complex linear equation?

[usctrojanfan]

So the directions say : list all restrictions on solutions, solve and check the solution:

x/x-5=25/x^2-5x so i got it in simplest form which is x/x-5=25/x(x-5) so i see the denominators are almost alike the first one is missing just one x, but i'm confused as what i'm actually suppose to do on this problem do i set both denominators to =0? and is that the answer? For the first denominator I do x-5=0 so x=5 and that is a restriction because of non zero denominator rule so one restrction is x cant be 5 correct? what about the second denominator it's x^2-5x or x(x-5) what do i do with this denominator?

 

[Deleted member 4993]

So the directions say : list all restrictions on solutions, solve and check the solution:

x/x-5=25/x^2-5x so i got it in simplest form which is x/(x-5)=25/[x(x-5)] so i see the denominators are almost alike the first one is missing just one x, but i'm confused as what i'm actually suppose to do on this problem do i set both denominators to =0? and is that the answer? For the first denominator I do x-5=0 so x=5 and that is a restriction because of non zero denominator rule so one restrction is x cant be 5 correct? what about the second denominator it's x^2-5x or x(x-5) what do i do with this denominator?

You need to put those brackets[] and parentheses() to pose the problem correctly.

x/(x-5)=25/[x(x-5)]

Yes x \(\displaystyle \ne\) 5 is a restriction.

x \(\displaystyle \ne\) 0 is another restriction. At x = 0, the right-hand-side of the equation becomes undefined (in addition to being undefined at x = 5).
.

 

[stapel]

So the directions say : list all restrictions on solutions, solve and check the solution:

. . . . .\(\displaystyle \dfrac{x}{x\, -\, 5}\, =\, \dfrac{25}{x^2\, -\, 5x}\)

I think you mean the above to have been the original equation.

so i got it in simplest form which is

. . . . .\(\displaystyle \dfrac{x}{x\, -\, 5}\,=\, \dfrac{25}{x(x\, -\, 5)}\)

I think you mean here that you "got the right-hand side in factored form".

so i see the denominators are almost alike the first one is missing just one x, but i'm confused as what i'm actually suppose to do on this problem

You're supposed to follow the instructions in the manner which was explained and demonstrated in the book and in class:

Find all restrictions: What x-values would cause division by zero? (Hint: Set the denominators equal to zero, and solve.) Is division by zero allowed? (Hint: No.) So what x-values are not allowed? (Hint: The ones that would cause division by zero.) So what are the "restrictions"? (Hint: The values that are not allowed.)

Solve: Use whatever method you like for solving rational equations.

Check the solution: Compare the solution x-values with the list of restrictions. If there's a match, cross that off the "solution" list.