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The equation 3x+6=2x+10+x-4 is true for all real numbers. The equation 6x+2-2x=4x+1 has no solution. When do you think an equation has all real numbers as its solutions?​

1 Answer

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Answer:

when it simplifies to 0=0

Explanation:

An equation has "all real numbers" as its solution when it can be simplified to ...

0 = 0 . . . has "all real numbers" as a solution set

__

Example:

3x +6 = 2x +10 +x -4

Subtract (3x+6) from both sides:

(3x +6) -(3x +6) = (2x +10 +x -4) -(3x +6)

0 = x(2 +1 -3) +(10 -4 -6)

0 = 0x +0

0 = 0 . . . . has "all real numbers" as a solution set (True for every value of x.)

Example 2:

6x +2 -2x = 4x +1

Subtract (4x+1) from both sides:

(6x +2 -2x) -(4x +1) = (4x +1) -(4x +1)

x(6 -2 -4) +(2 -1) = 0

0x +1 = 0

1 = 0 . . . . . does not have "all real numbers" as a solution set. There is no solution. (No value of x will make this true.)

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