What an Equation Says

An equation joins two expressions with an equals sign and claims they have the same value. That makes it a different kind of object from everything in chapters 1 through 3. An expression has a value. An equation makes a statement, and a statement can be true or false. Start by judging a few.

Problem

Each equation below makes a claim.

7 + 8 = 15 3 \cdot 6 = 21 40 - 13 = 27 \frac{72}{8} = 8

Check each one by computing its left side. Enter how many of the four claims are true.

Show a hint

Work out each left side and compare it to the number on the right. The claim is true only when they match exactly.
$3 \cdot 6$ is $18$ , not $21$ , so that claim is false. Settle the other three the same way.

Show the full solution

7 + 8 = 15

and

40 - 13 = 27

are true, while

3 \cdot 6

18

, not

21

, and

72 \div 8

9

, not

8

. So

2

of the four claims are true. Writing an equation down does not make it true. The symbols make the claim, and arithmetic settles it.

Problem

To compute

7 \times 8 + 6

, a student writes

7 \times 8 = 56 + 6 = 62

. The final number is right, but the first equals sign makes a false claim. It says

7 \times 8

equals

56 + 6

. Enter the difference between those two values.

Show a hint

Read the chain literally. Everything left of the first equals sign is one side of its claim, and everything between the two equals signs is the other.
The false claim is $7 \times 8 = 56 + 6$ , so compare $56$ with $62$ .

Show the full solution

The first equals sign claims

7 \times 8 = 56 + 6

, which says

56 = 62

. The difference is

62 - 56 = 6

. Every equals sign in a line makes its own claim, so a running chain of them asserts things you never meant. Write the steps on separate lines instead.

Every claim so far could be settled outright because both sides were numbers. Put a variable in and that changes. $n + 3 = 10$ is true if $n$ is $7$ and false if $n$ is anything else, so the equation is not simply true or false. Its truth depends on the value of the variable.

Problem

Consider the equation

5 n + 8 = 43

. Substitute

n = 5

n = 6

, and

n = 7

in turn, evaluating the left side each time. One of the three values makes the claim true. Enter it.

Show a hint

Replace $n$ with each value and work out $5 n + 8$ . You want the value that makes the left side come out to $43$ .
At $n = 5$ the left side is $33$ , and each step of $1$ in $n$ adds $5$ more.

Show the full solution

n = 5

the left side is

33

, at

n = 6

it is

38

, and at

n = 7

it is

5 \cdot 7 + 8 = 43

, which matches the right side. The claim is true at

n = 7

and false at the other two.

Problem

A student claims

y = 6

is a solution of

4 y - 7 = 2 y + 5

. Check the claim by substituting

6

for

y

on each side and evaluating the two sides separately. Both come out to the same number. Enter it.

Show a hint

Treat the two sides as two separate expressions and evaluate each at $y = 6$ , the way 1.6 evaluated expressions.
The left side is $4 \cdot 6 - 7$ . Compute it, then compute $2 \cdot 6 + 5$ and compare.

Show the full solution

The left side is

4 \cdot 6 - 7 = 17

and the right side is

2 \cdot 6 + 5 = 17

. Both sides equal

17

, so the claim is true at

y = 6

and

6

is a solution. A check evaluates each side on its own and only compares the results at the end.

Problem

The same student claims

w = 3

is a solution of

4 w + 3 = 2 w + 19

. Substitute

w = 3

and evaluate both sides. They do not match. Enter how much larger the right side is than the left.

Show a hint

Evaluate $4 w + 3$ and $2 w + 19$ separately at $w = 3$ , then subtract the smaller result from the larger.
The left side comes out to $15$ . Now evaluate the right side and take the difference.

Show the full solution

w = 3

the left side is

4 \cdot 3 + 3 = 15

and the right side is

2 \cdot 3 + 19 = 25

, so the right side is larger by

25 - 15 = 10

. The mismatch means

w = 3

is not a solution, and one failed check settles that completely. The equation does have a solution, since checking

w = 8

gives

35

on both sides.

Substituting a value settles the claim for that value.

x = 2

makes the left side

8

and

x = 6

makes it

20

, so the claim is false at both.

x = 4

makes both sides

14

, so

4

is a solution. On the number line, every tested value is either a solution or not, and solving means finding every value where the claim is true.

Problem

One of

x = - 2

and

x = - 5

is a solution of

x^{2} - 3 x = 10

. Check each candidate, keeping every substitution inside parentheses. Enter the one that works.

Show a hint

Evaluate $x^{2} - 3 x$ at each candidate. Written out with parentheses, the first check is $(- 2)^{2} - 3 (- 2)$ .
Watch the second term. Subtracting $3$ times a negative number adds a positive amount.

Show the full solution

x = - 2

the left side is

(- 2)^{2} - 3 (- 2) = 4 + 6 = 10

, which matches. At

x = - 5

it is

(- 5)^{2} - 3 (- 5) = 25 + 15 = 40

, which does not. The solution is

x = - 2

. It is not the only one.

x = 5

gives

25 - 15 = 10

as well, so this equation has two solutions, and a check can confirm one without ruling out others.

So far the equations tested have had one solution, or two. Those are not the only possibilities. An equation can also have no solutions at all, or be true for every value of its variable. The next two problems build one of each kind, and both kinds keep showing up for the rest of the course.

Problem

Try to find a solution of

m + 10 = m + 2

. Substitute a few values of

m

and evaluate both sides. Whatever you pick, the left side exceeds the right side by the same fixed amount. Enter that amount.

Show a hint

Try $m = 1$ , then $m = 5$ . Compare the two sides each time and watch the gap between them.
At $m = 1$ the sides are $11$ and $3$ . The gap cannot depend on $m$ , since both sides grow by the same amount when $m$ does.

Show the full solution

Both sides contain the same

m

, so the comparison is really between the

10

and the

2

added to it. The left side is always larger by

10 - 2 = 8

. No value of

m

can close that gap, so the equation has no solutions. That is a legitimate outcome, since the equation makes a claim and the claim happens to be false for every value.

Problem

Now try

3 (t + 4) = 3 t + 12

. Test all four values

t = - 6

t = 0

t = 2

, and

t = 9

. Enter how many of the four make the equation true.

Show a hint

Evaluate the two sides separately at each value, parentheses around $- 6$ included.
At $t = 0$ both sides are $12$ and at $t = 2$ both are $18$ . Keep going, and think about what 2.1 says about these two sides.

Show the full solution

All four values work. At

t = - 6, 0, 2, 9

both sides come out to

- 6, 12, 18, 39

, so the count is

4

. No luck is involved. By the distributive property from 2.1,

3 (t + 4)

and

3 t + 12

are the same expression in two forms, so the equation is true for every value of

t

, not just these four.

Testing values proved everything so far, and testing values has a limit. A check can confirm a solution you already suspect, but it cannot by itself find one, and it cannot promise that none were missed. The last three problems push guessing as far as it goes and show where it stops working.

Problem

The equation

9 x + 7 = 52

is true for exactly one value of

x

. You do not need to find that value yet. Read the structure of the left side and decide what the term

9 x

must equal for the equation to hold. Enter that number.

Show a hint

The left side is a sum of two parts, $9 x$ and $7$ , and together they must reach $52$ .
Whatever the $7$ does not cover, the $9 x$ part must make up. How far is $7$ from $52$ ?

Show the full solution

The left side adds

7

to the term

9 x

, and the total has to be

52

. The part the

7

does not cover is

52 - 7

, so

9 x = 45

. Whichever number

x

is, nine copies of it must total

45

. Lesson 4.2 turns this kind of reading into a method that finds

x

every time.

Problem

Whole-number guesses fail on

6 x + 5 = 8

x = 0

leaves the left side at

5

and

x = 1

pushes it to

11

, so any solution sits between them. Of the candidates

\frac{1}{3}

\frac{1}{2}

, and

\frac{3}{2}

, enter the one that is a solution.

Show a hint

Substitute each candidate into the left side and see which one makes it $8$ .
At $x = \frac{1}{2}$ the first term is $6 \cdot \frac{1}{2} = 3$ .

Show the full solution

Checking

x = \frac{1}{2}

gives

6 \cdot \frac{1}{2} + 5 = 3 + 5 = 8

, a true claim, so the solution is

\frac{1}{2}

. The other candidates fail, since

x = \frac{1}{3}

makes the left side

7

and

x = \frac{3}{2}

makes it

14

. Solutions are not required to be whole numbers, so guessing whole numbers cannot be the whole story.

Problem

One more, with the variable on both sides. Exactly one of

a = 7

a = 11

, and

a = 15

is a solution of

2 a + 11 = 3 a

. Check each candidate on both sides and enter the one that works.

Show a hint

Substitute each candidate into both sides separately, the way every check in this lesson has run.
At $a = 7$ the sides are $25$ and $21$ . Keep going up the list.

Show the full solution

a = 7

the sides are

25

and

21

, no match. At

a = 11

both sides equal

33

, a match. At

a = 15

they are

41

and

45

, no match. The solution is

a = 11

. Equations with the variable on both sides check exactly like any other, and 4.3 studies them directly.

An equation claims its two sides have the same value, a solution is a value of the variable that makes the claim true, and substitution can check any candidate. What checking cannot do is produce the solutions in the first place. 4.2 builds the tool for that, moves that trade an equation for a simpler one with exactly the same solutions.

Practice these ideas

Practice

For the equation

5 x - 6 = 37

, evaluate the left side at

x = 7

Show the solution

Substituting gives

5 \cdot 7 - 6 = 35 - 6 = 29

. The right side says

37

, so the two sides disagree at

x = 7

and the claim is false there.

Practice

One of the values

2

5

9

is a solution of

x + 8 = 13

. Enter it.

Show the solution

Checking

x = 5

gives

5 + 8 = 13

, a true claim, so the solution is

5

. The other candidates give

10

and

17

, and both of those claims are false.

Practice

Enter the solution of

x - 9 = 14

Show the solution

The equation says

x

9

more than

14

, so

x = 23

. Substituting back gives

23 - 9 = 14

, which is true, so

23

really is the solution.

Practice

Enter the solution of

6 x = 54

Show the solution

Six copies of

9

total

54

, and substituting confirms

6 \cdot 9 = 54

, so

x = 9

Practice

For

8 x + 7 = 31

, evaluate left side minus right side at

x = 3

Show the solution

x = 3

the left side is

8 \cdot 3 + 7 = 31

, and

31 - 31 = 0

. A difference of

0

means the sides agree, so

3

is a solution of the equation.

Practice

For

2 x + 9 = 25

, evaluate left side minus right side at

x = 7

Show the solution

x = 7

the left side is

2 \cdot 7 + 9 = 23

, and

23 - 25 = - 2

. The difference is not

0

, so

7

is not a solution, the left side lands

2

short of the right.

Practice

How many of the values

1

2

3

4

are solutions of

5 x - 3 = 12

Show the solution

The left side comes out

2

7

12

17

x = 1, 2, 3, 4

, and only

x = 3

matches the right side, so

1

of the values is a solution.

Practice

Exactly one of

1

2

3

is a solution of

7 x + 6 = 27

. Enter it.

Show the solution

Checking

x = 3

gives

7 \cdot 3 + 6 = 27

, a match, so the solution is

3

. The candidates

1

and

2

give

13

and

20

, both short of the right side.

Practice

Enter the solution of

x + 9 = 5

Show the solution

The check

- 4 + 9 = 5

comes out true, so

x = - 4

. Nothing in the definition of a solution requires it to be positive, a solution is any value that makes the claim true.

Practice

For

10 x + 1 = 16

, evaluate the left side at

x = \frac{3}{2}

Show the solution

Substituting gives

10 \cdot \frac{3}{2} + 1 = 15 + 1 = 16

. That matches the right side, so

\frac{3}{2}

is a solution. Solutions are not required to be whole numbers.

Practice

Enter the solution of

4 x = 10

as a fraction in lowest terms.

Show the solution

Four copies of

\frac{5}{2}

total

\frac{20}{2} = 10

, so the check passes and

x = \frac{5}{2}

. No whole number works here, but the equation still has exactly one solution.

Practice

How many of the values

- 4

0

3

10

are solutions of

6 (x + 7) = 6 x + 42

Show the solution

Every check passes. At

x = - 4

both sides are

18

, at

0

both are

42

, at

3

both are

60

, and at

10

both are

102

, so the count is

4

. Distributing explains it,

6 (x + 7)

expands to

6 x + 42

, so the two sides are equivalent expressions and the claim is true for every value of

x

Practice

How many of the values

- 3

0

8

are solutions of

x = x + 7

Show the solution

The checks give

- 3

against

4

, then

0

against

7

, then

8

against

15

, all false, so the count is

0

. The right side is always

7

more than the left, so no number anywhere makes this claim true. An equation can have no solutions at all.

Practice

How many of the values

0

1

2

are solutions of

7 (x + 2) = 7 x + 2

Show the solution

The checks give

14

against

2

, then

21

against

9

, then

28

against

16

, all false, so the count is

0

. Distributing shows why, the left side is

7 x + 14

, always

12

more than

7 x + 2

, so no value of

x

closes the gap.

Practice

How many of the values

- 6

- 1

6

are solutions of

x \cdot x = 36

Show the solution

Both

x = - 6

and

x = 6

work, since

(- 6) (- 6) = 36

and

6 \cdot 6 = 36

, while

x = - 1

gives

1

, so

2

of the values are solutions. An equation can have more than one solution, and solving means finding every one of them.

Practice

For

5 x - 4 = 3 x + 1

, evaluate left side minus right side at

x = 4

Show the solution

x = 4

the left side is

5 \cdot 4 - 4 = 16

and the right side is

3 \cdot 4 + 1 = 13

, so left minus right is

16 - 13 = 3

. The difference is not

0

, so

4

is not a solution of the equation.

Practice

For one particular number

k

, every one of the values

2

5

8

11

14

is a solution of

5 (x + 6) = 5 x + k

. Enter that

k

Show the solution

x = 2

the left side is

5 \cdot 8 = 40

and the right side is

10 + k

, and the claim is true only when

10 + k = 40

, so

k = 30

. With

k = 30

the right side is exactly what

5 (x + 6)

distributes to, so the sides are equivalent expressions and every number on the list, along with every other number, is a solution.