Linear Equations and Applications: College Algebra Study Notes

Linear Equations and Word Problems

Strategy for Solving Word Problems

Solving word problems using linear equations is a fundamental skill in College Algebra. The process involves translating real-world scenarios into mathematical language and solving for unknowns.

• Step 1: Read the problem carefully, identifying all given information and what is being asked.

• Step 2: Assign variables to unknown quantities and express other unknowns in terms of these variables.

• Step 3: Write an equation using the relationships described in the problem.

• Step 4: Solve the equation and check the solution in the context of the original problem.

Example: If you invest $15,000 in two accounts paying different interest rates, and the total interest earned is $710, you can set up a system of equations to find how much was invested in each account.

Solving Formulas for a Variable

Isolating Variables in Formulas

In algebra, you often need to solve a formula for a specific variable. This involves rearranging the equation using algebraic operations.

• Key Point: Use inverse operations to isolate the desired variable.

• Key Point: Apply the distributive property and combine like terms as needed.

Example: Solve the formula $P = 2l + 2w$ for $w$:

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

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

Simple Interest Applications

Simple Interest Formula

Simple interest is calculated using the formula:

• $I = Prt$

Where:

• I = Interest earned

• P = Principal (initial amount)

• r = Interest rate (as a decimal)

• t = Time (in years)

Example: If $x$ is invested at 2% and $15,000 - x$ at 9%, and the total interest is $710$:

• Set up the equation: $0.02x + 0.09(15,000 - x) = 710$

• Solve for $x$ to find the amount invested at each rate.

Practice Problems: Linear Equations

Solving for Unknowns

Practice problems reinforce the process of setting up and solving linear equations.

• Example 1: $x + y = 32$, $y = x + 20$ Substitute $y$ into the first equation: $x + (x + 20) = 32$ $2x + 20 = 32$ $2x = 12$ $x = 6$, $y = 26$

• Example 2: $M = 42 - 2x$, $B = x + 42$, $M + B = 123$ Substitute and solve: $42 - 2x + x + 42 = 123$ $84 - x = 123$ $x = -39$

Solving for a Variable in Geometric Formulas

Perimeter of a Rectangle

Given the perimeter formula $P = 2l + 2w$, you can solve for any variable as needed.

• Example: If $P = 382$ and $l = 7 + 3w$, substitute and solve for $w$:

• $382 = 2(7 + 3w) + 2w$

• $382 = 14 + 6w + 2w$

• $382 = 14 + 8w$

• $368 = 8w$

• $w = 46$

Summary Table: Key Linear Equation Concepts

Concept	Formula/Method	Example
Simple Interest	$I = Prt$	$0.02x + 0.09(15,000-x) = 710$
Perimeter of Rectangle	$P = 2l + 2w$	$w = \frac{P - 2l}{2}$
Solving for a Variable	Isolate variable using algebraic operations	$P = C + mC \rightarrow C = \frac{P}{1 + m}$
Word Problem Strategy	Assign variables, write equation, solve	Let $x$ = unknown, set up equation

Additional info: These notes cover foundational skills in solving linear equations, manipulating formulas, and applying algebraic reasoning to word problems, which are essential for success in College Algebra.