2.3 Models and Applications

Linear Equations and Applications

Linear equations let you take a real-world situation and turn it into math you can actually solve. Whether it's figuring out costs, distances, or interest earned, the process is the same: identify what you don't know, build an equation around it, and solve.

Setting Up Linear Models

The hardest part of applied problems isn't usually the algebra. It's translating the words into an equation. Here's a reliable process:

1. Read the problem carefully and figure out what you're solving for.
2. Assign a variable to the unknown quantity. Be specific: not just "let $x$ = apples," but "let $x$ = the number of apples purchased."
3. Identify the relationship between quantities. Look for phrases like "total," "combined," "per," or "each," which signal mathematical operations.
4. Write the equation using that relationship.
5. Solve for the unknown, then check that your answer makes sense in context.

Example: A store sells apples for $2 each. You spent $14. How many apples did you buy?

• Let $x$ = number of apples
• Relationship: total cost = price per apple × number of apples
• Equation: $2x = 14$
• Solve: $x = 7$ apples

Always ask yourself: does this answer make sense? A negative number of apples, for instance, would tell you something went wrong.

Working with Formulas

Many application problems give you a standard formula and ask you to plug in what you know, then solve for what you don't. You should be comfortable with these common formulas:

The key skill here is solving a formula for a different variable than the one that's isolated. For example, if you know the area and length of a rectangle but need the width:

1. Start with $A = lw$
2. Divide both sides by $l$: $w = \frac{A}{l}$
3. Substitute: if $A = 15$ cm² and $l = 5$ cm, then $w = \frac{15}{5} = 3$ cm

Simple interest example: You invest $1,000 at 4% annual interest for 3 years. How much interest do you earn?

• $I = Prt = (1000)(0.04)(3) = 120$
• You earn $120 in interest.

Notice that the rate must be written as a decimal (4% = 0.04). This is one of the most common mistakes on exams.

Solving Word Problems

Word problems trip people up because the math is hidden inside sentences. The trick is having a consistent approach so you don't freeze up.

Translating Words into Equations

Certain English phrases map directly to math operations. Knowing these translations speeds everything up:

• "sum," "more than," "increased by," "combined" → addition
• "difference," "less than," "decreased by," "fewer" → subtraction
• "of," "times," "per," "each" → multiplication
• "per," "ratio," "divided by," "out of" → division
• "is," "equals," "results in," "gives" → equals sign

Example: "Five more than twice a number is 17."

• Let $x$ = the number
• "Twice a number" → $2x$
• "Five more than" → $2x + 5$
• "Is 17" → $2x + 5 = 17$

Solving: subtract 5 from both sides to get $2x = 12$, then divide by 2 to get $x = 6$.

A Step-by-Step Process for Word Problems

1. Read the entire problem before writing anything. Read it twice if needed.
2. Identify the unknown(s) and assign variables with clear definitions.
3. Pull out the given information: numbers, rates, totals, relationships.
4. Write the equation(s) that connect the knowns to the unknowns.
5. Solve using standard algebraic techniques (combine like terms, isolate the variable with inverse operations).
6. Check your solution by substituting back into the original equation and by confirming it makes sense in the real-world context.

Example: A theater sold 200 tickets. Adult tickets cost $8 and child tickets cost $5. Total revenue was $1,180. How many of each type were sold?

• Let $x$ = number of adult tickets. Then $200 - x$ = number of child tickets (since they must add to 200).
• Revenue equation: $8x + 5(200 - x) = 1180$
• Distribute: $8x + 1000 - 5x = 1180$
• Combine like terms: $3x + 1000 = 1180$
• Subtract 1000: $3x = 180$
• Divide by 3: $x = 60$ adult tickets, so $200 - 60 = 140$ child tickets
• Check: $8(60) + 5(140) = 480 + 700 = 1180$ ✓

Notice the strategy of using one variable instead of two. Since adult + child = 200, defining child tickets as $200 - x$ lets you write a single equation. This is a very common technique in these problems.

Advanced Modeling Techniques

Systems of Equations for Complex Problems

When a problem has two unknowns and two separate relationships, you'll need two equations solved together. The theater problem above could also be set up as a system:

• $x + y = 200$ (total tickets)
• $8x + 5y = 1180$ (total revenue)

You can solve this using substitution (solve one equation for a variable, plug into the other) or elimination (add/subtract equations to cancel a variable). Both methods give the same answer.

Optimization and Graphical Interpretation

Some problems ask you to maximize or minimize a quantity (like profit or cost) subject to constraints. At this level, the main idea is:

• The objective is what you're trying to maximize or minimize.
• The constraints are limits on your variables (budgets, capacity, time).
• Plotting the equations on a coordinate plane can help you visualize which values satisfy all constraints and which give the best result.

You'll explore these ideas more deeply in later courses, but for now, practice graphing linear equations and reading solutions from the graph. The point where two lines intersect represents the solution to that system of equations.