﻿ ﻿Single-Variable Linear Equations - Equations and Inequalities - High School Algebra I Unlocked (2016)

## High School Algebra I Unlocked (2016)

### Chapter 4. Equations and Inequalities

GOALS

By the end of this chapter, you will be able to:

•Explain the properties of linear equations

•Solve single-variable linear equations and inequalities

•Construct linear equations from word problems

•Solve two-variable linear equations

•Rewrite equations in slope-intercept form

•Determine the slope of a line using multiple methods:

•Find the x- and y-intercepts of linear equations

•Identify equivalent equations

•Graph exponential equations and graphs

### Lesson 4.1. Single-Variable Linear Equations

REVIEW

BEFORE BEGINNING THIS CHAPTER, YOU SHOULD BE FAMILIAR WITH:

•general number properties and operations

•the parts of an equation or expression

•solving simple numeric and variable expressions

•substituting in numbers for variables

•manipulating polynomial expressions

Now that we’ve covered the basic mathematical properties, let’s dive deeper into the main purpose of this book: algebra. If you’ve made it this far, you’ve already been working with algebraic expressions and equations. But let’s start with the basics: single-variable equations, or equations that deal with only one variable.

What’s a Linear Equation?

Linear means “line.” A
linear equation is an
algebraic equation in
which each variable has
an exponent of 1, and
the graph of the equation
is a straight line.

Solving Single-Variable Equations

One basic type of algebraic problem involves solving for a single variable; these are often referred to as “solving for x” questions.

Let’s take a look at a sample question.

EXAMPLE

If 4x + 12 = 16, then x = ?

Before delving into the solution, let’s discuss the problem. Here you are given an equation with one variable, x, and asked to find the value of x. In order to do this, you need to manipulate the equation and isolate the variable on one side of the equation.

In the equation 4x + 12 = 16, x is already on the left side of the equation, so move the remaining terms to the right side of the equation. First, deal with 12. Whenever you manipulate an equation, you need to make sure that the equation stays balanced. Therefore, in order to make the 12 disappear from the left side of the equation, you must subtract 12 from both sides:

You’re left with the equation 4x = 4. Now you need to tackle the coefficient, 4, which is being multiplied by x. To remove the coefficient, divide both sides of the equation by 4:

4x = 4

Now reduce the fractions, noting that you can cancel out all the 4s in the equation.

x = 1

Therefore, in this equation x = 1.

But wait! Not so fast. There’s an alternative way to solve this question. Did you notice that all of the terms in the equation 4x + 12 = 16 are divisible by 4? So, before you do anything else, you can divide both sides of the equation by 4.

4x + 12 = 16

x + 3 = 4

Finally, subtract 3 from both sides to find x.

x + 3 = 4

x = 1

Again, you find that x = 1.

Remember, whatever you do to one side of the equation, you MUST do the other side of the equation as well. Keep the equation balanced!

It would be nice if every question were as straightforward as the previous example, but unfortunately, such is not the case. Let’s try a question that is a bit more complex but still involves solving for a variable.

EXAMPLE

If x = 4, y = 0, and 6x − 5 = 2y2 + z, then z = ?

If you noticed that there are three variables in the given equation, you may be wondering why this question is included in our discussion of single-variable equations. Well, you’re told that x = 4 and y = 0. When you plug these values into the equation, it quickly becomes a single-variable linear equation:

6x − 5 = 2y2 + z

6(4) − 5 = 2(0)2 + z

24 − 5 = 0 + z

Now the equation no longer has three variables; instead, you have a single-variable linear equation that just needs to be simplified! So go ahead and solve for z:

24 − 5 = 0 + z

19 = 0 + z

19 = z

Done. You’re able to find the value of z just by substituting in the given values and simplifying the equation. Although the original equation had three variables, you used the information in the question to substitute in numbers for two of the variables and find that z = 19.

Writing Single-Variable Equations

Now that we have covered solving single-variable equations, we need to talk a little bit about writing these equations, or translating information from English to math. Here are some English words and their math equivalents:

 English Word Math Equivalent is, are, were, did, does = of × out of ÷ what variable, such as a, b, or c

So how does this translating business work? Let’s say you’re asked the following: What is one-half of one hundred? Using your translating skills, you know that what means a variable, let’s say a; of means ×; and is means =. Therefore, the question can be translated into math as a = 1/2 × 100. Now you can solve for a, finding that a = 50.

Let’s see how you can apply these skills to create equations from a word problem.

EXAMPLE

If it takes Dmitri two hours to bake banana bread, and it takes Pavel six hours to bake banana bread, then which of the following is an expression for how many more hours it takes Pavel to bake banana bread than it does Dmitri?

A) 6 − 2

B) 6 + 2

C) 6 × 2

D) 6 ÷ 2

With word problems, always break the question down into smaller parts to make it more manageable. When you attack a question in stages, you’ll be less likely to make a mistake. Start with what you are given: The question states that it takes Dmitri two hours to bake banana bread, and it takes Pavel six hours to bake banana bread. You’re asked to find how many more hours it takes Pavel to make banana bread than it does Dmitri. Using the information you know already, you can find that it takes Pavel 6 − 2 = 2 hours longer to bake banana bread than it does Dmitri. Since the question asks you for an expression, use 6 − 2, which is (A).

Here’s a more complex question for you to try.

EXAMPLE

In a card game involving cats, a white cat is worth 2 points, a black cat is worth 3 points, and a gray cat is worth 5 points. If a player collects more than five gray cats, a 20-point bonus is awarded. Which of the following equations expresses the number of points, p, Tashi will receive if he collects 8 white cats, 15 black cats, and 6 gray cats?

A) p = 8 + 15 + 6

B) p = 8 + 15 + 6 + 20

C) p = (8 × 2) + (15 × 3) + (6 × 5)

D) p = (8 × 2) + (15 × 3) + (6 × 5) + 20

As we did for the previous example, break down this question into smaller parts. First, gather the information you know. You’re told that a white cat is worth 2 points, a black cat is worth 3 points, and a gray cat is worth 5 points, and that a player collects 20 bonus points if she collects more than five gray cats. You’re asked to find the equation that expresses the number of points Tashi will receive if she collects eight white cats, 15 black cats, and six gray cats.

Start with the white cats. If Tashi collects eight white cats that are worth 2 points each, she receives a total of 8 × 2 = 16 points for white cats. Thus, the first part of your equation will be written as

p = (8 × 2)

Next, you’re told that Tashi collects 15 black cats worth 3 points each, which is added to your equation as follows:

p = (8 × 2) + (15 × 3)

Then Tashi collects six gray cats worth 5 points each, which is also added to the equation:

p = (8 × 2) + (15 × 3) + (6 × 5)

Finally, add a 20-point bonus, since Tashi collected more than five gray cats.

p = (8 × 2) + (15 × 3) + (6 × 5) + 20

Therefore, the equation that represents the Tashi’s points is p = (8 × 2) + (15 × 3) + (6 × 5) + 20, or (D).

﻿