Variables on Both Sides

In 4.1 a check proved that $y = 6$ solves $4 y - 7 = 2 y + 5$ . The check never produced the $6$ , though, and 4.2's moves only reached equations with the variable on one side. One observation closes the gap. Once the variable has a value, a term like $2 y$ is a number, and 4.2 already lets you subtract any number from both sides.

Problem

The equation

8 k = 5 k + 21

has its variable on both sides, which 4.2 never faced. At the solution, though,

5 k

is a number like any other. Subtract

5 k

from both sides, finish with 4.2's moves, and check in the original. Enter the solution.

Show a hint

The balance move from 4.2 subtracts equal amounts from both sides, and the amount subtracted is allowed to contain the variable.
Subtracting $5 k$ from both sides leaves $3 k = 21$ .

Show the full solution

Subtract

5 k

from both sides,

3 k = 21

, then divide by

3

k = 7

. Check,

8 \cdot 7 = 56

and

5 \cdot 7 + 21 = 56

. Subtracting

5 k

is a valid balance move because at the solution

5 k

stands for a single number.

Two

x

tiles come off each pan together, the same balance move as lifting off a number weight, and the beam stays level. What remains is 4.2's picture, the variable on one side only. Putting the

2 x

back rebuilds the first scale, so

5 x + 40 = 2 x + 58

and

3 x + 40 = 58

are true at exactly the same value.

Problem

Solve

9 m + 16 = 4 m + 61

. Collect the variable terms with one subtraction, then unwind the 4.2 way. Check in the original before entering the solution.

Show a hint

Which single subtraction leaves the variable on only one side of the equation?
Subtracting $4 m$ from both sides gives $5 m + 16 = 61$ .

Show the full solution

Subtract

4 m

5 m + 16 = 61

, subtract

16

5 m = 45

, divide by

5

m = 9

. Check,

9 \cdot 9 + 16 = 97

and

4 \cdot 9 + 61 = 97

Problem

Solve

2 y + 59 = 7 y + 19

twice. First subtract

2 y

from both sides and finish. Then start over and subtract

7 y

instead, letting the coefficient go negative along the way. Enter the value both routes give.

Show a hint

Both routes are legal. One subtracts $2 y$ from both sides, the other subtracts $7 y$ .
Route one gives $59 = 5 y + 19$ . Route two gives $- 5 y + 59 = 19$ .

Show the full solution

Route one, subtract

2 y

59 = 5 y + 19

, so

5 y = 40

and

y = 8

. Route two, subtract

7 y

- 5 y + 59 = 19

, so

- 5 y = - 40

and

y = 8

again. Either way

y = 8

, and both sides check to

75

. Subtracting the smaller variable term keeps the coefficient positive, which is the friendlier road.

Problem

Solve

4 a + 57 = 12 a + 9

. The smaller variable term is on the left this time, so subtracting it collects the variable on the right. Enter the solution.

Show a hint

Subtracting the smaller variable term keeps the remaining coefficient positive.
Subtracting $4 a$ from both sides gives $57 = 8 a + 9$ .

Show the full solution

Subtract

4 a

57 = 8 a + 9

, subtract

9

8 a = 48

, so

a = 6

. Check,

4 \cdot 6 + 57 = 81

and

12 \cdot 6 + 9 = 81

. A variable on the right is no problem, you can swap the two sides of an equation whenever you like.

Some equations need cleanup before anything crosses the equals sign. If a side is a pile of like terms, combine them the 2.4 way, and if parentheses are in the way, distribute once the 2.1 way. Simplifying a side rewrites it as an equal expression, so nothing on the other side changes. Collect once the sides are clean.

Problem

Solve

9 g + 8 - 2 g + 7 = 3 g + 35

. Combine like terms on the left the 2.4 way before collecting anything across the equals sign. Enter the solution.

Show a hint

Clean each side first. Balance moves come after the sides are simplified.
The left side combines to $7 g + 15$ .

Show the full solution

The left side combines to

7 g + 15

, so

7 g + 15 = 3 g + 35

. Subtract

3 g

4 g + 15 = 35

, so

4 g = 20

and

g = 5

. Check,

45 + 8 - 10 + 7 = 50

and

3 \cdot 5 + 35 = 50

Problem

Solve

7 (w - 2) = 4 w + 22

. Distribute once the 2.1 way, then collect the variable terms. Enter the solution. Check in the original before entering it.

Show a hint

Parentheses come off before any term moves across the equals sign.
Distributing gives $7 w - 14 = 4 w + 22$ .

Show the full solution

Distribute,

7 w - 14 = 4 w + 22

. Subtract

4 w

3 w - 14 = 22

, add

14

3 w = 36

, so

w = 12

. Check,

7 (12 - 2) = 70

and

4 \cdot 12 + 22 = 70

Negative and fraction solutions were normal outcomes in 4.2, and they stay normal here. Collecting the variable terms changes where the variable sits, not what kind of number the answer is allowed to be, so run the same moves and let the value be whatever it is.

Problem

Solve

5 - 6 x = 4 x + 25

. The variable term on the left is subtracted, so adding

6 x

to both sides removes it. Enter the solution.

Show a hint

A subtracted term is removed by adding it to both sides, the same balance principle.
Adding $6 x$ to both sides gives $5 = 10 x + 25$ .

Show the full solution

Add

6 x

to both sides,

5 = 10 x + 25

, subtract

25

10 x = - 20

, so

x = - 2

. Check,

5 - 6 (- 2) = 17

and

4 (- 2) + 25 = 17

. Negative solutions are ordinary outcomes here, exactly as in 4.2.

Problem

Solve

5 x + 2 = 2 x + 4

. Nothing about collecting requires a whole-number answer. Enter the solution as a fraction in lowest terms.

Show a hint

Collect first, then divide at the end, and keep the result exact.
Subtracting $2 x$ and then $2$ leaves $3 x = 2$ .

Show the full solution

Subtract

2 x

3 x + 2 = 4

, then subtract

2

3 x = 2

, and divide by

3

x = \frac{2}{3}

. Check, both sides come out

\frac{16}{3}

. A fractional solution is a normal outcome, and exact fractions keep the check honest.

Problem

Solve

5 (4 x + 3) = 20 x + 8

. Distribute, then try to collect. The variable cancels entirely and a number statement is left behind. Decide whether that statement is true, then enter how many solutions the equation has.

Show a hint

When every variable term cancels, the leftover statement answers the question by itself.
Distributing gives $20 x + 15 = 20 x + 8$ .

Show the full solution

Distribute,

20 x + 15 = 20 x + 8

. Subtracting

20 x

from both sides leaves

15 = 8

, which is false, so the equation has

0

solutions. The left side is always

7

more than the right, and no value of

x

closes that gap.

Problem

Solve

13 x + 18 - 6 x = 7 x + 18

. Combine the left side, then try to collect. The variable cancels and a true statement remains. Enter how many of the values

- 7

0

5

15

are solutions.

Show a hint

A true leftover statement means the two sides were the same expression all along.
The left side combines to $7 x + 18$ , identical to the right side.

Show the full solution

The left side combines to

7 x + 18

, so the equation reads

7 x + 18 = 7 x + 18

, true at every value, and all

4

of the listed values are solutions. An equation whose two sides are equivalent expressions, like this one, is called an identity, and every value of

x

satisfies it.

Problem

Solve

11 x - 4 = 6 x - 4

. It looks like the last two problems and is neither one. Collect the variable terms, finish the 4.2 way, and check your value in the original before entering the solution.

Show a hint

Did the variable actually cancel, or is a variable term still there after the collecting?
Subtracting $6 x$ from both sides gives $5 x - 4 = - 4$ , so $5 x = 0$ .

Show the full solution

Subtract

6 x

5 x - 4 = - 4

, add

4

5 x = 0

, so

x = 0

. Check,

11 \cdot 0 - 4 = - 4

and

6 \cdot 0 - 4 = - 4

. Having the solution

0

is not the same as having

0

solutions, one value works and that value happens to be zero.

A check belongs to the original equation. Substituting a candidate into one of your own later lines only tests the work that came after that line, and a line with a mistake in it will confirm an answer carrying the same mistake. The original equation is the claim you set out to solve, so the original is where a check means something.

Problem

A student solves

7 x + 44 = 3 x + 8

, subtracting

3 x

and then

44

, but writes

4 x = 36

instead of

4 x = - 36

and gets

x = 9

. Checking against the line

4 x = 36

passes. Run the check where it counts. Substitute

x = 9

into both sides of the original equation and enter how much larger the left side is than the right.

Show a hint

A check against a derived line only confirms that line, not the original equation.
At $x = 9$ the left side is $63 + 44$ .

Show the full solution

x = 9

the left side is

7 \cdot 9 + 44 = 107

and the right side is

3 \cdot 9 + 8 = 35

, so the left side is larger by

107 - 35 = 72

. The correct line is

4 x = - 36

, so

x = - 9

, and both sides of the original then equal

- 19

. A check against

4 x = 36

tested the wrong claim.

One subtraction turns a both-sides equation into a one-side equation, and a variable that cancels is an answer rather than a failure, either no solutions or all of them. 4.4 takes on equations that hide this shape behind fractions and heavier parentheses, and 4.5 puts equations to work on word problems.

Practice these ideas

Practice

Enter the solution of

6 x + 7 = 4 x + 15

Show the solution

Subtract

4 x

2 x + 7 = 15

, then subtract

7

2 x = 8

, so

x = 4

. Check, both sides equal

31

Practice

Enter the solution of

10 x + 26 = 3 x + 89

Show the solution

Subtract

3 x

7 x + 26 = 89

, then subtract

26

7 x = 63

, so

x = 9

. Check, both sides equal

116

Practice

Enter the solution of

4 x + 54 = 9 x + 4

. The variable collects on the right, which is fine.

Show the solution

Subtract

4 x

from both sides,

54 = 5 x + 4

, so

5 x = 50

and

x = 10

. Check,

4 (10) + 54 = 94

and

9 (10) + 4 = 94

. An equation reads the same in either direction, so a variable on the right needs no extra step.

Practice

Enter the solution of

6 x + 27 = 15 x

Show the solution

Subtract

6 x

from both sides,

27 = 9 x

, so

x = 3

. Check,

6 (3) + 27 = 45

and

15 (3) = 45

Practice

Enter the solution of

5 x = 3 x + 34

Show the solution

Subtract

3 x

from both sides,

2 x = 34

, so

x = 17

. Check,

5 (17) = 85

and

3 (17) + 34 = 85

Practice

Enter the solution of

7 x + 41 = 2 x + 6

Show the solution

Subtract

2 x

from both sides,

5 x + 41 = 6

, then

5 x = - 35

and

x = - 7

. Check,

7 (- 7) + 41 = - 8

and

2 (- 7) + 6 = - 8

Practice

Combine like terms on the left first, then solve

8 x - 1 + 2 x = 7 x + 38

Show the solution

Combine first,

10 x - 1 = 7 x + 38

. Subtract

7 x

3 x - 1 = 38

, add

1

3 x = 39

, so

x = 13

. Check, both sides equal

129

Practice

Distribute first, then solve

4 (2 x - 5) = 6 x + 2

Show the solution

Distribute,

8 x - 20 = 6 x + 2

. Subtract

6 x

2 x - 20 = 2

, then

2 x = 22

and

x = 11

. Check,

4 (2 \cdot 11 - 5) = 4 \cdot 17 = 68

and

6 (11) + 2 = 68

Practice

Solve

6 x + 3 = 2 x + 17

. Enter the solution as a fraction in lowest terms.

Show the solution

Subtract

2 x

4 x + 3 = 17

, then

4 x = 14

and

x = \frac{14}{4} = \frac{7}{2}

. Check,

6 \cdot \frac{7}{2} + 3 = 24

and

2 \cdot \frac{7}{2} + 17 = 24

Practice

Enter the solution of

48 - 3 x = 4 x + 6

. Adding

3 x

to both sides is the collect move.

Show the solution

Add

3 x

to both sides,

48 = 7 x + 6

, then

7 x = 42

and

x = 6

. Check,

48 - 3 (6) = 30

and

4 (6) + 6 = 30

Practice

How many solutions does

6 x + 37 = 6 x + 50

have? Enter the count.

Show the solution

Subtracting

6 x

from both sides leaves

37 = 50

, which is false, so the equation has

0

solutions. Whatever

x

is, the two sides differ by

13

, so no value can ever make them equal.

Practice

How many of the values

- 8

- 1

0

6

20

are solutions of

7 (2 x + 8) = 9 x + 5 x + 56

Show the solution

Distributing gives

14 x + 56

on the left, and the right combines to

14 x + 56

as well, so the equation is an identity and every value works. All

5

listed values are solutions. By 2.1 the two sides are equivalent expressions, so no substitution was ever needed.

Practice

Solve

7 x + 5 = 3 x + 5

. Enter the solution.

Show the solution

Subtract

3 x

4 x + 5 = 5

, then subtract

5

4 x = 0

, so

x = 0

. Check, both sides equal

5

. Zero is a legitimate solution, an equation whose sides agree only at

0

is nothing like an identity.

Practice

Solve

5 x + 1 = 9 x + 6

. Enter the solution as a fraction in lowest terms.

Show the solution

Subtract

5 x

1 = 4 x + 6

, then

4 x = - 5

and

x = - \frac{5}{4}

. Check,

5 (- \frac{5}{4}) + 1 = - \frac{21}{4}

and

9 (- \frac{5}{4}) + 6 = - \frac{21}{4}

. A negative solution changes nothing about the steps, collect the variable terms on one side and divide as usual.

Practice

A student solves

9 x - 7 = 2 x + 42

and gets

x = 3

. Substitute

x = 3

into both sides of the original and enter how much larger the right side is than the left.

Show the solution

x = 3

the left side is

9 (3) - 7 = 20

and the right side is

2 (3) + 42 = 48

, so the right side is larger by

48 - 20 = 28

. The correct run is

7 x = 49

, so

x = 7

, where both sides equal

56

Practice

Solve

2 (6 x + 7) = 5 (2 x + 9)

, one distribute on each side. Enter the solution as a fraction in lowest terms.

Show the solution

Distribute on both sides,

12 x + 14 = 10 x + 45

. Subtract

10 x

2 x + 14 = 45

, then

2 x = 31

and

x = \frac{31}{2}

. Check,

2 (6 \cdot \frac{31}{2} + 7) = 2 (100) = 200

and

5 (2 \cdot \frac{31}{2} + 9) = 5 (40) = 200

Practice

Simplify both sides first, then solve

9 - 2 x + 8 = 4 x - 1 + 2 x

. Enter the solution as a fraction in lowest terms.

Show the solution

The left combines to

17 - 2 x

and the right to

6 x - 1

. Add

2 x

to both sides,

17 = 8 x - 1

, then

8 x = 18

and

x = \frac{18}{8} = \frac{9}{4}

. Check,

9 - 2 \cdot \frac{9}{4} + 8 = \frac{25}{2}

and

4 \cdot \frac{9}{4} - 1 + 2 \cdot \frac{9}{4} = \frac{25}{2}