# If 0 is equal to 0

### Special cases when solving equations

When solving an equation, these special cases can occur:

As a solution set are

**all rational numbers**possible. $$ L = {QQ} $$The equation is at

**no inserted number**correct. $$ L = {$$ $$} $$**0**is the solution to the equation. $$ L = {0} $$

### 1. All rational numbers are possible as a solution set. $$ L = {QQ} $$

**Example:**

$$ 2 * x + 2 = 2 * x + 2 $$ You remove two $$ x $$ boxes.

$$2=2$$

It creates a **true statement** on the last line of the solved equation.

You can now put any weight you want in the $$ x $$ box. Since you're doing it on both sides of the scale, the scales will get stuck in equilibrium.

Write down the solution set as follows: $$ L = {QQ} $$

is the unknown weight box.

stands for 1 kg.

If you do another equivalent conversion, you get $$ 0 = 0 $$.

### 2. The equation is not correct for any number inserted. $$ L = {$$ $$} $$

**Example:**

$$ 2 * x + 2 = 2 * x + 4 $$ You remove two $$ x $$ boxes.

$$ 2 = 4 $$ That's one **wrong statement**.

The equation is not solvable. That is, the solution set is empty.

Write down the solution set as follows: $$ L = {$$ $$} $$

**Summary of the two special cases:**

Whenever the **$$ x $$****ceases to exist**, has the equation

- either
**no**Solution $$ L = {$$ $$} $$ - or
**infinitely many**Solutions $$ L = {QQ} $$.

The scale model is crossed out because the scale **out of balance** hangs.

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### 3. 0 is the solution to the equation.

$$ L = {0} $$

**Example:**

$$ 5 * x = 7 * x $$ $$ | -7 * x $$

$$ - 2 * x = 0 $$ $$ |: (- 2) $$

$$ x = 0 $$

$$ L = {0} $$

If each $$ x $$ box weighs $$ 0 $$ kg, the scales are in equilibrium.

This reshaping is **not permitted**:

$$ 5 x = 7 x $$ $$ |: x $$

$$5=7$$

Here you would assume that $$ x $$ is not $$ 0 $$, because you cannot divide by 0. The 0 is just the solution.

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