Problem 32 - Fundamentals of Mathematics

Question

Translate each phrase or sentence to a mathematical expression or equation. When four thirds of a number is increased by twelve, the result is five.

Step-by-Step Solution

Answer

The number is \(-5.25\).

Step 1: Identify the Variable

To translate the given sentence into a mathematical expression, we start by identifying the unknown number in the sentence. Let's denote this unknown number as \( x \).

Step 2: Translate 'Four Thirds of a Number'

The phrase "four thirds of a number" refers to the fraction \( \frac{4}{3} \times x \). This is the starting point of our mathematical expression.

Step 3: Incorporate 'Increased by Twelve'

Next, the phrase "increased by twelve" means we need to add 12 to the previous expression. So, \( \frac{4}{3}x + 12 \) now represents this part of the sentence.

Step 4: Express 'The Result is Five' as an Equation

The final part "the result is five" means the expression equals 5. This gives us the complete equation: \( \frac{4}{3}x + 12 = 5 \).

Step 5: Solve the Equation

To find \( x \), solve the equation \( \frac{4}{3}x + 12 = 5 \).1. Subtract 12 from both sides: \[ \frac{4}{3}x + 12 - 12 = 5 - 12 \] \[ \frac{4}{3}x = -7 \]2. Multiply both sides by the reciprocal of \( \frac{4}{3} \) which is \( \frac{3}{4} \): \[ x = -7 \times \frac{3}{4} \] \[ x = -\frac{21}{4} \] \[ x = -5.25 \]

Key Concepts

Solving Equations

Solving equations is like solving a mystery! You're on a quest to find the value of a variable that makes a mathematical expression true. Let's break it down:

When we're given an equation like \( \frac{4}{3}x + 12 = 5 \), our job is to isolate the variable \( x \), which means we want \( x \) to be on one side of the equation all by itself. Here's how:

• Start by getting rid of any numbers added or subtracted with \( x \). In this case, subtract 12 from both sides of the equation. Now, we're left with \( \frac{4}{3}x = -7 \).
• Since \( x \) is multiplied by \( \frac{4}{3} \), we do the opposite operation to isolate \( x \). Multiply both sides by the reciprocal, which is \( \frac{3}{4} \). This gives us \( x = -\frac{21}{4} \).

Now, you've successfully solved for \( x \)! Each step works to "undo" what's being done to \( x \) until it's isolated. Remember, whatever you do to one side of the equation, also do to the other to keep it balanced.

Fractions in Algebra

Fractions play a big role in algebra, helping us express parts of a whole as in many real-world problems. Dealing with fractions might seem tricky at first, but it's manageable!

Here, "four thirds of a number" translates into a fraction \( \frac{4}{3} \), which means you're multiplying 4 by a number and dividing by 3. In terms of our variable, it becomes \( \frac{4}{3} \times x \).

When working with fractions in equations:

• Keep an eye on denominators. They tell you how many equal parts the whole is divided into.
• To solve an equation involving fractions, you often need to "clear" the fraction by multiplying through by its reciprocal. This helps eliminate the fraction and simplify the equation.

By being careful with fraction operations like multiplication and division, you can easily include them in algebraic expressions and equations.

Variable Identification

Identifying variables is the first step in writing algebraic expressions. Variables stand in for unknown values and can be any letter. In word problems, find what you need to solve for, and that's usually your variable!

In our example, "a number" is the clue that tells us we need to choose a variable, often denoted by \( x \), \( y \), or any letter you choose.

• Determine the unknown quantity you're looking for in the problem. Here, it's the mysterious number increased by four thirds in the equation.
• Pick a letter to represent this unknown quantity. In this case, we chose \( x \).

This simple identification step transforms real-world situations into solvable math problems, making complicated sentences much clearer. It’s like giving a shape to something unknown, easy to handle and manipulate algebraically.