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:

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

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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When There Are Two Unknowns: Use Two Equations

Some word problems give you two unknown amounts and two facts about them. Then you need two equations, called a system. Each fact becomes one equation.

Example: A school sold 10 tickets for $65. Adult tickets cost $8 and child tickets cost $5. How many of each were sold?

1. Name the variables: x = adult tickets, y = child tickets.
2. Write one equation for the count: x + y = 10.
3. Write one equation for the money: 8x + 5y = 65.
4. Solve the first equation for y: y = 10 − x.
5. Substitute: 8x + 5(10 − x) = 65, so 8x + 50 − 5x = 65, and 3x = 15. That gives x = 5.
6. Find y: y = 10 − 5 = 5.
7. Check: 5 + 5 = 10 tickets, and 8(5) + 5(5) = 40 + 25 = 65 dollars.

Another example: Two numbers add up to 30 and their difference is 6. Write x + y = 30 and x − y = 6. Add the equations to get 2x = 36, so x = 18 and y = 12.

Words to watch for

Frequently Asked Questions

How do you find x and y in a word problem?

Decide what x and y stand for, write each fact as an equation, then solve. If a plumber charges a $40 visit fee plus $60 per hour, the total cost is y = 60x + 40, where x is hours. To find the cost for 3 hours, substitute x = 3: y = 60 × 3 + 40 = $220. If you know y instead, solve the equation for x.

What are two-variable word problems?

They are word problems where two quantities are linked, so a change in one changes the other. Examples include hours worked and pay, or miles driven and rental cost. You write them as an equation like y = mx + b, where m is the rate and b is the starting amount. On the GED, these often show up as cost, distance or savings situations.

How do you solve equations with variables on both sides in word problems?

Set the two expressions equal, then collect the variable terms on one side. Gym A costs $20 plus $5 per visit and Gym B costs $35 plus $2 per visit. When do they cost the same? 20 + 5x = 35 + 2x, so 3x = 15 and x = 5 visits. Check it: both gyms cost $45 at 5 visits.

How do you write a variable expression from a word problem?

Choose a letter for the unknown and translate the words piece by piece. Twice a number is 2n, and 8 less than twice a number is 2n − 8, not 8 − 2n. Words like sum and more than mean add, difference and less than mean subtract, and product and of mean multiply. Read the phrase again to check the order.

How do you solve word problems with two variables?

When there are two unknowns, write two equations and solve them together. Twelve coins, all quarters and dimes, are worth $2.25. Then q + d = 12 and 25q + 10d = 225. Substitute d = 12 − q: 25q + 120 − 10q = 225, so 15q = 105 and q = 7. That means 7 quarters and 5 dimes, which is $1.75 + $0.50 = $2.25.

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