HomeVerbal Descriptions to Direct-Variation Equations y = kx Free original lessons
Verbal Descriptions to Direct-Variation Equations y = kx
Learn to translate word scenarios into y = kx, find the constant k, and answer follow-up questions with clear steps and practice.
Free lesson
Four exercises
Worked answers
Algebra and Function Practice
Direct variation describes a relationship where one quantity is a constant multiple of another. In this lesson, you will convert verbal descriptions into equations of the form y = kx, determine the constant of variation k, and use the equation to answer questions. The focus is on translation and solving, not on graphing or interpreting tables. You will practice identifying the two variables, writing the equation, solving for k, and checking your work with simple arithmetic.
Practice materials: Printable downloads are not available for this lesson yet. Use the instructions and four exercises below.
Identifying Variables and the Constant k
A direct-variation relationship can be written as y = kx, where k is a nonzero constant. In a verbal description, look for phrases such as 'varies directly as,' 'is proportional to,' or 'is a constant multiple of.' The quantity that depends on the other is usually y, and the independent quantity is x. For example, 'the cost C varies directly as the number n of items' gives C = kn. The constant k represents the cost per item.
To find k, substitute a known pair of values from the description into y = kx and solve. Keep units consistent: if y is in dollars and x is in hours, then k is in dollars per hour. Always define your variables in words before writing the equation. This prevents confusion when the description uses letters other than x and y.
- Read the description and identify the two quantities that are related.
- Assign one quantity to y (the dependent quantity) and the other to x (the independent quantity).
- Write the general equation y = kx using your variable names.
- Substitute a known pair of values to solve for k.
- Rewrite the equation with the numerical value of k.
- Use the equation to answer any follow-up question.
Solving for k and Answering Follow-Up Questions
Once you have the equation with k, you can answer questions about other values. For instance, if y = 3x and you need y when x = 7, substitute 7 for x to get y = 21. If you need x when y = 15, solve 15 = 3x to get x = 5. Always check that your answer makes sense in the context of the description.
When the description gives a rate directly, such as 'earns $12 per hour,' k is that rate. If the description gives a pair of values, such as 'when x = 4, y = 20,' then k = 20/4 = 5. Be careful to use the correct units and to keep the same order when dividing. A common error is to divide in the wrong order, giving 4/20 instead of 20/4.
- Substitute the known values into y = kx.
- Divide both sides by the value of x to isolate k.
- Simplify the result and note the units of k.
- Write the final equation with k.
- For a follow-up question, substitute the given value and solve for the unknown.
- Check your answer by substituting back into the equation.
Checking Your Work and Avoiding Common Errors
A quick check is to substitute your answer back into the original equation. If the equation is y = 2.5x and you found y = 10 when x = 4, then 2.5 times 4 equals 10, so the answer is correct. Another check is to see whether the relationship is proportional: if x doubles, y should double. This helps catch mistakes in finding k.
Pay attention to the wording. 'Varies directly as' means y = kx, not y = k/x. 'Is proportional to' also means direct variation. If the description says 'y is 3 more than x,' that is not direct variation because it is not a constant multiple. Only use y = kx when the description indicates a constant ratio between the two quantities.
- Substitute your answer back into the equation to verify.
- Check that doubling x doubles y.
- Confirm that the description uses direct-variation language.
- Ensure units are consistent and k has the correct units.
- Review the division order when solving for k.
Worked example
A recipe uses 3 cups of flour for every 2 cups of sugar. Write a direct-variation equation relating cups of flour F to cups of sugar S. Then find how many cups of flour are needed for 5 cups of sugar.
Worked answer
F = 1.5S; for S = 5, F = 7.5 cups of flour.
Why: The description gives a constant ratio: 3 cups flour to 2 cups sugar. Let F be flour and S be sugar. Since F varies directly as S, write F = kS. Substitute F = 3 when S = 2: 3 = k(2), so k = 1.5. The equation is F = 1.5S. For S = 5, F = 1.5(5) = 7.5. Check: 7.5/5 = 1.5, which matches k.
Four practice exercises
Exercise 1
The distance D a car travels varies directly as the time t in hours. In 3 hours, the car travels 150 miles. Write the equation relating D and t, and find the distance traveled in 5 hours.
Show answer and check
D = 50t; in 5 hours, D = 250 miles.
Why: Let D be distance in miles and t be time in hours. Write D = kt. Substitute D = 150 when t = 3: 150 = k(3), so k = 50. The equation is D = 50t. For t = 5, D = 50(5) = 250 miles.
Check: Substitute t = 5 into D = 50t to get 250. Check the ratio: 250/5 = 50, which matches k.
Exercise 2
The cost C of buying n notebooks varies directly as the number of notebooks. If 4 notebooks cost $10, write the equation and find the cost of 7 notebooks.
Show answer and check
C = 2.5n; 7 notebooks cost $17.50.
Why: Let C be total cost in dollars and n be the number of notebooks. Write C = kn. Substitute C = 10 when n = 4: 10 = k(4), so k = 2.5. The equation is C = 2.5n. For n = 7, C = 2.5(7) = 17.5 dollars.
Check: Substitute n = 7 into C = 2.5n to get 17.5. Check the ratio: 17.5/7 = 2.5, which matches k.
Exercise 3
The amount of water W in gallons used by a shower varies directly as the number of minutes m. A 6-minute shower uses 12 gallons. Write the equation and find the water used in a 9-minute shower.
Show answer and check
W = 2m; a 9-minute shower uses 18 gallons.
Why: Let W be water in gallons and m be minutes. Write W = km. Substitute W = 12 when m = 6: 12 = k(6), so k = 2. The equation is W = 2m. For m = 9, W = 2(9) = 18 gallons.
Check: Substitute m = 9 into W = 2m to get 18. Check the ratio: 18/9 = 2, which matches k.
Exercise 4
The number of pages P printed varies directly as the number of minutes t. A printer prints 45 pages in 3 minutes. Write the equation and find how many minutes it takes to print 120 pages.
Show answer and check
P = 15t; it takes 8 minutes to print 120 pages.
Why: Let P be pages and t be minutes. Write P = kt. Substitute P = 45 when t = 3: 45 = k(3), so k = 15. The equation is P = 15t. To find t when P = 120, solve 120 = 15t, so t = 8 minutes.
Check: Substitute t = 8 into P = 15t to get 120. Check the ratio: 120/8 = 15, which matches k.
Common mistakes
- Dividing in the wrong order when finding k, such as computing x/y instead of y/x. Always divide y by x: k = y/x. Write the equation y = kx first, then substitute and solve for k by dividing both sides by x.
- Confusing direct variation with inverse variation or with a linear relationship that has a nonzero intercept. Direct variation must be of the form y = kx with no added constant. If the description says 'varies inversely,' use y = k/x. If it says '3 more than,' it is not direct variation.
- Forgetting to define variables in words before writing the equation. Write a sentence such as 'Let C be the total cost in dollars and n be the number of items.' Then write C = kn. This keeps the meaning clear.
- Using inconsistent units, such as mixing hours and minutes. Convert all quantities to the same unit before finding k. For example, if time is given in minutes but the rate is per hour, convert minutes to hours first.
Self-check rubric
- Defines both variables in words before writing the equation.
- Writes the general form y = kx using the chosen variable names.
- Substitutes the known values correctly and solves for k.
- States the final equation with the numerical value of k.
- Answers the follow-up question by substituting and solving.
- Shows a check by substituting back or verifying the ratio.
- Uses consistent units and labels k with appropriate units.
Questions and answers
What does the constant k represent in a direct-variation equation?
k is the constant ratio y/x. It represents the rate or multiplier that relates the two quantities, such as cost per item or speed in miles per hour.
How do I know if a verbal description is a direct variation?
Look for phrases like 'varies directly as,' 'is proportional to,' or 'is a constant multiple of.' These indicate y = kx. Descriptions with added amounts or inverse relationships are not direct variation.
Can k be negative or zero?
In direct variation, k is nonzero. It can be negative if one quantity decreases as the other increases, but the relationship is still a constant multiple. k = 0 would mean y is always 0, which is not considered direct variation.
What if the description gives a rate directly, like 'earns $15 per hour'?
Then k is that rate, 15, and the equation is y = 15x if y is total earnings and x is hours. You do not need to solve for k separately.