SOLUTION: decide what values of the variable cannot possibly be solutions for the equation. 1/x-3 + 1/x+2 = 1/x^2-x-6

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Question 1203824: decide what values of the variable cannot possibly be solutions for the equation. 1/x-3 + 1/x+2 = 1/x^2-x-6
Found 2 solutions by MathLover1, ikleyn:
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

decide what values of the variable cannot possibly be solutions for the equation
1%2F%28x-3+%29%2B+1%2F%28x%2B2+%29=+1%2F%28x%5E2-x-6%29

since denominator cannot be equal to zero, the values of the variable cannot possibly be solutions are:
%28x+%2B+2%29+=0=> x=-2
or
%28x+-+3%29=0=> x=3


Answer by ikleyn(52800) About Me  (Show Source):
You can put this solution on YOUR website!
.

Notice that calculations in the post by @MathLover1 are incorrect.


She incorrectly added fractions  1%2F%28x-3%29  and  1%2F%28x%2B2%29  and obtained the sum  1%2F%28%28x%2B2%29%2A%28x-3%29%29.


The correct sum is  %282x-1%29%2F%28%28x%2B2%29%2A%28x-3%29%29.


But if the point of the solution is to define the domain of the equation
and then to exclude it from the possible solution set,
then there is no need to make any calculations.


The prohibited values of x are those that make the denominators equal to zero.

These prohibited values are  x= 3 and x= -2.

Solved.

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Actually, the true meaning of this assignment is not clear from the post.


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        It is typical how @MathLover1 works at this forum.
        She comes,  write some gibberish and comes out.
        Nothing more is interesting for her.

        Nevertheless,  she receives the greatest and hottest approval from her admirers at this forum.

        It shows the level of her admirers,  which is much below par.

        Their level of knowledge is zero - but instead, they do have their own opinion.

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