learning algebra.

real-life applications of algebra

Algebra is not just about solving abstract problems; it has real-life applications. For instance, our exercise deals with rent calculations. Such problems can help you:

• Budget: You can determine how much you need to save for monthly expenses.
• Comparison: Compare different financial scenarios, such as determining who pays more or less.
• Planning: Make informed decisions about renting, buying, or investing in properties.

By understanding word problems and solving them using algebra, students can apply these skills in daily life. It makes the subject not just about numbers and letters, but about making meaningful connections in the world around us.