Equations in Disguise

4.3 ended with a promise, equations that hide the familiar both-sides shape behind fractions and heavier parentheses. This lesson keeps it. Nothing below is new mathematics. Every equation here becomes a 4.2 or 4.3 equation after one cleanup move you already own, and the lesson is learning to see which move each shape calls for.

Problem

Nothing in 4.3 had parentheses wrapped around both sides. Solve

5 (2 x - 3) = 3 (3 x + 4)

anyway. Distribute on each side the 2.1 way, then collect the 4.3 way, and check in the original before entering the solution.

Show a hint

The parentheses are a 2.1 job, not a new rule. Nothing crosses the equals sign until both sides are clean.
Distributing on both sides gives $10 x - 15 = 9 x + 12$ .

Show the full solution

Distribute on each side,

10 x - 15 = 9 x + 12

. Subtract

9 x

x - 15 = 12

, then add

15

x = 27

. Check,

5 (54 - 3) = 255

and

3 (81 + 4) = 255

. The parentheses only made the equation look unfamiliar, every move was an old one.

Problem

Solve

7 x - 2 (3 x - 8) = 25

. The subtracted

2

distributes over both terms of

3 x - 8

, signs included. Enter the solution.

Show a hint

The minus in front of the parentheses belongs to everything inside, both terms.
$- 2 (3 x - 8)$ is $- 6 x + 16$ , not $- 6 x - 16$ .

Show the full solution

Distribute,

7 x - 6 x + 16 = 25

, so

x + 16 = 25

and

x = 9

. Check,

63 - 2 (27 - 8) = 63 - 38 = 25

. Dropping the sign flip on the

16

is the single most common slip in this shape.

Fractions are the main event. $\frac{x}{3} + \frac{x}{5} = 16$ could be worked the 2.5 way, combining the left side into one fraction, but that drags fractions through every line. 4.2 allows multiplying both sides by any nonzero number, and one well-chosen multiplication removes every fraction at once. Pick a number every denominator divides.

Problem

Solve

\frac{x}{3} + \frac{x}{5} = 16

by multiplying both sides by

15

, then check in the original before entering the solution.

Show a hint

At the solution both sides are the same number, and multiplying both copies by $15$ keeps them equal.
Multiplying both sides by $15$ gives $5 x + 3 x = 240$ .

Show the full solution

Multiply both sides by

15

5 x + 3 x = 240

, so

8 x = 240

and

x = 30

. Check,

\frac{30}{3} + \frac{30}{5} = 10 + 6 = 16

. Multiplying by

15

cleared both fractions in one step.

One multiplication reaches all three terms, the whole

14

included. Each denominator divides

20

, so every piece comes out whole and the fractions are gone in a single line. The survivors collect to

7 x = 280

, so

x = 40

, and the check in the original gives

10 + 4 = 14

Problem

Solve

\frac{x}{6} + \frac{x}{9} = 10

. The product

54

clears both denominators, and so does the LCD

18

, with smaller numbers. Use

18

and enter the solution.

Show a hint

$18$ works because both $6$ and $9$ divide it.
Multiplying both sides by $18$ gives $3 x + 2 x = 180$ .

Show the full solution

Multiply both sides by

18

3 x + 2 x = 180

, so

5 x = 180

and

x = 36

. Check,

6 + 4 = 10

. The

54

route gives

9 x + 6 x = 540

and the same answer, with numbers three times the size.

Problem

Solve

\frac{x}{6} + 5 = \frac{x}{4}

. The multiplier

12

hits every term, the lone

5

included. Enter the solution.

Show a hint

The $5$ is a full term of the left side, and the multiplication distributes over the whole side.
Clearing with $12$ gives $2 x + 60 = 3 x$ .

Show the full solution

Multiply both sides by

12

2 x + 60 = 3 x

, so subtracting

2 x

leaves

x = 60

. Check,

10 + 5 = 15

and

\frac{60}{4} = 15

. Leaving the

5

unmultiplied is the error this lesson's checks keep catching.

Problem

Solve

\frac{x + 9}{2} - \frac{x}{5} = 6

. The fraction bar groups its numerator, so when you multiply by

10

the whole

x + 9

travels together, in parentheses. Enter the solution.

Show a hint

$10 \cdot \frac{x + 9}{2}$ is $5 (x + 9)$ , never $5 x + 9$ .
Clearing with $10$ gives $5 (x + 9) - 2 x = 60$ .

Show the full solution

Multiply both sides by

10

5 (x + 9) - 2 x = 60

, so

3 x + 45 = 60

3 x = 15

, and

x = 5

. Check,

\frac{14}{2} - 1 = 7 - 1 = 6

. The bar is a grouping symbol, doing the same job parentheses do.

When each side of an equation is a single fraction and nothing else, two multiplications clear everything, one for each denominator, and each denominator cancels on its own side. Watch what remains after each cancellation, since the same pattern appears every time this shape does.

Problem

Solve

\frac{2 x + 3}{9} = \frac{x + 7}{6}

. Multiply both sides by a common multiple of the denominators so each one cancels. Enter the solution.

Show a hint

The least common multiple of $9$ and $6$ is $18$ , so multiply both sides by $18$ .
Clearing the denominators leaves $2 (2 x + 3) = 3 (x + 7)$ .

Show the full solution

Multiply both sides by

18

2 (2 x + 3) = 3 (x + 7)

, so

4 x + 6 = 3 x + 21

and

x = 15

. Check,

\frac{33}{9}

and

\frac{22}{6}

both equal

\frac{11}{3}

Problem

Use the shortcut on

\frac{2 x + 5}{12} = \frac{x - 3}{9}

. The shortcut applies here, since each side is a single fraction and nothing else. Enter the solution as a fraction in lowest terms.

Show a hint

Each numerator multiplies the other denominator, with whole numerators kept in parentheses.
The cleared equation is $9 (2 x + 5) = 12 (x - 3)$ .

Show the full solution

Cross-multiply,

9 (2 x + 5) = 12 (x - 3)

, so

18 x + 45 = 12 x - 36

6 x = - 81

, and

x = - \frac{27}{2}

. Check, both sides come out

- \frac{11}{6}

. A negative fraction solution is a normal outcome, exactly as in 4.3.

A decimal coefficient is a count of tenths or hundredths, so a decimal equation is a fraction equation. Tenths clear when both sides are multiplied by $10$ , hundredths need $100$ , and the finest decimal place present picks the power. The move is 4.2's same multiplication, and it hits every term, whole-number constants included.

Problem

Solve

0.8 x - 3.1 = 1.7

by multiplying both sides by

10

first, so every count of tenths becomes a whole number. Enter the solution.

Show a hint

Every term is a count of tenths, so multiplying by $10$ makes every coefficient whole.
The cleared equation is $8 x - 31 = 17$ .

Show the full solution

Multiply both sides by

10

8 x - 31 = 17

, so

8 x = 48

and

x = 6

. Check,

4.8 - 3.1 = 1.7

Problem

Solve

0.06 x + 4 = 0.11 x + 1.5

. Hundredths call for

100

, and the multiplication turns the whole

4

into

400

. Enter the solution.

Show a hint

Choose the multiplier by the finest decimal place present, and remember it reaches the plain $4$ too.
Clearing with $100$ gives $6 x + 400 = 11 x + 150$ .

Show the full solution

Multiply both sides by

100

6 x + 400 = 11 x + 150

, so

250 = 5 x

and

x = 50

. Check,

3 + 4 = 7

and

5.5 + 1.5 = 7

. Multiplying by

10

instead would leave

1.1 x

behind, which is why the finest place picks the power.

Problem

A student solves

\frac{x}{2} - 7 = \frac{x}{6} + 4

by multiplying only the two fractions by

6

, writes

3 x - 7 = x + 4

, and gets

x = \frac{11}{2}

. That candidate fails the 4.1 check in the original. Multiply every term by

6

and enter the correct solution.

Show a hint

A legal multiplication reaches all four terms, the $7$ and the $4$ included.
The correct cleared equation is $3 x - 42 = x + 24$ .

Show the full solution

Multiply every term by

6

3 x - 42 = x + 24

, so

2 x = 66

and

x = 33

. Check,

16.5 - 7 = 9.5

and

5.5 + 4 = 9.5

. The student's check against the flawed line

3 x - 7 = x + 4

would have passed, which is why a check belongs to the original.

Every multiplier so far was a known nonzero number. In $\frac{44}{x + 15} = 4$ the natural multiplier is $x + 15$ , which contains the variable. 4.2's rule still holds wherever $x + 15$ is not zero, ruling out only $x = - 15$ , where the original divides by zero anyway. Clear it, solve, and check in the original, which also shows the denominator was not zero.

Problem

Solve

\frac{44}{x + 15} = 4

. Multiply both sides by

x + 15

, finish with 4.2's moves, and check your value in the original before entering it.

Show a hint

After the one multiplication, the equation is a shape 4.2 solved many times.
Clearing gives $44 = 4 (x + 15)$ , so $44 = 4 x + 60$ .

Show the full solution

Multiply both sides by

x + 15

44 = 4 (x + 15)

, so

44 = 4 x + 60

4 x = - 16

, and

x = - 4

. Check,

- 4 + 15 = 11

and

\frac{44}{11} = 4

, so the value works and the denominator was not zero. The check did logical work here, not just hygiene.

Every equation in this lesson stopped looking unfamiliar after one move. Distribute, clear the denominators, or clear the decimals, and each landed in 4.2's or 4.3's hands. 4.5 starts from equations with no algebra written down at all, plain English sentences, and turning those into equations is the next skill.

Practice these ideas

Practice

Solve

4 (2 x + 1) = 3 (x + 8)

Show the solution

Distribute on both sides,

8 x + 4 = 3 x + 24

. Subtract

3 x

, then subtract

4

, so

5 x = 20

and

x = 4

. Check, both sides equal

36

Practice

Solve

9 x - 4 (2 x - 7) = 31

Show the solution

Distribute,

9 x - 8 x + 28 = 31

, so

x + 28 = 31

and

x = 3

. Check,

27 - 4 (- 1) = 31

. Dropping the sign and writing

- 28

is the slip to watch here.

Practice

Solve

\frac{4 x + 3}{5} = 7

Show the solution

Multiply both sides by

5

4 x + 3 = 35

, so

4 x = 32

and

x = 8

. Check,

\frac{35}{5} = 7

Practice

Solve

\frac{x}{3} + \frac{x}{7} = 20

Show the solution

Multiply both sides by

21

7 x + 3 x = 420

, so

10 x = 420

and

x = 42

. Check,

14 + 6 = 20

Practice

Solve

\frac{x}{2} - \frac{x}{6} = 13

Show the solution

Multiply both sides by

6

3 x - x = 78

, so

2 x = 78

and

x = 39

. Check,

\frac{39}{2} - \frac{13}{2} = 13

Practice

Solve

0.9 x - 2.5 = 3.8

by multiplying both sides by

10

first.

Show the solution

Multiply both sides by

10

9 x - 25 = 38

, so

9 x = 63

and

x = 7

. Check,

6.3 - 2.5 = 3.8

Practice

Solve

\frac{x}{6} = \frac{15}{9}

Show the solution

Cross-multiply,

9 x = 6 \cdot 15 = 90

, so

x = 10

. Check, both sides equal

\frac{5}{3}

Practice

Solve

\frac{x}{9} + 4 = \frac{x}{6}

. The multiplier reaches the

4

too.

Show the solution

Multiply both sides by

18

2 x + 72 = 3 x

, so

x = 72

. Check,

8 + 4 = 12

and

\frac{72}{6} = 12

. Multiplying only the fractions is the error the check would have caught.

Practice

Solve

\frac{x + 8}{7} = \frac{x - 4}{3}

Show the solution

Cross-multiply,

3 (x + 8) = 7 (x - 4)

, so

3 x + 24 = 7 x - 28

, then

4 x = 52

and

x = 13

. Check, both sides equal

3

Practice

Solve

0.13 x = 0.05 x + 2

Show the solution

Multiply both sides by

100

13 x = 5 x + 200

, so

8 x = 200

and

x = 25

. Check,

3.25 = 1.25 + 2

Practice

Solve

\frac{x}{3} + 8 = \frac{x}{2} + 5

Show the solution

Multiply both sides by

6

2 x + 48 = 3 x + 30

. Subtract

2 x

, then subtract

30

, so

x = 18

. Check,

6 + 8 = 14

and

9 + 5 = 14

Practice

Solve

\frac{x}{8} + \frac{3}{4} = \frac{x}{2}

. Three fractions, so cross-multiplying does not apply. One multiplier clears all three.

Show the solution

Multiply both sides by

8

x + 6 = 4 x

, so

3 x = 6

and

x = 2

. Check,

\frac{1}{4} + \frac{3}{4} = 1

Practice

Solve

\frac{x}{2} + \frac{x}{3} = - 15

Show the solution

Multiply both sides by

6

3 x + 2 x = - 90

, so

5 x = - 90

and

x = - 18

. Check,

- 9 - 6 = - 15

Practice

Solve

\frac{2 x + 7}{5} = \frac{x + 1}{2} + 3

. The extra

+ 3

means the cross-multiplying shortcut does not apply.

Show the solution

Multiply both sides by

10

2 (2 x + 7) = 5 (x + 1) + 30

, so

4 x + 14 = 5 x + 35

and

x = - 21

. Check,

\frac{- 35}{5} = - 7

and

\frac{- 20}{2} + 3 = - 7

Practice

A student solves

0.5 x - 2 = 0.25 x + 4.5

by multiplying only the

x

terms by

100

, writes

50 x - 2 = 25 x + 4.5

, and gets a candidate that fails the check in the original. Multiply every term by

100

and enter the correct solution.

Show the solution

Multiply every term by

100

50 x - 200 = 25 x + 450

, so

25 x = 650

and

x = 26

. Check,

13 - 2 = 11

and

6.5 + 4.5 = 11

. The student's line skipped the

2

and the

4.5

, so the two sides were scaled unequally and the candidate fails the original.

Practice

How many solutions does

\frac{6 x + 9}{3} = 2 x + 5

have? Enter the count.

Show the solution

The left side is

\frac{6 x + 9}{3} = 2 x + 3

, so the equation reads

2 x + 3 = 2 x + 5

. Subtracting

2 x

leaves

3 = 5

, which is false, so there are

0

solutions. The fraction was hiding a 4.3 outcome, not changing it.

Practice

Solve

\frac{91}{x - 6} = 7

, and check that your value is not the excluded

6

Show the solution

Multiply both sides by

x - 6

91 = 7 (x - 6)

, so

x - 6 = 13

and

x = 19

. Check,

\frac{91}{13} = 7

, and

19

is not the excluded

6

, so the denominator was never zero.

Practice

Solve

\frac{3 x - 2}{4} - \frac{x - 3}{6} = 2

. Enter the solution as a fraction in lowest terms.

Show the solution

Multiply both sides by

12

3 (3 x - 2) - 2 (x - 3) = 24

, so

9 x - 6 - 2 x + 6 = 24

, then

7 x = 24

and

x = \frac{24}{7}

. Check,

\frac{29}{14} - \frac{1}{14} = \frac{28}{14} = 2

. Distributing the

- 2

over both terms of

x - 3

is where this one is usually lost.