Question 25281
<pre><font face = "lucida console" size = 4><b>5x - 5 = 3x - 7 + 2(x + 1)

First remove the parentheses on the RHS by 
multiplying each term inside the parentheses 
by the outer multiplier 2:

5x - 5 = 3x - 7 + 2x + 2

Combine the 3x and the 2x on the RHS as 5x; 
also combine the -7 and the +2 as -5

5x - 5 = 5x - 5

Add 5 to both sides

5x - 5 + 5 = 5x - 5 + 5

        5x = 5x

Subtract 5x from both sides

   5x - 5x = 5x - 5x

         0 = 0

The 0 on the left should be thought of as 0x

        0x = 0

Now we see that any number can replace x and
the equation will be true, that is

      0(1) = 0

      0(0) = 0

     0(-7) = 0

   0(9999) = 0

  0(11/17) = 0

etc., etc., etc.

Therefore we say x can be any real number.
This kind of equation is called an <i><u>identity</u></i>.
Every number is a solution to it.

Edwin
AnlytcPhil@aol.com</pre>