Solving Linear Equations and Applications in College Algebra

Solving Linear Equations and Applications

Outcomes When Solving Linear Equations

When solving a linear equation in one variable, there are three possible outcomes:

• One Solution (Conditional Equation): The equation is true for exactly one value of the variable.

• No Solution (Contradiction/Inconsistent): There is no value that makes the equation true. The solution set is the empty set.

• Infinitely Many Solutions (Identity): The equation is true for all real numbers; every value of the variable satisfies the equation.

Solving Formulas for a Specified Variable

It is often necessary to rearrange a formula to solve for a particular variable. This process is called solving for a variable or literal equations.

• Example: The distance formula is $D = r t$. To solve for time $t$:

$t = \frac{D}{r}$

Practice Problems

• 1. Solve $P = 2l + 2w$ for $l$:

  • Subtract $2w$ from both sides: $P - 2w = 2l$

  • Divide both sides by 2: $l = \frac{P - 2w}{2}$

• 2. Solve $A = \frac{1}{2}bh$ for $b$:

  • Multiply both sides by 2: $2A = bh$

  • Divide both sides by $h$: $b = \frac{2A}{h}$

• 3. Solve $A = \frac{1}{2}bh$ for $h$:

  • Multiply both sides by 2: $2A = bh$

  • Divide both sides by $b$: $h = \frac{2A}{b}$

• 4. Solve $ac + b = 3(x - a)$ for $x$:

  • Expand right side: $ac + b = 3x - 3a$

  • Add $3a$ to both sides: $ac + b + 3a = 3x$

  • Divide both sides by 3: $x = \frac{ac + b + 3a}{3}$

• 5. Solve $V = \frac{1}{3}\pi r^2 h$ for $h$:

  • Multiply both sides by 3: $3V = \pi r^2 h$

  • Divide both sides by $\pi r^2$: $h = \frac{3V}{\pi r^2}$

Applications of Linear Equations

Linear equations are used to solve a variety of real-world problems, including percentage, interest, money, and geometry problems.

Percentage Problems

• Example 1: After a 30% reduction, a television costs $336. What was the original price?

Let $x$ be the original price. After a 30% reduction, the price is 70% of the original: $0.70x = 336$ $x = \frac{336}{0.70} = 480$

• Example 2: Including a 10.5% hotel tax, a room costs $216.58 per night. What is the cost before tax?

Let $x$ be the pre-tax cost. The total cost is $x$ plus 10.5% of $x$: $x + 0.105x = 216.58$ $1.105x = 216.58$ $x = \frac{216.58}{1.105} \approx 196.00$

Interest Problems

• Example 3: A sum is split into two accounts: one at 2% interest, the other (which is $3000 more) at 3%. The total annual interest is $690. How much was invested at each rate?

Let $x$ be the amount at 2%, so $x + 3000$ is at 3%. $0.02x + 0.03(x + 3000) = 690$ $0.02x + 0.03x + 90 = 690$ $0.05x = 600$ $x = 12000$ Amount at 2%: $12,000; Amount at 3%: $15,000

Money Problems

• Example 4: A repair shop charges $448: $63 for parts, the rest for labor at $35/hour. How many hours of labor?

Labor cost: $448 - 63 = 385$ $35h = 385$ $h = \frac{385}{35} = 11$

• Example 5: Comparing two streaming services:

Company	Membership Fee	Cost per Movie
A	$22	$6
B	$28	$4

Let $x$ be the number of movies for which the total cost is the same: $22 + 6x = 28 + 4x$ $2x = 6$ $x = 3$ Total cost: $22 + 6 \times 3 = 40$

Geometry Problems

• Example 6: The perimeter of a rectangular field is 200 ft, and the width is 20 ft more than the length. Find the dimensions.

Let $l$ = length, $w = l + 20$. Perimeter: $2l + 2w = 200$ $2l + 2(l + 20) = 200$ $2l + 2l + 40 = 200$ $4l = 160$ $l = 40$, $w = 60$

• Example 7: A rectangular pool is three times as long as it is wide. Perimeter is 320 ft. Find dimensions.

Let $w$ = width, $l = 3w$. Perimeter: $2l + 2w = 320$ $2(3w) + 2w = 320$ $6w + 2w = 320$ $8w = 320$ $w = 40$, $l = 120$

Summary Table: Types of Linear Equation Solutions

Type	Description	Example
Conditional	One solution	$2x + 3 = 7$
Contradiction	No solution	$x + 2 = x + 5$
Identity	All real numbers	$2(x + 1) = 2x + 2$

Key Reminders

• Homework and quiz due dates are important for success.

• Attendance and participation may affect your grade.

• Utilize office hours for additional help.