Problem 65 - Introductory and Intermediate Algebra for College Students 4th

Question

Use the given information to write an equation. Let \(x\) represent the number described in each exercise. Then solve the equation and find the number. Negative three times a number, increased by \(15,\) is \(-6\)

Step-by-Step Solution

Answer

So, the solution for the equation is \(x = 7\).

Step 1: Translate words into a mathematical equation

First, it's noticed that 'a number' mentioned in the problem is represented by \(x\). Now, 'Negative three times a number' can be transcribed as \(-3x\). 'Increased by 15' means we add 15 to \(-3x\), becoming \(-3x + 15\). Lastly, the statement mentions that this is equal to \(-6\), allowing us to set up the complete equation as \(-3x + 15 = -6\).

Step 2: Solve for the variable \(x\)

To solve for \(x\), we can start by subtracting 15 from both sides of the equation, resulting in \(-3x = -6 - 15 = -21\). Then, divide both sides by -3 to solve for \(x\), leading to \(x = -21 / -3 = 7\).

Key Concepts

Algebraic Expressions

Algebraic expressions are mathematical phrases that can include numbers, variables, and operation symbols but do not have an equality sign like equations do. When we see word problems, the first step is to convert the written words into an algebraic expression. Just like a language translator, we're converting from English to the 'language' of algebra.

In the given problem, we identify the key parts: a number (which we call 'x'), three times this number (which becomes '3x'), and an instruction to increase that quantity by 15, resulting in the expression ewline 3x + 15ewline. This expression alone doesn't tell us the value of 'x', but it's an essential step towards creating an equation that we can solve.

Solving Linear Equations

Once we have an equation, such as ewline 3x + 15 = -6ewline, the goal is to find the value of 'x' that makes the equation true. Solving linear equations involves applying a series of mathematical operations to isolate the variable on one side of the equation. We aim to get 'x' by itself, which means undoing any addition, subtraction, multiplication, or division that's been applied to it.

In our case, we start by undoing the addition of 15. We do this by subtracting 15 from both sides:ewline 3x = -6 - 15ewline. Next, we address the multiplication by 3. To cancel this, we divide both sides by 3:ewline x = (-6 - 15) / 3ewline. These steps systematically reduce the equation until 'x' stands alone, giving us our solution.

Variable Representation

In algebra, variables are symbols that represent unknown quantities. They're placeholders for values we want to find or understand in context. A key to mastering word problems is deciphering what the variable represents and how it fits into the equation we're trying to solve.

In this problem, 'x' represents 'a number.' But it's not just any number—it's a number that, when multiplied by -3 and increased by 15, equals -6. But why use 'x'? Well, 'x' is simply a tradition, but the choice of the letter is arbitrary. We could just as well use another symbol, but 'x' is familiar and widely accepted in algebra. Recognizing variables and their role in equations is crucial in translating word problems into solvable mathematical challenges.