The roots of linear equations with one variable.
If the coefficient a ≠ 0, then the solution to this equation is unique and equal to x = b/a.
If a = 0, and b ≠ 0, then the equation has no roots, since 0*x ≠ b for any b ≠ 0.
It remains to consider the case a = 0 and b = 0. In this case, any value of x will be a root of the equation, since the equality 0*x = 0 is true for any value x.
Thus, the linear equation ax = b has a unique root for a ≠ 0, has no roots for a = 0 and b ≠ 0, and has an infinite number of roots for a = 0 and b = 0.
Examples of solving linear equations with one variable.
Example 1.
Solve the equation x = -x.
Solution.
x = -x <=> x + x=0 <=> 2x=0 <=> x=0/2 <=> x=0.
Answer: x = 0.
Example 2.
Solve the equation 3(2.5 - 2x) = 5 - 6x.
Solution.
3(2.5 - 2x) = 5 - 6x <=> 7.5 - 6x = 5 - 6x <=> -6x + 6x = 5 - 7.5 <=> 0*x = -2.5.
But 0 ≠ -2.5 for any x. Therefore, the equation has no roots.
Answer: there are no roots.
Example 3.
Solve the equation 8x = 6 + 2(4x - 3).
Solution.
8x = 6 + 2(4x - 3)x <=> 8x = 6 + 8x - 6 <=> 8x - 8x = 6 - 6 <=> 0*x = 0. We have obtained an identity that is true for any values of x. Therefore, the root of the equation is any number.
Answer: x - any number.
Example 4.
Solve the equation 3x - (10 + 5x) = 54.
Solution.
3x - (10 + 5x) = 54 <=> 3x - 10 - 5x = 54 <=> -2x = 54 + 10 <=> -2x = 64 <=> x = -32.
Answer: x = -32.
Example 5.
Solve the equation 6x - (x - 1) = 4 + 5x.
Solution.
6x - (x - 1) = 4 + 5x <=> 6x - x + 1 = 4 + 5x <=> 5x - 5x = 4 - 1 <=> 0 = 3.
But 0 ≠ 3 for any x. Therefore the original equation has no roots.
Answer: there are no roots.
