Appendix A
Linear Equations

This chapter is adapted from Applied Finite Mathematics by Rupinder Sekhon and Roberta Bloom, available at https://math.libretexts.org/Bookshelves/Applied_Mathematics/Applied_Finite_Mathematics_(Sekhon_and_Bloom), licensed under CC BY 4.0. Content has been modified and remixed for this book.

Learning Outcomes

• Graph linear equations.

• Find the slope of a line.

• Determine equations of lines.

• Solve linear systems.

• Apply linear equations to real-world problems.

Chapter Overview

In this chapter¹, we focus on linear equations, their graphical representations, finding slopes, determining line equations from given information, and solving systems of linear equations. We also examine applications of these concepts in real-world scenarios, such as cost modeling, demand-supply equilibrium, and break-even analysis.

A.1 Graphing a Linear Equation

Learning Outcomes

• Graph a line from its equation.

• Graph a line given its equation in parametric form.

• Graph and find equations of vertical and horizontal lines.

Equations whose graphs are straight lines are called linear equations. Examples include:

A line is determined by two points. To graph a linear equation, we often choose values for one variable and solve for the other.

Example A.1: Graphing a Line

Graph the line: $y=3x+2$.

Solution: Choose convenient $x$-values:

Plot these points and draw the line.

Example A.2: Graphing Another Line

Graph the line: $2x+y=4$.

Solution: Choose points:

Plotting these points gives:

Intercepts

The $x$-intercept occurs where $y=0$, and the $y$-intercept occurs where $x=0$.

Example A.3: Finding Intercepts

Find the intercepts of the line $2x-3y=6$ and graph it.

Solution: For the $x$-intercept, set $y=0$:

For the $y$-intercept, set $x=0$:

Plot these intercepts and draw the line.

Parametric Form

Lines can also be given in parametric form, e.g. $x=3+2t,y=1+t$.

Example A.4: Parametric Equations

Graph the line $x=3+2t,y=1+t.$

Solution: Let $t=0,1,2$:

Plot these points and draw the line.

Horizontal and Vertical Lines

Example A.5: Horizontal and Vertical Lines

Graph $x=-2$ and $y=3$.

Solution: The line $x=-2$ is vertical through $(-2,0)$. The line $y=3$ is horizontal through $(0,3)$.

A.2 Slope of a Line

Learning Outcomes

• Find the slope of a line given two points.

• Graph a line if a point and slope are given.

The slope $m$ of a line passing through $(x_1,y_1)$ and $(x_2,y_2)$ is:

Example A.6: Finding Slope

Find the slope of the line through $(-2,3)$ and $(4,-1)$.

Solution:

Vertical lines have undefined slope; horizontal lines have slope 0.

Example A.7: Vertical and Horizontal Slopes

a) For points $(2,3)$ and $(2,-1)$:

b) For points $(-1,-4)$ and $(3,-4)$:

Example A.8: Graphing a Line from a Point and Slope

Graph the line passing through $(1,2)$ with slope $-\frac{3}{4}$.

Solution: Starting from $(1,2)$, a slope of $-\frac{3}{4}$ means down 3, right 4 to get another point $(5,-1)$. Plot and draw the line.

If a line is written as $y=mx+b$, the coefficient $m$ is the slope and $b$ is the $y$-intercept.

A.3 Determining the Equation of a Line

Learning Outcomes

• Find an equation of a line given a point and slope.

• Find an equation of a line given two points.

Forms of a Line

Example A.9: Equation Given Slope and Intercept

Find the equation of a line with slope $m=5$ and $y$-intercept $b=3$.

Solution: $y=5x+3.$

Example A.10: Equation Given a Point and Slope

Find the equation of the line passing through $(2,7)$ with slope $3$.

Solution:

Example A.11: Equation Given Two Points

Find the equation of the line passing through $(-1,2)$ and $(1,8)$.

Solution:

Use one point:

Example A.12: Using Intercepts

Find the equation of the line with $x$-intercept 3 and $y$-intercept 4.

Solution: Intercepts give points $(3,0)$ and $(0,4)$:

A.4 Applications of Linear Equations

Learning Outcomes

• Use linear functions to model real-world applications.

Linear equations commonly model cost, revenue, and population changes.

Example A.13: Cost Function

A taxi charges $0.50 per mile plus a $5 flat fee.

Solution: Let $x=$ miles, $y=$ cost:

At $x=20$: $y=0.50(20)+5=15.$

Example A.14: Cost from Two Points

It costs $750 to make 25 items and $1000 to make 50 items. Assume linearity.

Solution: Points $(25,750)$ and $(50,1000)$:

Using $(25,750)$:

Cost of 100 items: $y=10(100)+500=1500.$

Example A.15: Temperature Conversion

Freezing point of water: $(C,F)=(0,32)$; boiling point: $(100,212)$.

Solution:

For $C=30$: $F=1.8(30)+32=86.$

A.5 More Applications: Systems of Linear Equations

Learning Outcomes

• Solve linear systems in two variables.

• Find equilibrium points (supply = demand).

• Find break-even points (cost = revenue).

Solving Systems of Equations

The intersection of two lines can be found by setting their equations equal or by using the elimination method.

Example A.16: Solving a System

Solve:

Solution: Add them:

Substitute $x=2$ into $2x+y=7\Rightarrow 4+y=7\Rightarrow y=3.$

Solution: $(2,3)$.

Supply, Demand, and Equilibrium

The equilibrium price occurs where supply equals demand.

Example A.17: Equilibrium Point

Supply: $y=3.5x-14$, Demand: $y=-2.5x+34$.

Solution: Set equal:

At $x=8$: $y=3.5(8)-14=14$ or $y=-2.5(8)+34=14$.

Equilibrium: Price = $8, Quantity = 14 items.

Break-Even Point

For cost $C$ and revenue $R$, the break-even point is where $C=R$.

Example A.18: Break-Even Analysis

$R=5x$, $C=3x+12$.

Solution:

At $x=6$, $R=C=30$.

Break-even at $(6,30)$.

Chapter Review

This chapter introduced the fundamental concepts of linear equations, their graphs, slopes, forms of equations, and their real-world applications. You can now model scenarios involving costs, pricing, and market equilibrium using linear models and solve these models for key insights.