Linear Equations, Functions, Zeros, and Applications – Precalculus Study Notes

Section 1.5: Linear Equations, Functions, Zeros, and Applications

Objectives

• To solve linear equations.

• To solve applied problems using linear models.

• To find zeros of linear functions.

Equations and Solutions

Definitions and Key Concepts

An equation is a statement that two expressions are equal. To solve an equation in one variable means to find all values of the variable that make the equation true. Each such value is called a solution of the equation. The set of all solutions is called the solution set.

• Equation in one variable: An equation involving only one variable, e.g., $2x + 3 = 5$.

• Solution: A value of the variable that makes the equation true.

• Solution set: The set of all solutions to the equation.

Examples of equations in one variable:

• $2x + 3 = 5$

• $3(x - 1) = 4x + 5$

• $\frac{x - 3}{x + 4} = 1$

• $x^2 - 3x + 2 = 0$

Linear Equations

Definition

A linear equation in one variable is an equation that can be written in the form:

$mx + b = 0$

where m and b are real numbers and $m \neq 0$.

Equivalent Equations

Definition and Examples

Equivalent equations are equations that have the same solution set.

• Example: $2x + 3 = 5$ and $x = 1$ are equivalent because $x = 1$ is the solution to both.

• Non-example: $x^2 - 3x + 2 = 0$ and $x = 1$ are not equivalent because $x^2 - 3x + 2 = 0$ has solutions $x = 1$ and $x = 2$, but $x = 1$ only has one solution.

Equation-Solving Principles

Properties

• Addition Principle: If $a = b$ is true, then $a + c = b + c$ is also true for any real number $c$.

• Multiplication Principle: If $a = b$ is true, then $ac = bc$ is also true for any real number $c$.

Solving Linear Equations: Examples

Example 1: Solving a Fractional Linear Equation

Solve $\frac{3}{4}x - 1 = \frac{7}{5}$.

• Multiply both sides by the least common denominator (LCD) to clear fractions.

• Steps:

  • Multiply both sides by 20 (LCD of 4 and 5): $20\left(\frac{3}{4}x - 1\right) = 20\left(\frac{7}{5}\right)$

  • $15x - 20 = 28$

  • $15x = 48$

  • $x = \frac{48}{15} = \frac{16}{5}$

• Check: Substitute $x = \frac{16}{5}$ back into the original equation to verify the solution.

Example 2: Special Cases

• No Solution: $-24x + 7 = 17 - 24x$ leads to $7 = 17$, which is false for any $x$. Thus, the equation has no solution.

• Infinitely Many Solutions: $3 - \frac{1}{3}x = -\frac{1}{3}x + 3$ simplifies to $3 = 3$, which is true for all real $x$. Thus, the solution set is all real numbers: $(-\infty, \infty)$.

Applications Using Linear Models

Overview

Mathematical techniques, especially linear equations and functions, are used to model and solve real-world problems.

Steps for Problem Solving

1. Familiarize yourself with the problem. Read carefully, make drawings, and list knowns and unknowns.

2. Assign variables to unknown quantities.

3. Organize information in a chart or table if helpful.

4. Find further information if needed (formulas, references, etc.).

5. Estimate or guess the answer and check your estimate.

6. Translate the problem into mathematical language (equations).

7. Solve the equation(s).

8. Check if the solution makes sense in the context of the problem.

9. State the answer clearly in a complete sentence.

Application Example 1: Motion Formula

The distance $d$ traveled by an object moving at a constant rate $r$ for time $t$ is given by:

$d = rt$

Example: Airplane Overtaking Problem

• A Saab 340B (290 mph) takes off. One hour later, a B737/800 (517 mph) takes off on the same route. How long until the B737/800 overtakes the Saab?

• Let $t$ = time (in hours) the B737/800 travels before overtaking.

• The Saab travels $t + 1$ hours before being overtaken.

Plane	Rate (mph)	Time (hr)	Distance (mi)
B737/800	517	$t$	$517t$
Saab 340B	290	$t+1$	$290(t+1)$

• Set distances equal: $517t = 290(t + 1)$

• Solve: $517t = 290t + 290 \implies 227t = 290 \implies t \approx 1.28$ hours

• Conclusion: About 1.28 hours after the B737/800 takes off, it overtakes the Saab 340B.

Application Example 2: Simple Interest Formula

The simple interest $I$ on a principal $P$ at interest rate $r$ for $t$ years is:

$I = Prt$

• $I$ = interest earned ($)

• $P$ = principal ($)

• $r$ = annual interest rate (decimal)

• $t$ = time (years)

Example: Student Loan Interest

• Damarion has two loans totaling $28,000. One is at 5% interest, the other at 3%. After 1 year, he owes $1,040$ in interest. Find the amount of each loan.

• Let $x$ = amount at 5%, $28,000 - x$ = amount at 3%.

Loan	Amount ($)	Interest Rate	Interest ($)
5% loan	$x$	0.05	$0.05x$
3% loan	$28,000 - x$	0.03	$0.03(28,000 - x)$

• Equation: $0.05x + 0.03(28,000 - x) = 1,040$

• Solve: $0.05x + 840 - 0.03x = 1,040 \implies 0.02x = 200 \implies x = 10,000$

• Conclusion: $10,000$ at 5%, $18,000$ at 3%.

Zeros of Linear Functions

Definition

A zero of a function $f$ is a value $c$ such that $f(c) = 0$. For a linear function $f(x) = mx + b$ with $m \neq 0$, there is exactly one zero.

Example: Finding the Zero of a Linear Function

• Find the zero of $f(x) = 5x - 9$.

• Set $5x - 9 = 0 \implies 5x = 9 \implies x = 1.8$

• Graphical Solution: The x-intercept of the graph $y = 5x - 9$ is at $x = 1.8$.

Additional info: In practice, zeros of linear functions correspond to the x-intercepts of their graphs, and are important in applications such as finding break-even points in business or roots in physical problems.