Math 8 · 8.EE.C.7A

No solution, one solution, or every number

Solving an equation usually ends with the unknown equal to one number. What has happened when the unknown disappears instead?

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Equations with variables on both sidesMath 8

Where you are 11 in Linear equations

Algebra9

Linear equations11

▴ 6 above Multi-step equations Absolute value equations Equations with variables on both sides Literal equations: solving for one variable

No solution, one solution, or every number

Why this exists

Most equations you solve end the same way: the unknown alone on one side, a number on the other, x = 4. Every step keeps the two sides equal, by the properties of equality, so the last line names the one number that makes the first line true.

Sometimes that last line never arrives. You gather the variable terms on one side, as in solving with the unknown on both sides, and they all disappear, because 6x - 6x is 0. What is left has no letter in it, such as 15 = 1 or -12 = -12. That is not a slip in the working. The equation is telling you that either no number makes it true, or every number does. Knowing which of the three cases you are in, and why, finishes the job of solving linear equations.

This lesson is taught over two days. Part 1 is what the vanished variable means. Part 2 is predicting the case before you solve, building equations with a chosen number of solutions, and seeing the three cases as two lines on a graph.

Definition

A solution of an equation is a number that makes it true when it replaces the unknown (what a solution is). A linear equation in one unknown is one where, after each side is simplified, the unknown appears only to the first power: each side has the form ax + b. Such an equation has exactly one of three kinds of answer.

Exactly one solution. 5x - 1 = 2x + 8 is true for x = 3 and for no other number.
No solution. No number makes it true. x + 5 = x + 2 is false whatever x is, because a number plus 5 is never that number plus 2. Such an equation is called a contradiction.
Every number is a solution, so there are infinitely many. 2(x + 1) = 2x + 2 is true whatever x is, because the two sides are one expression written two ways, by the distributive property. Such an equation is called an identity.

The symbol \neq means “is not equal to”, so 15 \neq 1 says that 15 and 1 are different numbers.

Intuition

Part 1: two expressions that never or always agree

Each side of an equation is an expression: a rule that turns a number into a value. The equation asks for which numbers the two rules give the same value. A table of values lets you watch.

x	2x + 7	2x + 3	4x + 1	2(x + 1) + 1
0	7	3	1	3
1	9	5	5	5
2	11	7	9	7
3	13	9	13	9
4	15	11	17	11

2x + 7 = 4x + 1. The gap between the columns is 6, then 4, 2, 0, and then 4x + 1 is ahead. The faster rule catches the slower one once, at x = 3. One solution.
2x + 7 = 2x + 3. The gap is 4 in every row. Both rules add 2 for every step in x, so the lead never changes. No solution.
2x + 3 = 2(x + 1) + 1. The two columns are identical. Every number is a solution.

A table shows only a few rows; algebra speaks for every row at once. Subtracting 2x from both sides of 2x + 7 = 2x + 3 removes the part the two sides share and leaves the gap: 7 = 3. That is false, so the gap is never zero.

Part 2: two lines

Draw each side as a line, such as y = 2x + 7 and y = 4x + 1. A solution is an x where the lines are at the same height, which is where they meet. The coefficient of x is the line’s Slope and the constant is where it crosses the y-axis. Different slopes cross once; equal slopes with different starting heights are parallel; equal slopes and heights are one line. Example 6 draws all three.

Results

Part 1: what is left when the variable vanishes

Solve as usual. If the variable terms cancel, the last line is a statement with no letter in it.

If it is false, such as 15 = 1, the equation has no solution.
If it is true, such as -12 = -12, every number is a solution.

If a variable term survives, finishing the solving gives exactly one solution.

Sample numbers put into the original equation support the answer: with no solution the two sides differ by the same amount every time, and with every number they agree every time. One number that works proves nothing on its own, since an equation with exactly one solution also passes at that number.

Part 2: predicting from the coefficients

Simplify each side completely, distributing and combining like terms, so that the equation reads ax + b = cx + d. Then compare.

The simplified sides	Solutions	As two lines
a \neq c	exactly one, x = \dfrac{d - b}{a - c}	cross at one point
a = c and b \neq d	none	parallel
a = c and b = d	every number	the same line

The constants decide anything only when the coefficients are equal. When a \neq c there is one solution whatever b and d are; the constants move it but cannot remove it.

Why it is true

Start from the simplified form

ax + b = cx + d

Subtract cx from both sides, then subtract b. Each step can be undone by adding back, so the new equation has exactly the same solutions as the old one:

(a - c)x = d - b

Now there are two cases.

If a \neq c, the number a - c is not zero, so both sides can be divided by it: x = \dfrac{d - b}{a - c}. Every step can be run backwards, so this number really is a solution, and it is the only one.

If a = c, then a - c = 0 and the left side is 0 \cdot x, which is 0 whatever x is. The equation now says 0 = d - b, that is, b = d. There is no x left in it, so its truth cannot depend on x. If b = d it is true for every x; if b \neq d it is false for every x.

Put another way, the difference of the two sides is (a - c)x + (b - d). If a = c it is the fixed number b - d, so it is zero always or never. If a \neq c it changes steadily with x and passes through zero at exactly one place.

This rules out every other count. A linear equation never has exactly two solutions, or exactly ten. If two different numbers both make one true, it is in neither the one-solution case nor the no-solution case, so every number makes it true.

Worked examples

Examples 1 to 4 go with Part 1. Examples 5 and 6 go with Part 2.

Example 1

A false statement at the end

Solve 3(2x + 5) = 6x + 1.

Solution

Distribute the 3 over the bracket, by the distributive property:

6x + 15 = 6x + 1

Subtract 6x from both sides, by the subtraction property of equality:

15 = 1

No x is left, and 15 \neq 1, so this statement is false. Each step kept the same solutions, so the original equation has no solution. The second line shows why: both sides contain 6x, and the left adds 15 where the right adds 1, so the left side is always 14 more than the right.

Check with sample numbers. At x = 0 the sides are 3(5) = 15 and 1. At x = 10 they are 3(25) = 75 and 61. Different by 14 both times.

Example 2

A true statement at the end

Solve 4(x - 3) + x = 5x - 12.

Solution

Distribute the 4:

4x - 12 + x = 5x - 12

Combine the like terms on the left, 4x + 1x = (4 + 1)x = 5x, by the distributive property:

5x - 12 = 5x - 12

Subtract 5x from both sides:

-12 = -12

This is true, and there is no x in it, so it is true whatever x was. Every number is a solution: the equation is an identity. The line 5x - 12 = 5x - 12 already said so, since its two sides are the same expression.

Check with sample numbers. At x = 2 the left side is 4(-1) + 2 = -2 and the right side is 10 - 12 = -2. At x = -1 they are 4(-4) + (-1) = -17 and -5 - 12 = -17.

Example 3

Two gyms that never cost the same

Fionnuala is choosing between two gyms. Gym A charges 45 dollars to join and then 22 dollars a month. Gym B charges 25 dollars to join, then 18 dollars a month plus a locker fee of 4 dollars a month. Is there a number of months after which she would have paid the same total at either gym?

Solution

Let m be the number of months. The totals are 45 + 22m dollars at Gym A and 25 + 18m + 4m dollars at Gym B. “The same total” is the equation

45 + 22m = 25 + 18m + 4m

Combine the like terms on the right, 18m + 4m = 22m:

45 + 22m = 25 + 22m

Subtract 22m from both sides:

45 = 25

This is false, so the equation has no solution. Once the locker fee is counted, both gyms charge 22 dollars every month, so the difference in joining fees never closes: Gym A is always 20 dollars dearer. At m = 6, for instance, Gym A comes to 45 + 132 = 177 dollars and Gym B to 25 + 108 + 24 = 157. The question was “when?”, and “never” is a complete answer.

Example 4

One perimeter, two ways

Figure 1 shows an L-shaped plot, with lengths in meters. The step is (2x + 3) - x = x + 3 long and the inner edge is (x + 4) - 3 = x + 1 tall, as labeled. Hiroko finds the perimeter by adding the six sides. Cosmin notices that the step and the inner edge, slid outwards, would fill in the dashed rectangle exactly, so he writes the perimeter as 2(2x + 3) + 2(x + 4). Write the equation that says their perimeters agree, and solve it. Which values of x make sense?

Figure 1Hiroko adds the six sides. Cosmin uses the dashed rectangle. Their two expressions agree for every x.

Solution

Hiroko’s sum, with the x terms and the constants gathered (the commutative and associative properties let a sum be reordered and regrouped):

\begin{aligned} &(2x + 3) + 3 + (x + 3) + (x + 1) + x + (x + 4) \\ &\quad = (2x + x + x + x + x) + (3 + 3 + 3 + 1 + 4) \\ &\quad = 6x + 14 \end{aligned}

Cosmin’s expression, distributed:

2(2x + 3) + 2(x + 4) = 4x + 6 + 2x + 8 = 6x + 14

So the equation simplifies to 6x + 14 = 6x + 14, and subtracting 6x from both sides leaves

14 = 14

This is true, so every number is a solution: Cosmin’s shortcut is not luck. The step x + 3 and the top x make up the full width 2x + 3, and the inner edge x + 1 and the short side 3 make up the full height x + 4.

In the figure, though, x is the length of the top edge, so here “every number” means every positive x. At x = 5 the six sides are 13, 3, 8, 6, 5 and 9, which add to 44, and Cosmin’s rectangle gives 2(13) + 2(9) = 44.

Example 5

Predicting the case before solving

Without solving, decide how many solutions each equation has. Then solve the one that has exactly one.

(a) 7x - 2(x + 3) = 5x - 6 \qquad (b) 8 - 3x = -3(x - 2) \qquad (c) x + 4(x - 1) = 4x + 9

Solution

Simplify each side to the form ax + b, then compare the coefficients of x first and the constants only if the coefficients match.

(a) Distributing -2 over x + 3 gives -2x - 6, so the left side is 7x - 2x - 6 = 5x - 6, the same as the right side. Same coefficient and same constant, so every number is a solution.

(b) The right side is -3x + 6. The left side, reordered by the commutative property, is -3x + 8. The coefficients are both -3 but the constants are 8 and 6, so there is no solution: the left side is always 2 more than the right.

(c) The left side is x + 4x - 4 = 5x - 4, and the right side is 4x + 9. The coefficients 5 and 4 differ, so there is exactly one solution. Subtract 4x from both sides, then add 4:

\begin{aligned} 5x - 4 &= 4x + 9 \\ x - 4 &= 9 \\ x &= 13 \end{aligned}

Check. 13 + 4(12) = 61 and 4(13) + 9 = 61.

In (c) the constants played no part in deciding how many solutions there were. They decided only where the one solution sits.

Example 6

Building an equation to order

Write an expression on the right of 5x - 3(x - 1) - 1 = \;? so that the equation has (a) no solution, (b) every number as a solution, (c) exactly one solution, x = 3. Then describe each equation as a pair of lines.

Solution

First simplify the left side, distributing -3 over x - 1:

5x - 3x + 3 - 1 = 2x + 2

(a) For no solution the right side needs the same coefficient, 2, and a different constant. For example 2x + 5: then 2x + 2 = 2x + 5 leads to 2 = 5, which is false. Any 2x + c with c \neq 2 works.

(b) For every number, the right side must be the same expression as the left: 2x + 2, or anything equal to it, such as 2(x + 1).

(c) For exactly one solution the coefficient must differ from 2, and the right side must equal the left side when x = 3. The left side at x = 3 is 2(3) + 2 = 8. Choosing the coefficient -1, the right side is -x + c with -3 + c = 8, so c = 11, and the right side is -x + 11. Solving confirms it: 2x + 2 = -x + 11 gives 3x = 9, so x = 3.

Figure 2The two sides of each equation in Example 6, drawn as lines. Parallel lines, no solution. One line, every number. Crossing lines, one solution.

Figure 2 draws each side as a line. In (a), y = 2x + 2 and y = 2x + 5 both have slope 2 and cross the y-axis at 2 and 5: they are parallel, always 3 apart, and share no point. In (b), the two sides draw the same line, and every point on it is shared. In (c), the slopes 2 and -1 differ, so the lines cross once, at (3, 8): the solution is the x-coordinate 3, and 8 is the value both sides take there. This is a first look at solving a system of equations by graphing.

Boundaries

Linear equations only. The three cases rest on each side simplifying to ax + b. Once the unknown is squared, other counts appear: x^2 = 4 has exactly two solutions, 2 and -2.

Every number means every number for which both sides mean something. The equation \dfrac{x}{x} = 1 is true for every x except 0, where the left side divides by zero and has no value. An unknown in a denominator takes an equation outside this lesson.

Compare the simplified sides. In 3(x + 2) = 3x + 2 the left constant is 6, not 2, so there is no solution. And the formula x = \dfrac{d - b}{a - c} has no value when a = c, since it would divide by zero.

A setting can narrow the answer. When the unknown is a length, “every number” means every positive number, as in Example 4; when it counts months, only 0, 1, 2, 3 and so on make sense.

One number is not a proof. A sample number that works fits both the one-solution case and the every-number case. For a linear equation, two different numbers that both work do settle it.

Where this goes next

The picture in Example 6 is the start of systems of equations. There the two lines are two separate equations in x and y, and the same three outcomes return: lines that cross give one solution, parallel lines give none, and two equations for the same line give infinitely many. Solving by graphing and graphing a line from its equation are where that happens.

Identities grow too: many useful facts in algebra, such as (a + b)^2 = a^2 + 2ab + b^2, are equations true for every value, and what an identity is takes that further.

Exercises

Part 1: solve, and name the case

3(x - 4) = 3x - 12
2(4x + 1) = 8x + 5
7x - 3 = 4x + 9
-(x - 8) + 3x = 2x + 8
9 - 4x = -2(2x - 3)

Part 1: checking with numbers

Show that x = 4 and x = -2 both make 2(x + 6) - x = x + 12 true. Then solve the equation and say which case it is.
For 4(x + 2) = 4x + 3, work out both sides at x = 0, x = 1 and x = 10. What do you notice? Then solve the equation.
Cosmin tries x = 2 in 3x + 4 = x + 8, finds that both sides equal 10, and says every number is a solution. Is he right?

Part 1: in context

Delphine-Ava compares two gyms. Gym C charges 35 dollars to join and 19 dollars a month. Gym D charges 10 dollars to join, 15 dollars a month, and 4 dollars a month for classes. After how many months are the two totals equal?
Gym E charges 50 dollars to join and 16 dollars a month. Gym F charges nothing to join and 21 dollars a month. After how many months are the totals equal, and what is the total then?
A rectangle has length 2x + 5 and width x - 1. Hiroko writes its perimeter as (2x + 5) + (x - 1) + (2x + 5) + (x - 1) and Ezekiel writes 2(2x + 5) + 2(x - 1). Show that the equation saying the two agree has every number as a solution. Which values of x make sense for the rectangle?

Part 2: predict before solving

Without solving, say how many solutions each equation has. (a) 9x - 4 = 9x + 4 (b) 2(5x - 3) = 10x - 6 (c) 7x + 1 = 7 + x (d) -3(x + 2) = 6 - 3x
Which of these has exactly one solution, and what is it? 5 - 2x = -2x + 5, \;5 - 2x = 2x + 5, \;5 - 2x = -2x - 5

Part 2: build an equation

Write an expression for the right side of 4(x + 3) - x = \;? so that the equation has (a) no solution, (b) every number as a solution, (c) exactly one solution, x = 2.
For which value of a does ax - 4 = 5x + 1 have no solution? Can any value of a make every number a solution?
Delphine-Ava’s puzzle: think of a number, add 5, multiply by 4, subtract 8, divide by 4, and take away the number you thought of. Calling the starting number n, show that everyone ends at 3. Then change one number in the puzzle so that everyone ends at 7.

Part 2: as two lines

The lines y = 3x - 4 and y = 6 - 2x cross at one point. Find it by solving an equation.
Show that y = 2(x - 3) + 4 and y = 2x - 2 are the same line, and say what this means for the equation 2(x - 3) + 4 = 2x - 2.

Reasoning

Explain why a linear equation in one unknown cannot have exactly two solutions.
Hiroko says: “If x = 0 and x = 1 both make a linear equation true, then every number does.” Is she right? Would one number be enough? Does her rule work for x^2 = x?

Answers

Every number. Distributing gives 3x - 12 = 3x - 12, and subtracting 3x leaves -12 = -12, which is true.
No solution. 8x + 2 = 8x + 5, and subtracting 8x leaves 2 = 5, which is false.
Exactly one, x = 4. Subtract 4x: 3x - 3 = 9, so 3x = 12. Check: 28 - 3 = 25 and 16 + 9 = 25.
Every number. The left side is -x + 8 + 3x = 2x + 8, the same as the right.
No solution. The right side is -4x + 6; adding 4x to both sides leaves 9 = 6, which is false.
Every number. At x = 4 both sides are 16; at x = -2 both are 10. The left side is 2x + 12 - x = x + 12, so subtracting x leaves 12 = 12. The checks suggested it; the algebra proves it.
No solution. The sides are 8 and 3, then 12 and 7, then 48 and 43: the left side is 5 more every time. Solving, 4x + 8 = 4x + 3 leaves 8 = 3, which is false.
No: there is exactly one solution, x = 2. At x = 0 the sides are 4 and 8. Subtracting x gives 2x + 4 = 8, so x = 2.
Never: no solution. The right side of 35 + 19m = 10 + 15m + 4m is 10 + 19m, and subtracting 19m leaves 35 = 10, false. Gym C stays 25 dollars dearer.
After 10 months, at 210 dollars each. 50 + 16m = 21m gives 50 = 5m, so m = 10. Check: 50 + 160 = 210 and 21(10) = 210.
Every number, and for the rectangle every x > 1. Both expressions simplify to 6x + 8, and subtracting 6x leaves 8 = 8. The width x - 1 must be positive.
(a) None: coefficients 9 and 9, constants -4 and 4. (b) Every number: both sides are 10x - 6. (c) Exactly one, since the coefficients 7 and 1 differ (it is x = 1). (d) None: the sides are -3x - 6 and -3x + 6.
The second, 5 - 2x = 2x + 5, with solution x = 0: adding 2x gives 4x = 0. The first has every number as a solution and the third has none. A solution of 0 is still exactly one solution.
The left side simplifies to 3x + 12. (a) For example 3x + 1: any 3x + c with c \neq 12. (b) 3x + 12, or anything equal to it, such as 3(x + 4). (c) For example x + 16: the left side at x = 2 is 18, and 2 + 16 = 18. Any coefficient other than 3 works with the matching constant.
a = 5 gives no solution, since subtracting 5x leaves -4 = 1. No value of a gives every number, because the constants -4 and 1 always differ. Any other a gives exactly one solution, \displaystyle x = \frac{5}{a - 5}; for instance a = 7 gives \displaystyle x = \frac{5}{2}.
4(n + 5) - 8 = 4n + 12, dividing by 4 gives n + 3, and taking away n leaves 3 for every n. To end at 7, change “add 5” to “add 9”: 4(n + 9) - 8 = 4n + 28, then n + 7, then 7.
(2, 2). Setting 3x - 4 = 6 - 2x and adding 2x and 4 gives 5x = 10, so x = 2, and y = 3(2) - 4 = 2. Check: 6 - 2(2) = 2.
Every number is a solution. 2(x - 3) + 4 = 2x - 6 + 4 = 2x - 2, so both lines have slope 2 and cross the y-axis at -2: they are one line, and every point is shared.
Simplify both sides to ax + b = cx + d and subtract cx and b: (a - c)x = d - b. If a \neq c, dividing by a - c gives exactly one solution. If a = c, the left side is 0 for every x, so the equation says 0 = d - b, true for every x or false for every x. The possible counts are one, none and every number, so two is impossible.
She is right for linear equations. Two different solutions rule out the one-solution case, and having any solution rules out the no-solution case, so every number is a solution. One number is not enough: 3x + 4 = x + 8 works at x = 2 and nowhere else (Exercise 8). Her rule fails for x^2 = x: both 0 and 1 work, but x = 2 gives 4 \neq 2. That equation is not linear, so the three cases do not apply to it.