Choosing a Coefficient to Fix the Solution Count

Match the x terms and the constants decide.

Gather the x terms with the letter still unknown

Sometimes a number in an equation is a letter still to be chosen, and the question asks which choice gives no solution, exactly one solution, or infinitely many.

Take ax + 3 = 5x + 7. The coefficient a is unknown, so treat it like any number. Subtract 5x from both sides: ax − 5x + 3 = 7. Subtract 3 from both sides: ax − 5x = 4. The two x terms have x in common, so ax − 5x = (a − 5)x, and the equation becomes (a − 5)x = 4.

Everything now depends on the bracket (a − 5), the number of x left after the clearing.

Make the x terms match: no solution

Choose a = 5. Then a − 5 = 0, and the equation reads 0x = 4. Zero lots of x is 0, whatever x is, and 0 is never 4. So there is no solution.

You can see it in the original equation too. With a = 5 it is 5x + 3 = 5x + 7. The two sides have the same x term, 5x, and different numbers, 3 and 7, so the right side is always 4 more than the left.

5x + 35x + 7both sides − 5x

With a = 5, take 5x off both pans: 3 is left against 7. Whatever x is, the right pan is 4 heavier, so the scales never level.

Any other choice: one solution

If a is any number other than 5, then a − 5 is not 0, and dividing by it is allowed. There is exactly one solution: x = 4 / (a − 5).

For example, choose a = 7. Then a − 5 = 2, so 2x = 4 and x = 2. Check in the original: 7 × 2 + 3 = 17 and 5 × 2 + 7 = 17.

7x + 35x + 7both sides − 5x

With a = 7, take 5x off both pans: 2x + 3 is left against 7. Those balance for one value of x, x = 2.

Infinitely many needs two things

For every number to be a solution, the equation must become a true statement once the x terms cancel. That needs two things: the x terms must match, so that they cancel, and the numbers left must agree, so that the statement is true.

In ax + 3 = 5x + 7 the numbers are 3 and 7. Even with a = 5 they disagree, so no choice of a gives infinitely many solutions.

Now take 5x + 3 = 5x + k. The x terms already match. Subtract 5x from both sides: 3 = k. If k = 3, this reads 3 = 3, which is true for every x, so there are infinitely many solutions. If k is any other number, the statement is false and there is no solution.

The usual mistakes

Choosing a to make x = 0. In (a − 5)x = 4, the question is about the coefficient of x becoming 0, not x itself. With a = 5 there is no solution at all, not the solution x = 0.

Matching the x terms and stopping. Equal x terms give either no solution or infinitely many. Which one depends on the numbers left, so always compare them.

Worked example: A Car Rental Price at the Airport Branch Set Against the City Branch

Question At its city branch a car rental firm charges $30 a day plus a cleaning fee of $45, so a rental of d days costs 30d + 45 dollars. At its airport branch it will charge k dollars a day and add an airport fee equal to two days' rent, so a rental of d days costs k(d + 2) dollars. (a) The firm wants the airport branch to charge the same amount more than the city branch on every rental, however long. Find k, and show that the equation k(d + 2) = 30d + 45 then has no solution. (b) The firm later decides that the two branches should charge the same for every rental. The airport keeps its price rule from (a). What cleaning fee should the city branch charge instead of $45?

  1. 1.Expand the airport's rule: k(d + 2) = kd + 2k. The airport price minus the city price is (kd + 2k) − (30d + 45) = (k − 30)d + 2k − 45 dollars.

    airport: k(d + 2) = kd + 2kcity: 30d + 45airport − city = (k − 30)d + 2k − 45
    airport: k(d + 2) = kd + 2kcity: 30d + 45airport − city = (k − 30)d + 2k − 45
    The airport charges k(d + 2) = kd + 2k dollars, which is (k − 30)d + 2k − 45 dollars more than the city.

  2. 2.For this difference to be the same on every rental, it must not change with d, so the d terms must match: k − 30 = 0, and k = 30. For any other k the difference changes by k − 30 dollars with each extra day.

    the gap must not change with dso the d terms match: k − 30 = 0k = 30, and the airport charges 30d + 60
    the gap must not change with dso the d terms match: k − 30 = 0k = 30, and the airport charges 30d + 60
    That difference is the same for every d only when the d terms match: k − 30 = 0, so k = 30.

  3. 3.With k = 30 the equation is 30(d + 2) = 30d + 45, that is 30d + 60 = 30d + 45. Subtract 30d from both sides: 60 = 45, which is false whatever d is.

    a 3-day rentalCity3 days: $90fee $45$135Airport3 days: $90fee $60$150a 5-day rentalCity5 days: $150fee $45$195Airport5 days: $150fee $60$21030(d + 2) = 30d + 4530d + 60 = 30d + 45subtract 30d: 60 = 45, which is false
    a 3-day rentalCity$90$45$135Airport$90$60$150a 5-day rentalCity$150$45$195Airport$150$60$21030(d + 2) = 30d + 4530d + 60 = 30d + 45subtract 30d: 60 = 45, which is false
    With k = 30: 30d + 60 = 30d + 45. Subtracting 30d from both sides leaves 60 = 45, which is false.

  4. 4.(a) k = 30. The equation has no solution: the two prices are never equal, and the airport always charges 60 − 45 = $15 more.

    a 3-day rentalCity3 days: $90fee $45$135Airport3 days: $90fee $60$150$15 morea 5-day rentalCity5 days: $150fee $45$195Airport5 days: $150fee $60$210$15 moreno solution: the prices are never equalthe airport always charges $15 more
    a 3-day rentalCity$90$45$135Airport$90$60$150$15 morea 5-day rentalCity$150$45$195Airport$150$60$210$15 moreno solution: the prices are never equalthe airport always charges $15 more
    (a) k = 30, and the equation has no solution: the airport always charges $15 more.

  5. 5.With a city fee of f dollars the equation is 30d + 60 = 30d + f. The d terms already match, so the constants decide: subtracting 30d from both sides leaves 60 = f.

    a 3-day rentalCity3 days: $90f = 60$150Airport3 days: $90fee $60$150a 5-day rentalCity5 days: $150f = 60$210Airport5 days: $150fee $60$21030d + 60 = 30d + fthe d terms match, so the constants decide: f = 60
    a 3-day rentalCity$90f = 60$150Airport$90$60$150a 5-day rentalCity$150f = 60$210Airport$150$60$21030d + 60 = 30d + fthe d terms match, so the constants decide: f = 60
    With a city fee of f dollars, 30d + 60 = 30d + f. The d terms match, so the constants decide: f = 60.

  6. 6.(b) The city branch should charge a cleaning fee of $60. The equation then reads 60 = 60 once the d terms cancel, which is true for every d: there are infinitely many solutions. Check with d = 3: 30 × 5 = 150 and 90 + 60 = 150.

    a 3-day rentalCity3 days: $90fee $60$150Airport3 days: $90fee $60$150a 5-day rentalCity5 days: $150fee $60$210Airport5 days: $150fee $60$21060 = 60 for every d: infinitely many solutionscheck, 3 days: 30 × 5 = 150 and 90 + 60 = 150
    a 3-day rentalCity$90$60$150Airport$90$60$150a 5-day rentalCity$150$60$210Airport$150$60$21060 = 60 for every d: infinitely many solutionscheck, 3 days: 30 × 5 = 150 and 90 + 60 = 150
    (b) A cleaning fee of $60. Then 60 = 60 for every d, and there are infinitely many solutions.

Answer: (a) k = 30: the prices are never equal, and the airport always charges $15 more; (b) $60

Common mistakes

  • Treating d as a single number and solving k(d + 2) = 30d + 45 for k. The firm's condition is about every length of rental, so it fixes the coefficient of d, not one value of k for one rental.
  • Writing the airport price as kd + 2. The fee is two days' rent, which is 2k dollars, so the bracket multiplies both terms: k(d + 2) = kd + 2k.

More equations and inequalities problems, worked step by step →

Practice Choosing a Coefficient to Fix the Solution Count in the app