2.2 Use a Problem Solving Strategy

Problem Solving Strategy

Problem-solving in algebra follows a repeatable process. Once you internalize the steps, even complicated word problems become straightforward. The key is translating English sentences into math equations, then solving those equations with the techniques you already know.

The Step-by-Step Approach

Every word problem can be worked through using this sequence:

1. Read and understand the problem. Identify what you know (prices, ages, distances, rates) and what you need to find.
2. Assign a variable to the unknown quantity. Pick something meaningful: $x$ for an unknown price, $a$ for an unknown age. Write down exactly what your variable represents so you don't lose track.
3. Translate into an equation. Turn the English relationships into math. For example, "John is 5 years older than Amy" becomes $J = A + 5$.
4. Solve the equation using inverse operations to isolate your variable.
5. Interpret and check your answer. Does it make sense in context? A negative age or a negative distance is a red flag. Substitute your answer back into the original equation to verify it works.
6. Estimate as a sanity check. Before you move on, ask yourself whether the answer is in a reasonable ballpark.

Translating Word Problems into Algebra

The hardest part for most students isn't the algebra itself; it's setting up the equation. Here's how to get better at it:

• Identify the problem type. Age problems, distance-rate-time problems, and mixture problems each follow recognizable patterns. Spotting the type early helps you know what equation structure to expect.
• Pull out the numbers and relationships. Read the problem carefully and note every quantity and how they relate. "John's age is 3 times Amy's age" tells you two things: there are two ages, and one is 3 times the other.
• Assign variables and write expressions. If you let $x$ represent Amy's current age, then John's age is $3x$. If the problem asks for the sum of their ages in 5 years, Amy's future age is $x + 5$ and John's is $3x + 5$.
• Solve and simplify. Combine like terms, use inverse operations, and isolate the variable.
• Check by substituting back. If you find $x = 10$, then Amy is 10 and John is 30. Does John's age equal 3 times Amy's? $3(10) = 30$. Yes, it checks out.

Drawing a quick diagram or table can also help you organize the information, especially for distance-rate-time or mixture problems where multiple quantities interact.

Strengthening Your Problem-Solving Skills

• Break complex problems into smaller pieces. Solve one relationship at a time rather than trying to handle everything at once.
• Work through each step in order. Skipping ahead (especially skipping the "define your variable" step) is where most mistakes happen.
• If you're stuck, try approaching the problem from a different angle. Can you set up the equation using a different variable? Sometimes choosing a different unknown makes the algebra simpler.

Percentages and Financial Applications

Percentages show up constantly in real life: sales tax, discounts, tips, loan interest. All of these boil down to the same core idea: a percentage is a fraction out of 100.

Working with Percentages

A percentage, a decimal, and a fraction are three ways of writing the same value:

• 25% = 0.25 = $\frac{25}{100}$ = $\frac{1}{4}$
• 75% = 0.75 = $\frac{3}{4}$

To use a percentage in a calculation, convert it to a decimal first (divide by 100 or move the decimal point two places left).

Discounts — finding a sale price:

$\text{Discount Amount} = \text{Original Price} \times \text{Discount Rate}$

For a $50 item at 20% off: $50 \times 0.20 = 10$, so the discount is $10 and the sale price is $50 - 10 = 40$ dollars.

Markups — finding a selling price:

$\text{Markup Amount} = \text{Cost} \times \text{Markup Rate}$

A product that costs $20 with a 50% markup: $20 \times 0.50 = 10$, so the selling price is $20 + 10 = 30$ dollars.

Sales tax — finding the total cost:

$\text{Tax Amount} = \text{Pre\text{-}tax Price} \times \text{Tax Rate}$

A $100 item with 8% tax: $100 \times 0.08 = 8$, so the total cost is $100 + 8 = 108$ dollars.

Simple Interest Calculations

Simple interest is the most basic way to calculate the cost of borrowing money or the return on an investment. It uses three components:

• Principal ($P$): the starting amount borrowed or invested
• Rate ($r$): the annual interest rate, written as a decimal
• Time ($t$): the length of the loan or investment in years

The formula is:

$I = P \times r \times t$

where $I$ is the interest earned or owed. To find the total amount after interest:

$A = P + I$

Example 1: You invest $2,000 at 4% annual interest for 3 years.

1. Convert the rate: 4% = 0.04
2. Calculate interest: $I = 2000 \times 0.04 \times 3 = 240$
3. Find the total: $A = 2000 + 240 = 2240$

You earn $240 in interest, for a total of $2,240.

Example 2: You borrow $5,000 at 6% annual interest for 2 years.

1. Convert the rate: 6% = 0.06
2. Calculate interest: $I = 5000 \times 0.06 \times 2 = 600$
3. Find the total owed: $A = 5000 + 600 = 5600$

You'll pay $600 in interest on top of the original $5,000.

Note that if time is given in months, convert to years first. For instance, 18 months = $\frac{18}{12} = 1.5$ years.