Solving Linear Equations and Applications in Precalculus

Equations and Inequalities

Problem Solving with Linear Equations

Linear equations are fundamental tools for modeling and solving real-world problems in algebra and precalculus. The process of solving word problems using linear equations involves a systematic approach:

• Step 1: Let x represent one of the unknown quantities.

• Step 2: Express other unknowns in terms of x.

• Step 3: Write an equation in x that models the problem's conditions.

• Step 4: Solve the equation and answer the question.

• Step 5: Check the solution in the original context of the problem.

Example 1: Median Starting Salaries

This example demonstrates how to set up and solve a linear equation based on relationships between unknowns.

• Given: The median starting salary of a computer science major exceeds that of an education major by $21,000. The median starting salary of an economics major exceeds that of an education major by $14,000. Combined, their median starting salaries are $140,000.

• Let x = median starting salary of an education major.

• Then, computer science: x + 21; economics: x + 14.

• Equation: $x + (x + 21) + (x + 14) = 140$

• Solution: $x = 35$ (education), $x + 21 = 56$ (computer science), $x + 14 = 49$ (economics).

• Check: $35 + 56 + 49 = 140$ (verifies the solution).

Example 2: Comparing Texting Plans

This example illustrates how to model a cost comparison problem with linear equations.

• Plan A: $15 + 0.08x$

• Plan B: $3 + 0.12x$

• Equation: $0.08x + 15 = 0.12x + 3$

• Solution: $x = 300$ (the number of texts where costs are equal).

• Check: Both plans cost $39 for 300 texts.

Example 3: Price Reduction Problem

This example shows how to find the original price before a percentage reduction using a linear equation.

• Let x = original price.

• After 30% reduction: $x - 0.3x = 0.7x$

• Equation: $0.7x = 840$

• Solution: $x = 1200$ (original price).

• Check: $1200 - 0.3 \times 1200 = 840$ (verifies the solution).

Example 4: Investment Allocation

This example involves dividing an inheritance between two investments with different interest rates to achieve a specified total interest.

• Let x = amount invested at 9%.

• Amount at 11%: $5000 - x$

• Equation: $0.09x + 0.11(5000 - x) = 487$

• Solution: $x = 3150$ at 9%, $5000 - x = 1850$ at 11%.

• Check: $0.09 \times 3150 + 0.11 \times 1850 = 487$ (verifies the solution).

Example 5: Basketball Court Dimensions

This example demonstrates how to use perimeter formulas and linear equations to find the dimensions of a rectangle given a relationship between length and width.

• Let x = width; length = $x + 44$

• Perimeter: $2x + 2(x + 44) = 288$

• Solution: Width = 50 ft, Length = 94 ft.

• Check: $2 \times 50 + 2 \times 94 = 288$ (verifies the solution).

Solving a Formula for a Variable

Isolating a Variable in a Formula

Solving a formula for a variable means rewriting the equation so that the chosen variable is alone on one side. This is essential for rearranging equations in algebra and applied mathematics.

• Example: Solve $P = 2l + 2w$ for $w$.

• Solution: $w = \frac{P - 2l}{2}$

Solving a Formula for a Variable That Occurs Twice

Sometimes, the variable to be isolated appears in more than one term. Factorization and algebraic manipulation are used to solve for the variable.

• Example: Solve $P = C + MC$ for $C$.

• Solution: $C = \frac{P}{1 + M}$