# Module equations. Modules

**Module(absolute value)** of a positive number or zero is the number itself and module of a negative number is called its contrary number i.e.

|a| = -a if a < 0

From the definition is clear that the absolute value of every rational number different from zero is a positive number. Therefore contrary numbers have equal modules. We will observe the following equations |ax + b| = c

**Problem 1** Solve the equation:

A) |x| = 5

B) |3x + 4| = 7

C) |1 / 3x + 4| = 0

D) |2 - 5x| = - 3

E) –|3x – 1| = - 11

F) |3x - 3(x - 1)| = 3

**Solution**:

For solving these equations we will use the definition for module of a rational number.

A) If |x| = 5, then x = 5 or x = - 5, because both 5 and -5 have module 5 .

Besides there are no other numbers with such module;

B) From |3x + 4| = 7 we get that 3x + 4 = 7 or 3x + 4 = -7

From the first equation we find 3x = 7 - 4 <=> 3x = 3 <=> x = 1,

and from the second 3x = - 7 - 4 <=> 3x = -11 <=> x = -11/3

C) |^{1}/_{3}x + 4| = 0 means that

^{1}/_{3}x + 4 = 0 <=>

^{1}/_{3}x = -4 <=> x = -12

D) |2 - 5x| = -3 has no solution, because from the theory we find that there is no number which has negative number for module.

E) -|3x – 1| = - 11 <=> |3x - 1| = 11,

which gets 3x - 1 = 11 or 3x - 1 = -11

From solving the last two equations we find

x = 4 or x = -10/3

F) |3x - 3x + 3| = 3 <=>|3| = 3, which is identity.

Therefore every x is solution

**Problem 2** Solve the equation:

A) 3|5x|+ 4|5x| = 35

B) |2x|/3 + 3|2x|/2 = 1/2

C) 3.7|x| – 2.2|x| = 22.5

D) |(x + 1)/3| = 5

**Solution**:

A) 3|5x| + 4|5x| = 35 <=>

(3 + 4)|5x| = 35 <=>

7 |5x| = 35 <=>

|5x| = 35/7 <=> |5x| = 5

From the last equation we get 5x = 5 or 5x = - 5.

And we find x = 1 or x = -1

B) |2x|/3 + 3|2x|/2 = 1/2 <=>

2|2x| + 9|2x| = 3 <=>

11|2x| = 3 is equal to <=> |2x| = 3/11

Therefore 2x = 3/11 or 2x = - 3/11,

from where x = 3/22 or x = - 3/22

C) 3.7|x| – 2,2|x| = 22.5 <=>

(3.7 - 2,2)|x| = 22.5 <=>

1.5|x| = 22.5 <=>

|x| = 22.5/1.5 <=> |x| = 15,

from where x = 15 or x = - 15

D) |(x + 1)/3| = 5 we get (x + 1)/3 = 5 or (x + 1)/3 = -5.

Therefore x + 1 = 15 <=> x = 14 or x + 1 = -15 <=> x = -16

**Problem 3** Proof that the equation has no solution:

A) -|(2x + 3)/14| = 5

B) |8x – 4(2x + 3)| = 15

**Solution**:

A) -|(2x + 3)/14| = 5 <=> |(2x + 3)/14| = -5

which has no solution because there is no number with negative number for module.

B) |8x - 4(2x + 3)| = 15 <=> |8x - 8x - 12| = 15 <=>

|-12| = 15 <=> 12 = 15, which shows that it is impossible for any x

**Problem 4** Solve the equation:

A) 2|x – 1| + 3 = 9 – |x – 1|;

B) 3|x| – (x + 1)^{2} = 4|x| – (x^{2} -1) – 2(x - 5);

C) |-3 - 5x| = 3;

D) 2|x – 1| = 9 – |x – 1|;

E) |x| – (3 – x)/4 = (2x - 1)/8

**Solution**:

A) 2|x – 1| + |x -1| = 9 - 3 <=> (2 + 1)|x -1| = 6 <=>

3|x – 1| = 6 <=> |x - 1| = 2

Therefore x - 1 = 2 or x - 1 = - 2,

from where x = 3 or x = - 1

B) 3|x| – (x + 1)^{2} = 4|x| – (x^{2} - 1) - 2(x - 5)<=>

x^{2} - 1 + 2(x – 5) – (x + 1)^{2} = 4|x| – 3 |x| <=>

x^{2} - 1 + 2x - 10 – (x^{2} + 2x + 1) = (4 - 3)|x| <=>

x^{2} + 2x - 11 – x^{2} - 2x - 1 = |x| <=>

-12 = |x|, which has no solution;

C) from |-3 - 5x| = 3 we get -3 - 5x = 3 or -3 - 5x = - 3.

Therefore -3 - 3 = 5x <=> x = - 6/5 or -3 + 3 = 5x <=>

0 = 5x <=> x = 0;

D) 2 |x – 1| = 9 – |x – 1| <=>

2 |x – 1| + |x – 1| = 9 <=>

(2 + 1)|x – 1| = 9 <=> 3|x – 1| = 9 <=>

|x – 1| = 3 we get x - 1 = 3 or x - 1 = -3,

i.e. x = 4 or x = - 2

E) |x| = (2x - 1)/8 + (3 – x)/4 <=>

|x| = [2x - 1 +2(3 – x)]/8 <=>

|x| = 5/8, from where x = 5/8 or x = -5/8

**Problem 5** Solve the equation:

A) |4 – |x|| = 2

B) |9 + |x|| = 5

**Solution:**

A) |4 – |x|| = 2 we get 4 – |x| = 2 or 4 – |x| = -2

We find 4 - 2 = |x| <=>

|x| = 2 or 4 + 2 = |x| <=> |x| = 6

Therefore the solutions are x = 2, -2; 6, -6

B) |9 + |x|| = 5 we get 9 + |x| = 5 or 9 + |x| = - 5

We find |x| = -4 or |x| = -13, but for the last two equations has no solution.

**Problem 6** Solve the equation:

|(2x + 1)^{2} - 4x^{2} - 2| - 3|4x – 1| = - 6

**Solution**:

|(2x + 1)^{2} - 4x^{2} - 2| – 3|4x -1| = - 6 <=>

|4x^{2} + 4x + 1 - 4x^{2} - 2 | - 3|4x - 1| = - 6 <=>

|4x – 1| - 3|4x – 1| = - 6 <=> -2|4x – 1| = - 6 <=>

|4x – 1| = 3 <=> 4x - 1 = 3 or 4x - 1 = -3

Therefore x = 1 or x = -1/2

**Problem 7** Solve the equation:

A) |2x – (3x + 2)| = 1

B) |x|/3 – 2|x|/2 = - 1

C) |3x – 1| = 2|3x – 1| - 2

**Solution**:

A) |2x – 3x – 2| = 1 <=> |-x – 2| = 1 <=>

-x - 2 = 1 or –x - 2 = -1

From the first equation we get -2 - 1 = x <=> x = -3,

and from the second -2 + 1 = x <=> x = -1

B) |x|/3 – 2 |x|/2 = -1 We reduce to a common denominator and we get

2|x| – 3.2.|x| = - 6 <=>

2|x| - 6|x| = - 6 <=>

- 4|x| = -6 <=> |x| = 3/2 <=>

x = 3/2 or x = - 3/2

C) |3x – 1| = 2|3x – 1| – 2 <=>

2 = 2|3x – 1| – |3x – 1| <=>

2 = |3x – 1| <=>

3x - 1 = 2 or 3x - 1 = - 2,

from where 3x = 3 <=> x = 1 or 3x = - 1 <=> x = - 1/3

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