Chapter 2 Equations in one Variable

Section 2.1 Simple Equations

2.1.2 Conditions in Transformations

Multiplication, division, and taking reciprocals are equivalent transformations only if the factors or terms are non-zero. In the last step of example 2.1.7, the reader can see that both sides of the equation are always non-zero. Therefore the transformation is allowed. If the variables themselves are used in the transformation, we need to make a note somewhere to remind us that the respective term must be non-zero. The solution at the end of the transformations is then only valid for variable values satisfying the transformation conditions. All other values have to be checked separately, typically by inserting the value into the equation:
Example 2.1.8
In this example, the necessary transformation conditions are not problematic:

Start: 9x  =  81 x2     ‖                :x , transformation allowed if   x≠0    ⇔ 9  =  81x    ‖                :81   and exchange sides     ⇔ x  =   1 9      and this value satisfies the condition   x≠0 .

The value x=0, initially rejected by the transformation condition, has to be checked separately. The equation 9x=81 x2 is also satisfied for x=0, hence x=0 is also a solution of the equation. In set notation, the solution set is L={0; 1 9 }.

In any case, values violating a condition have to be checked separately. It may or may not turn out that they are part of the solution.
Example 2.1.9


Start: x2 -2x  =  2x-4    ‖                 factor out on both sides     ⇔ x·(x-2)  =  2·(x-2)    | :(x-2) , transformation only allowed if   x≠2    ⇔ x  =  2 .

This value of x violates the condition x≠2. Hence, this is possibly no solution. Inserting x=2 into the initial equation gives 22 -2·2=0 on the left-hand side and also 2·2-4=0 on the right-hand side. Hence, x=2 is indeed a solution, even though it violated the transformation condition.

Exercise 2.1.10
Find the solution of the equation (x-2)(x-3)= x2 -9 by transforming the right-hand side using the third binomial formula and then dividing by a common factor.
The solution is x = .

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