Solving Linear Equations

In 4.1 a check settled any candidate. Substitute, evaluate both sides, compare. What it never did was produce a candidate. This lesson builds the producer, a way to trade an equation for a simpler equation with exactly the same solutions, and the first trade is one you can already make.

Problem

The equation

w + 24 = 63

is true at exactly one value of

w

, and at that value its two sides are the same number. Take

24

away from that number on each side. On the left this leaves

w

alone. Enter the number the right side becomes.

Show a hint

At the solution both sides are one number. Taking the same amount from each copy of that number leaves equal amounts.
Subtracting $24$ from the left side leaves $w$ alone. Do the identical subtraction to the $63$ .

Show the full solution

63 - 24 = 39

, so the right side becomes

39

. The new equation reads

w = 39

, and the check

39 + 24 = 63

confirms it. That one subtraction solved the equation.

Lifting the

31

off each pan leaves the beam level, so whatever value of

x

balances the first scale balances the second. Putting the

31

back rebuilds the first scale, so the two equations are true at exactly the same value.

Problem

Solve

t - 38 = 26

. One balance move does it, adding

38

to both sides removes the subtraction and leaves

t

alone. Check your value in the original equation, the 4.1 way, before entering it.

Show a hint

Undo the subtraction with the opposite move, made on both sides.
Adding $38$ to both sides leaves $t$ alone. The right side becomes $26 + 38$ .

Show the full solution

Add

38

to both sides,

t = 26 + 38 = 64

. Check,

64 - 38 = 26

. The check costs seconds and catches every slip.

The same argument covers scaling. At a solution both sides are one number, tripling that number on each side keeps the sides equal, and dividing by $3$ undoes it. It looks like any multiplier works. One of them does not, and the next problem finds it.

Problem

The equation

6 d = 84

is true only at

d = 14

. Multiply both sides by

0

and every term collapses, leaving

0 = 0

. Test the values

- 3

0

8

14

, and

20

in the new equation. Enter how many of them are solutions of the new equation but not of the original.

Show a hint

The new equation never mentions $d$ , so ask which values could possibly fail it.
All five values satisfy $0 = 0$ , and the original is satisfied only at $14$ . Count the values that pass one equation but not the other.

Show the full solution

Every value makes

0 = 0

true, and only

14

satisfies

6 d = 84

, so

4

of the five are new. Multiplying by zero manufactured solutions, and nothing undoes it, since dividing by zero is not a move.

Problem

Solve

12 m = 132

. The left side is twelve copies of

m

, so divide both sides by

12

. Check the value before entering it.

Show a hint

Division is the reverse of the multiplication that built the left side.
Dividing both sides by $12$ leaves $m$ alone against $132$ divided by $12$ .

Show the full solution

Divide both sides by

12

m = \frac{132}{12} = 11

. Check,

12 \cdot 11 = 132

Problem

Solve

\frac{n}{15} = 8

by multiplying both sides by

15

. Check the value in the original equation before entering it.

Show a hint

The left side divided $n$ by $15$ , so undo it with the opposite scaling move.
Multiplying both sides by $15$ leaves $n$ alone. The right side becomes $8$ times $15$ .

Show the full solution

Multiply both sides by

15

n = 8 \cdot 15 = 120

. Check,

\frac{120}{15} = 8

In 4.1 you read $9 x + 7 = 52$ with no method at all. The term $9 x$ had to be $45$ , and dividing by $9$ finishes it, $x = 5$ . That reading was really two balance moves, subtract $7$ from both sides, then divide both sides by $9$ . The left side was built by multiplying and then adding, so it unwinds in the opposite order.

Problem

Solve

3 a + 29 = 98

. Subtract first to expose the term

3 a

, then divide. Check your value in the original equation before entering it.

Show a hint

The $29$ sits outside the product $3 a$ , so it is the last thing that happened and the first thing to undo.
Subtracting $29$ from both sides gives $3 a = 69$ .

Show the full solution

Subtract

29

from both sides,

3 a = 69

, then divide by

3

a = 23

. Check,

3 \cdot 23 + 29 = 98

Problem

Solve

10 y + 74 = 82

. The moves are the same as before, and the answer is not a whole number. Enter it as a fraction in lowest terms.

Show a hint

Nothing about the moves needs the answer to be whole. Unwind as usual.
Subtracting $74$ gives $10 y = 8$ . Divide and reduce.

Show the full solution

Subtract

74

10 y = 8

, divide by

10

y = \frac{8}{10} = \frac{4}{5}

. Check,

10 \cdot \frac{4}{5} + 74 = 8 + 74 = 82

. Fractional solutions are normal outcomes, not signs of a mistake.

Sometimes a side needs cleanup before any unwinding. When the variable side is a pile of like terms, combine them the 2.4 way first. After that the equation is two-step and the moves are the ones you already have.

Problem

Solve

16 q + 27 - 8 q + 13 = 100

. Combine like terms on the left before making any balance move. Enter the solution as a fraction in lowest terms.

Show a hint

The left side is not two-step yet. Collect the $q$ terms and the constants first.
$16 q - 8 q = 8 q$ and $27 + 13 = 40$ , so the equation is $8 q + 40 = 100$ .

Show the full solution

Combining gives

8 q + 40 = 100

, so

8 q = 60

and

q = \frac{60}{8} = \frac{15}{2}

. Check,

8 \cdot \frac{15}{2} + 40 = 60 + 40 = 100

Problem

Solve

\frac{3}{4} c = 51

. One move does it, multiply both sides by the reciprocal

\frac{4}{3}

, the 2.5 skill. Enter the solution.

Show a hint

A fractional coefficient clears in one move, scaling by the number that multiplies with it to $1$ .
$c = 51 \cdot \frac{4}{3}$ . Divide $51$ by $3$ before multiplying.

Show the full solution

Multiply both sides by

\frac{4}{3}

c = 51 \cdot \frac{4}{3} = 68

. Check,

\frac{3}{4} \cdot 68 = 51

Problem

Solve

28 - x = 93

. Subtracting

28

from both sides leaves

- x = 65

, a statement about the opposite of

x

. One more move finishes it. Enter the solution.

Show a hint

$- x$ means $(- 1) x$ , and $- 1$ is a nonzero number like any other.
Divide both sides of $- x = 65$ by $- 1$ , or read it directly, the opposite of $x$ is $65$ .

Show the full solution

Subtract

28

- x = 65

, then multiply both sides by

- 1

x = - 65

. Check,

28 - (- 65) = 28 + 65 = 93

. If the opposite of

x

65

, then

x

- 65

, the moves and the reading agree.

Problem

A student solves

5 x + 30 = 80

by erasing the

30

from the left side only, getting

5 x = 80

and then

x = 16

. Check that candidate the 4.1 way. Evaluate the left side at

x = 16

and enter how much it overshoots the right side.

Show a hint

A check needs no opinion about the moves. Substitute the candidate and evaluate the original left side.
At $x = 16$ the left side is $5 (16) + 30$ . Compare that with $80$ .

Show the full solution

The left side at

16

5 \cdot 16 + 30 = 110

, and

110 - 80 = 30

, exactly the

30

that left one side but not the other. Legal moves hit both sides. Subtracting

30

from both gives

5 x = 50

, so

x = 10

, and

5 \cdot 10 + 30 = 80

checks.

Problem

Solve

\frac{3}{5} h + 46 = 22

. Undo the addition first, then multiply by the reciprocal. The solution is negative, and the check still takes ten seconds. Enter the solution.

Show a hint

Last operation first. The $46$ was added after the coefficient did its work.
Subtracting $46$ gives $\frac{3}{5} h = - 24$ . Multiply by $\frac{5}{3}$ .

Show the full solution

Subtract

46

\frac{3}{5} h = - 24

, multiply by

\frac{5}{3}

h = - 24 \cdot \frac{5}{3} = - 40

. Check,

\frac{3}{5} (- 40) + 46 = - 24 + 46 = 22

Every equation in this lesson kept its variable on one side, and four reversible moves handled all of them. 4.3 starts where that stops, equations with the variable on both sides, and the balance principle carries over unchanged.

Practice these ideas

Practice

Enter the solution of

x + 34 = 59

Show the solution

Subtract

34

from both sides,

x = 59 - 34 = 25

. Check,

25 + 34 = 59

, so the value passes the 4.1 test.

Practice

Enter the solution of

y - 27 = 44

Show the solution

Add

27

to both sides,

y = 44 + 27 = 71

. Check,

71 - 27 = 44

Practice

Enter the solution of

7 k = 119

Show the solution

Divide both sides by

7

k = \frac{119}{7} = 17

. Check,

7 \cdot 17 = 119

Practice

Enter the solution of

\frac{t}{9} = 12

Show the solution

Multiply both sides by

9

t = 12 \cdot 9 = 108

. Check,

\frac{108}{9} = 12

Practice

Enter the solution of

w + 53 = 35

Show the solution

Subtract

53

from both sides,

w = 35 - 53 = - 18

. Check,

- 18 + 53 = 35

. Subtracting from both sides works the same even when the result is negative.

Practice

Enter the solution of

2 k + 41 = 79

Show the solution

Subtract

41

2 k = 38

, then divide by

2

k = 19

. Check,

2 \cdot 19 + 41 = 38 + 41 = 79

Practice

Enter the solution of

6 p + 61 = 13

Show the solution

Subtract

61

6 p = 13 - 61 = - 48

, then divide by

6

p = - 8

. Check,

6 (- 8) + 61 = - 48 + 61 = 13

Practice

Enter the solution of

\frac{k}{12} + 5 = 8

Show the solution

Subtract

5

\frac{k}{12} = 3

, then multiply both sides by

12

k = 36

. Check,

\frac{36}{12} + 5 = 3 + 5 = 8

Practice

Combine like terms first, then solve

13 w + 12 - 8 w + 73 = 175

. Enter the solution.

Show the solution

Combining gives

5 w + 85 = 175

, so

5 w = 90

and

w = 18

. Check,

13 \cdot 18 + 12 - 8 \cdot 18 + 73 = 234 + 12 - 144 + 73 = 175

. Combining is not a balance move at all, it rewrites one side into an equal expression, so nothing on the other side changes.

Practice

Solve

2 (u + 67) = 232

, distributing once the 2.1 way or dividing both sides by

2

first. Enter the solution.

Show the solution

Divide both sides by

2

u + 67 = 116

, then subtract

67

u = 49

. Check,

2 (49 + 67) = 2 \cdot 116 = 232

. Distributing first gives

2 u + 134 = 232

and lands on the same value.

Practice

Enter the solution of

\frac{2}{3} g = 58

Show the solution

Multiply both sides by

\frac{3}{2}

g = 58 \cdot \frac{3}{2} = 29 \cdot 3 = 87

. Check,

\frac{2}{3} \cdot 87 = 58

Practice

Enter the solution of

- 12 u = 72

Show the solution

Divide both sides by

- 12

u = \frac{72}{- 12} = - 6

. Check,

- 12 \cdot (- 6) = 72

. A positive product from a negative coefficient forces a negative solution.

Practice

Enter the solution of

66 - x = 145

Show the solution

Subtract

66

- x = 79

, then multiply both sides by

- 1

x = - 79

. Check,

66 - (- 79) = 66 + 79 = 145

Practice

Solve

8 n + 77 = 86

. Enter the solution as a fraction in lowest terms.

Show the solution

Subtract

77

8 n = 9

, then divide by

8

n = \frac{9}{8}

. Check,

8 \cdot \frac{9}{8} + 77 = 9 + 77 = 86

. A fractional answer is a normal outcome, not a sign of a slip.

Practice

To solve

3 x - 21 = 99

, a student divides the

3 x

and the

99

3

but leaves the

- 21

alone, writing

x - 21 = 33

, so

x = 54

. Check that candidate the 4.1 way. Evaluate the left side of the original equation at

x = 54

and enter how much it exceeds the right side.

Show the solution

The left side at

54

3 \cdot 54 - 21 = 162 - 21 = 141

, and

141 - 99 = 42

. Dividing both sides means every term,

x - 7 = 33

, or add

21

first to get

3 x = 120

. Either way

x = 40

, and

3 \cdot 40 - 21 = 99

checks.

Practice

Combine like terms first, then solve

3 m + 32 - 4 m + 57 = 96

. Enter the solution.

Show the solution

Combining gives

- m + 89 = 96

, so

- m = 7

and

m = - 7

. Check,

3 (- 7) + 32 - 4 (- 7) + 57 = - 21 + 32 + 28 + 57 = 96

. A net coefficient of

- 1

reads as the opposite of

m

, so

m

is the opposite of

7

Practice

Solve

\frac{8}{3} n = - 14

. Enter the solution as a fraction in lowest terms.

Show the solution

Multiply both sides by

\frac{3}{8}

n = - 14 \cdot \frac{3}{8} = - \frac{42}{8} = - \frac{21}{4}

. Check,

\frac{8}{3} \cdot (- \frac{21}{4}) = - \frac{168}{12} = - 14

. Multiplying by the positive number

\frac{3}{8}

does not change the sign, so the solution stays negative.