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What a solution is, and checking one

A sentence with a letter in it is neither true nor false until the letter has a value. Which values make it true, and how can you be sure a value you found really works?

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Why this exists

Wilma keeps coins in a jar. She drops in 15 more and counts 42. How many were in the jar before? Calling the unknown number c, the story becomes one sentence:

c + 15 = 42

As Expression, equation and inequality explains, an equation is a claim. But this claim has a letter in it, and a claim about “some number c” cannot be judged true or false until you say which number. Put in c = 10 and the claim is false, since 10 + 15 = 25. Put in c = 27 and it is true.

So the real question an equation asks is: which numbers make me true? Those numbers have a name, and there is a sure way to test any number you are handed or any number you work out: put it back in and see.

Definition

A solution of an equation or inequality in one letter is a value of the letter that makes the sentence true when it is put in place of the letter.

To check a value, substitute it for the letter, work out each side as in Evaluating expressions and formulas, and compare. For an equation the two sides must come out equal. For an inequality the comparison it states (> or <) must hold. While the comparison is still being worked out, the equals sign is written \stackrel{?}{=}, read “is it equal?”:

27 + 15 \stackrel{?}{=} 42 \qquad 42 = 42 \ \text{true}

To solve a sentence is to find all its solutions. Often a question hands you a short list of candidates written as a set, such as \{3, 4, 5, 6\}, and asks which of them are solutions; then the job is to check each one.

Intuition

Think of x + 7 = 12 as two bars that must be the same length. The top bar is 12 units long. The bottom bar is a piece of unknown length x joined to a piece 7 long.

12 x 7
Figure 1Only one length of the shaded piece makes the two bars match, and it is 5.

Make the shaded piece longer than 5 and the bottom bar overshoots 12; make it shorter and it falls short. Exactly one length fits, so the equation has exactly one solution, x = 5. You can find it by asking a plain question, “what number plus 7 is 12?”, and then check it: 5 + 7 = 12.

An inequality behaves differently. The sentence x > 4 is true for 5, for 6, for 100, and for 4.5 and 4.01 too. It does not pin down one number; it marks off a whole stretch of them.

x + 7 = 12
 0 1 2 3 4 5 6 7 8 9 10
x > 4
 0 1 2 3 4 5 6 7 8 9 10 ...
Figure 2The equation is true for one number; the inequality for every number past 4. Only whole numbers are drawn, but 4.5 and \displaystyle 7\frac{1}{4} make x > 4 true as well.

Results

Testing a value. A number is a solution exactly when substituting it makes the sentence true. For an equation, both sides give the same number; for an inequality, the stated comparison holds.

One solution or many. An equation in which the letter appears once, plus or times a number that is not zero, has exactly one solution. An inequality such as x > 4 or x < 9 has more solutions than anyone could list.

Context decides which values count. If the letter counts pots, coins or people, a solution must be a whole number. An equation can have a solution as a number and still have none that fits the story: 3n = 20 is true for \displaystyle n = 6\frac{2}{3}, but no whole number of pots works.

Why it is true

Take x + 7 = 12. If x stands for a number smaller than 5, then x + 7 is smaller than 5 + 7 = 12, so the sentence is false. If x is larger than 5, then x + 7 is larger than 12, false again. Only 5 is left, and it works. The same squeeze works whenever the letter appears once with a number added, or multiplied by a number greater than zero: a bigger value of the letter always gives a bigger side, so at most one value can land exactly on the other side.

For x > 4 the opposite happens. Whatever solution you name, adding 1 to it gives a larger number, which is also more than 4, so the solutions never run out. There are fresh ones between the whole numbers as well: 4.5, 4.25 and 4.125 are all more than 4.

For 3n = 20, every time n grows by 1 the product 3n grows by 3. At n = 6 it is 18; at n = 7 it is 21. It jumps straight over 20, so no whole number makes the sentence true, even though the number \displaystyle 6\frac{2}{3} does: \displaystyle 3 \times 6\frac{2}{3} = 18 + 2 = 20.

Worked examples

Example 1

Checking two values

Wilma’s jar held c coins. After she drops in 15 more it holds 42, so c + 15 = 42. Is c = 27 a solution? Is c = 37?

Solution

Substitute each value for c and work out the left side.

27 + 15 \stackrel{?}{=} 42 \qquad 42 = 42

The sides match, so c = 27 is a solution: the jar held 27 coins.

37 + 15 \stackrel{?}{=} 42 \qquad 52 \ne 42

The sides differ, so c = 37 is not a solution. In the story, 37 coins plus 15 would make 52, not 42.

Example 2

Testing every member of a set

A pottery class’s drying rack already holds 4 pots. The class adds k trays of 6 pots each, and now the rack holds 34 pots, so 6k + 4 = 34. Which values in \{3, 4, 5, 6\} are solutions?

Solution

Each value is substituted for k, and the left side is worked out with multiplication before addition.

k6k + 4equal to 34?
318 + 4 = 22no
424 + 4 = 28no
530 + 4 = 34yes
636 + 4 = 40no

Only k = 5 is a solution: five trays of six is 30 pots, and with the 4 already there that makes 34. Notice the left side grows by 6 for each extra tray, so once it passes 34 at k = 6 no larger k can bring it back.

Example 3

Fractions and decimals

Is \displaystyle m = \frac{3}{4} a solution of \displaystyle m + \frac{5}{8} = \frac{11}{8}? Is w = 0.8 a solution of 2.5w = 3?

Solution

To add \displaystyle \frac{3}{4} and \displaystyle \frac{5}{8}, write \displaystyle \frac{3}{4} in eighths as \displaystyle \frac{6}{8}:

\frac{6}{8} + \frac{5}{8} \stackrel{?}{=} \frac{11}{8} \qquad \frac{11}{8} = \frac{11}{8}

So \displaystyle m = \frac{3}{4} is a solution.

For the second, 2.5w means 2.5 times w:

2.5 \times 0.8 \stackrel{?}{=} 3 \qquad 2 \ne 3

So w = 0.8 is not a solution. (The solution is 1.2, since 2.5 \times 1.2 = 3.) Checking works the same way whatever kind of number is tried.

Example 4

An inequality has several solutions in one set

Which values in \displaystyle \{4, 6, 6\frac{1}{2}, 7, 8\} are solutions of 3b < 20?

Solution

Substitute each value and ask whether the product is less than 20.

b3bless than 20?
412yes
618yes
\displaystyle 6\frac{1}{2}\displaystyle 19\frac{1}{2}yes
721no
824no

Three values work: 4, 6 and \displaystyle 6\frac{1}{2}. The set only listed five candidates; plenty of numbers that were not offered are also solutions, such as 2, 5 and 6.6 (since 3 \times 6.6 = 19.8).

Example 5

Finding a solution by reasoning, then checking it

Yves takes 9 coins out of his jar and 26 are left. Write an equation for the number j he started with, find j, and check it.

Solution

Starting with j coins and taking away 9 leaves 26:

j - 9 = 26

Ask the plain question: what number, less 9, leaves 26? Putting the 9 coins back with the 26 that remain gives the starting amount, 26 + 9 = 35. Addition and subtraction undo each other, which is why the fact 26 + 9 = 35 also says 35 - 9 = 26.

Now check in the original equation:

35 - 9 \stackrel{?}{=} 26 \qquad 26 = 26

True, so j = 35. The check matters because it tests the answer against the sentence itself, not against the reasoning that produced it.

Example 6

No whole-number solution in the story

Brielle fires pots in 3 kiln loads with the same number n of pots in each load, 20 pots in all. Is there a solution of 3n = 20 that fits the story?

Solution

The letter counts pots, so only whole numbers make sense. Try the two near 20 \div 3:

3 \times 6 = 18 \qquad 3 \times 7 = 21

Six pots a load gives too few, seven too many, and each extra pot per load adds 3 to the total, so the total jumps from 18 to 21 without ever being 20. No whole number is a solution.

As a sentence about numbers, 3n = 20 does have a solution, \displaystyle n = 6\frac{2}{3}, because \displaystyle 3 \times 6\frac{2}{3} = 20. But two thirds of a pot cannot go into a kiln, so in the story the answer is that 20 pots cannot be split into 3 equal loads.

Example 7

The same two sides, an equation and an inequality

Which values in \{0, 1, 2, 3, 4\} are solutions of 2x + 3 = 3x + 1? Which are solutions of 2x + 3 > 3x + 1?

Solution

The letter appears on both sides, so substitute the same value into each side.

x2x + 33x + 1
031
154
277
3910
41113

The equation holds only where the two columns match: x = 2. The inequality holds where the left column is larger: x = 0 and x = 1.

The table also shows why nothing beyond the set helps. Each step up in x adds 2 to the left side and 3 to the right, so the right side gains 1 each time. Once it catches up at x = 2 it stays ahead, so no larger number solves either sentence. Numbers between 0 and 2 that were not in the set do solve the inequality: at x = 1.5 the sides are 6 and 5.5, and 6 > 5.5.

Boundaries

A solution belongs to one sentence. The value 5 solves x + 7 = 12 and says nothing about x + 7 = 15, even though the same letter appears.

Testing a set only sorts the candidates it lists. If no member works, the sentence may still have a solution outside the set; and for an inequality the set can never show all the solutions.

“Exactly one solution” needs the letter to appear once, added to a number or multiplied by a number that is not zero. Other equations can behave differently: 0 \times x = 5 has no solution, and x + 3 = 3 + x is true for every number by the commutative property. Those cases are taken up in No solution, one solution, or every number.

A value that makes part of the sentence undefined cannot be a solution. In \displaystyle \frac{12}{x} = 4, the value x = 3 works, but x = 0 cannot even be tested, because division by zero has no answer.

The story limits the solutions. A count must be a whole number, and a length or a weight cannot be less than zero, so a number that satisfies the bare sentence may still be thrown out.

Where this goes next

Testing values works, but it is slow when the solution could be any number. The next step is to find solutions directly by doing the same thing to both sides of an equation, which keeps every solution a solution: Inverse operations and the properties of equality explains why that is allowed, and One-step equations puts it to work. Because an inequality has so many solutions, they are drawn rather than listed, as in Inequalities on the number line.

Exercises

Testing one value

  1. Is x = 9 a solution of x + 14 = 23?
  2. Is k = 6 a solution of 7k = 48?
  3. Is p = 13 a solution of p - 5 = 18?
  4. Is t = 2.4 a solution of t + 1.7 = 4.1?
  5. Is \displaystyle r = \frac{2}{3} a solution of 6r = 4?
  6. Is y = 30 a solution of \displaystyle \frac{y}{5} = 7?

Testing a set

  1. Which values in \{8, 10, 12, 14\} are solutions of 4a - 6 = 42?
  2. Which values in \{1.5, 2, 2.5, 3\} are solutions of 2d + 1 = 6?
  3. Which values in \{0, 1, 2, 3\} are solutions of 5g = 4g + 3?
  4. Which values in \{5, 6, 7, 8\} are solutions of x + 11 = 30?

Inequalities

  1. Which values in \{2, 5, 5.5, 9\} are solutions of h > 5?
  2. Which values in \{0, 3, 3.5, 4, 7\} are solutions of 2m < 8?
  3. Which values in \displaystyle \{\frac{1}{2}, 1, \frac{3}{2}, 2\} are solutions of \displaystyle q + \frac{3}{4} > 2?
  4. Give four different solutions of v < 1.5, at least one of them not a whole number.

In context

  1. Esra’s coin jar held some coins. She added 18 and now has 65. Write an equation, test 57 and 47 in it, and say how many coins the jar held.
  2. Ansgar packs 30 mugs from the pottery class into boxes of 4, every box full. Write an equation for the number of boxes n, and decide whether it has a solution that fits the story.
  3. Wilma’s jar holds 85 coins and she adds h more. The lid only closes if the total stays below 120, so 85 + h < 120. (a) Which values in \{20, 34, 35, 40\} keep the lid closing? (b) How many whole numbers h work, counting h = 0?
  4. Each bowl in Esra’s class uses 0.75 kilograms of clay, and she has 6 kilograms. Is p = 8 a solution of 0.75p = 6, and what does the answer mean?

Reasoning

  1. Explain why 4 is not a solution of x > 4, even though 4 is the number written in the sentence.
  2. Explain why x + 9 = 15 cannot have two different solutions.
  3. Explain why the solutions of y > 3 could never all be written in a list.
  4. The equation 5n = 40 has a whole-number solution but 5n = 42 does not. Explain how you can tell without trying every whole number, and find the number that does make 5n = 42 true.
  5. Write an equation in x whose only solution is \displaystyle \frac{1}{2}, and an inequality that \displaystyle \frac{1}{2} makes true but 1 does not.
Answers
  1. Yes. 9 + 14 = 23.
  2. No. 7 \times 6 = 42, and 42 \ne 48.
  3. No. 13 - 5 = 8, and 8 \ne 18. (The solution is 23.)
  4. Yes. 2.4 + 1.7 = 4.1.
  5. Yes. \displaystyle 6 \times \frac{2}{3} = \frac{12}{3} = 4.
  6. No. 30 \div 5 = 6, and 6 \ne 7. (The solution is 35.)
  7. Only 12. The left side gives 26, 34, 42 and 50 for the four values, and only 4 \times 12 - 6 = 42 matches.
  8. Only 2.5. The left side gives 4, 5, 6 and 7, and only 2 \times 2.5 + 1 = 6 matches.
  9. Only 3. The two sides are 0 and 3, 5 and 7, 10 and 11, then 15 and 15.
  10. None of them. The left side gives 16, 17, 18 and 19, never 30. The solution, 19, is not in the set.
  11. 5.5 and 9. The value 5 is not a solution, because 5 > 5 is false: 5 equals 5.
  12. 0, 3 and 3.5, since 0, 6 and 7 are each less than 8. For 4 the product is 8, which is not less than 8, and for 7 it is 14.
  13. \displaystyle \frac{3}{2} and 2. The sums are \displaystyle \frac{5}{4}, \displaystyle \frac{7}{4}, \displaystyle \frac{9}{4} and \displaystyle \frac{11}{4}, and only the last two are more than \displaystyle 2 = \frac{8}{4}.
  14. For example 0, \displaystyle \frac{1}{2}, 1 and 1.25. Any number below 1.5 works.
  15. c + 18 = 65. Testing 57: 57 + 18 = 75, not a solution. Testing 47: 47 + 18 = 65, a solution. The jar held 47 coins.
  16. 4n = 30. No solution fits: 4 \times 7 = 28 and 4 \times 8 = 32, and the total jumps by 4 each time, so it is never 30. The number 7.5 makes the sentence true, but half a box would not be full.
  17. (a) 20 and 34: 85 + 20 = 105 and 85 + 34 = 119 are below 120, while 35 gives exactly 120 and 40 gives 125. (b) 35 whole numbers, 0 through 34.
  18. Yes. 0.75 \times 8 = 6, so 8 bowls use exactly all 6 kilograms of clay.
  19. Substituting gives 4 > 4, which is false: 4 is equal to 4, not greater. The sentence describes numbers past 4, and 4 itself is the edge, not one of them.
  20. A number below 6 gives a sum below 6 + 9 = 15, and a number above 6 gives a sum above 15. Only 6 lands on 15, so there is exactly one solution.
  21. Any solution plus 1 is another, larger solution, so they never run out. Even between 3 and 4 there is no end of them: 3.5, then 3.25 halfway back to 3, then 3.125, and so on.
  22. 5 \times 8 = 40 and 5 \times 9 = 45, and 5n goes up by 5 each time n goes up by 1, so it skips from 40 to 45 and never hits 42. The number 8.4 works: 5 \times 8.4 = 42.
  23. For example \displaystyle x + \frac{1}{2} = 1, whose only solution is \displaystyle \frac{1}{2}, and x < 1, which \displaystyle \frac{1}{2} makes true and 1 makes false. Many other answers are possible.
kg1128 · updated 2026-10-12 · CC-BY-SA-4.0