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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.

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

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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) |¹/₃x + 4| = 0 means that
¹/₃x + 4 = 0 <=>
¹/₃x = -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

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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

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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

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Problem 4 Solve the equation:
A) 2|x – 1| + 3 = 9 – |x – 1|;
B) 3|x| – (x + 1)² = 4|x| – (x² -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| – (x² - 1) - 2(x - 5)<=>
x² - 1 + 2(x – 5) – (x + 1)² = 4|x| – 3 |x| <=>
x² - 1 + 2x - 10 – (x² + 2x + 1) = (4 - 3)|x| <=>
x² + 2x - 11 – x² - 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

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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.

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Problem 6 Solve the equation:
|(2x + 1)² - 4x² - 2| - 3|4x – 1| = - 6

Solution:

|(2x + 1)² - 4x² - 2| – 3|4x -1| = - 6 <=>
|4x² + 4x + 1 - 4x² - 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

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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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