You will need

- the equation with the module.

Instruction

1

If the equation only one module, proceed as follows. Move all values that are not contained under the module on the right side. Then use the formula A=b => a=±b, where b≥0 (if b

2

In the same way to solve equations, in which x contains at the same time and under the module, and without module. Move all the pieces without the module to the right side and expand the module, transforming one equation into a system of two. Here already it is necessary to specify the DPG, as it will participate in finding the solution.

3

If the equation contains two modules are equal, do so. Expand the second module as if it were a regular number. Thus, you'll have a system of two equations, solve each separately and combine the solution. For example, the equation IX+3I=IX-7I. After the disclosure of the module you will get two equations: x+3=x-7 and x+3=-(x-7). The first equation has no solutions (3=-7), and from the second to obtain x=2. Thus, one solution x=2.

4

If in addition to the two modules in the equation is a number, the solution is slightly more complicated. To solve this equation, divide the range of permissible values for several intervals. To do this, locate the x values at which

**the modules**are reset (Paranaita**modules**to zero). Thus, you get a few intervals in which**modules**are disclosed with different signs. Then consider each case separately, revealing the module with the token, which is obtained by substituting the a value from the interval. As a result, you will get several solutions that need to be merged. For example, the equation IX+2I+IX-1I=5. Equating**modules**to zero, we get the boundaries of the intervals -2 and 1. Consider the first interval: x