How to Solve Word Problems by Finding Two-Variable Equations?

Many real-world situations can be described with an equation that connects two quantities — for example, total cost depending on the number of hours worked, or the distance traveled based on speed and time. On the GED, you need to translate a word problem into a two-variable equation and then use it to find unknown values. This lesson shows you exactly how to do that.

What Is a Two-Variable Equation?

A two-variable equation is an equation with two unknown quantities, usually called x (the independent variable — the input) and y (the dependent variable — the output). The equation describes how y changes as x changes. The most common form is:

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\(\color{blue}{y = \text{ mx } + b}\)

where m is the rate of change (how much y changes per unit of x) and b is the starting value (what y equals when \(\color{blue}{x = 0}\)).

How to Write a Two-Variable Equation from a Word Problem

1. Identify the two quantities

Decide which quantity is the input (x) and which is the output (y). The input is usually the thing you control or vary (hours, items, days). The output is usually the total or result.

2. Find the rate of change (m)

Look for a number that repeats or multiplies — a cost per hour, price per item, miles per hour. That is your slope, m.

3. Find the starting value (b)

Look for a fixed amount that applies no matter what — a flat fee, an initial balance, a base charge. That is your y-intercept, b.

4. Write the equation

Plug m and b into \(\color{blue}{y = \text{ mx } + b}\).

5. Solve for the unknown

Substitute the given value and solve for the missing variable.

Step-by-Step Summary

1. Read the problem twice. Underline the two quantities (input and output).
2. Find the rate (m) — the per-unit amount.
3. Find the fixed starting value (b).
4. Write \(\color{blue}{y = \text{ mx } + b}\).
5. Substitute the known value and solve for the unknown.
6. Check your answer in the original problem.

Watch: Variables and Equations Word Problem (Khan Academy)

This Khan Academy lesson uses a real-world yoga studio scenario to show how to set up and use a two-variable equation:

Worked Examples

Example 1: A plumber charges a $5 call-out fee plus $3 per hour. Write an equation for the total cost y after x hours. How much does 4 hours of work cost?

Rate \(\color{blue}{m = 3}\), fixed fee \(\color{blue}{b = 5}\). Equation: \(\color{blue}{y = 3x + 5}\).
At \(\color{blue}{x = 4}\): \(\color{blue}{y = 3(4) + 5 = 12 + 5}\) = $17.

Example 2: Using the plumber equation \(\color{blue}{y = 3x + 5}\), how many hours were worked if the bill was $20?

\(\color{blue}{20 = 3x + 5}\) → \(\color{blue}{15 = 3x}\) → \(\color{blue}{x = 15 \div 3}\) = 5 hours.

Example 3: A car rental company charges $10 per day plus a flat fee of $5. Write the equation and find the cost for 5 days.

\(\color{blue}{y = 10x + 5}\). At \(\color{blue}{x = 5}\): \(\color{blue}{y = 10(5) + 5 = 50 + 5}\) = $55.

Example 4: A savings account starts with $35 and its owner deposits $5 per week. When does the balance reach $60?

\(\color{blue}{y = 5x + 35}\). Set \(\color{blue}{y = 60}\): \(\color{blue}{60 = 5x + 35}\) → \(\color{blue}{25 = 5x}\) → x = 5 weeks.

More Practice: Two-Variable Equations and Their Graphs (Khan Academy)

See how the equation \(\color{blue}{y = \text{ mx } + b}\) connects to its graph, which reinforces understanding of slope and intercept in word problem contexts:

Exercises

1. A taxi charges $2.50 per mile plus a $3.00 base fare. Write the equation and find the cost for a 6-mile ride.
2. Using the taxi equation above, how far did a passenger travel if the fare was $15.50?
3. Maria earns $12 per hour. She has $40 in savings. Write an equation for her total savings y after working x hours.
4. Using Maria’s equation, how many hours must she work to have $160 total?
5. A phone plan costs $0.10 per text plus a $9 monthly fee. Find the monthly bill for 80 texts.
6. Use \(\color{blue}{y = 2x – 1}\). Find y when \(\color{blue}{x = 7}\), and find x when \(\color{blue}{y = 11}\).

Answers

1. \(\color{blue}{y = 2.50x + 3}\); at \(\color{blue}{x = 6}\): \(\color{blue}{y = 2.50(6) + 3 = 15 + 3}\) = $18.00
2. \(\color{blue}{15.50 = 2.50x + 3}\) → \(\color{blue}{12.50 = 2.50x}\) → x = 5 miles
3. \(\color{blue}{y = 12x + 40}\)
4. \(\color{blue}{160 = 12x + 40}\) → \(\color{blue}{120 = 12x}\) → x = 10 hours
5. \(\color{blue}{y = 0.10(80) + 9 = 8 + 9}\) = $17.00
6. \(\color{blue}{y = 2(7) – 1}\) = 13; \(\color{blue}{11 = 2x – 1}\) → \(\color{blue}{12 = 2x}\) → x = 6

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Frequently Asked Questions

How do I know which quantity is x and which is y?

The independent variable x is the one you choose or control — usually time, units bought, or hours worked. The dependent variable y is the result that depends on x — usually total cost, total savings, or total distance. Ask yourself: “Which quantity changes because of the other?” That one is y.

What if there is no starting value mentioned?

If there is no flat fee or initial amount, then \(\color{blue}{b = 0}\) and the equation simplifies to \(\color{blue}{y = \text{ mx }}\). For example, “cost is $4 per item” gives \(\color{blue}{y = 4x}\).

Can the same equation be used to find both x and y?

Yes. Once you have the equation, you can plug in any known value (whether x or y) and solve for the other. For \(\color{blue}{y = 3x + 5}\), you can find y when x is given, or find x when y is given — just use inverse operations.

Related Topics

• Using a Table to Write Down a Two-Variable Equation
• How to Complete a Table and Graph a Two-Variable Equation
• Independent and Dependent Variables in Tables and Graphs
• Word Problems Involving One-Step Equations
• Unit Prices with Decimals and Fractions