Solve for x: [3x - 25/x + 7] - 5 = 3/x

[Bladesofhalo]

Sorry about the numerous posts on this forum, this is my last question for today and your help is appreciated greatly:
Solve for x:
[3x - 25/x + 7] - 5 = 3/x

Heres how I tried to do it:
Multiplying both sides by x(x+7) yields:
[3x[sup:1shbg61x]2[/sup:1shbg61x] + 25x] - (5x[sup:1shbg61x]2[/sup:1shbg61x] + 35x) = 3x + 21
So adding and subtracting done, I get:
-2x[sup:1shbg61x]2[/sup:1shbg61x] + 57x -21
But, substituting into the quadratic formula yields two totally different answers. I guess my math teacher hasnt been doing her job :?

Thnaks in advance for any help

 

[stapel]

[3x - 25/x + 7] - 5 = 3/x

What you have posted inside the brackets means the following:

. . . . .3x - (25/x) + 7

Was that what you meant? Or did you mean something more like the following?

. . . . .(3x - 25) / (x + 7)

Also, what operation do the brackets represent? (Brackets are often used to represent the "greatest integer" function, or the floor or ceiling function.)

Thank you!

Eliz.

 

[Mrspi]

Sorry about the numerous posts on this forum, this is my last question for today and your help is appreciated greatly:
Solve for x:
[3x - 25/x + 7] - 5 = 3/x

Heres how I tried to do it:
Multiplying both sides by x(x+7) yields:
[3x[sup:2czybzko]2[/sup:2czybzko] + 25x] - (5x[sup:2czybzko]2[/sup:2czybzko] + 35x) = 3x + 21
So adding and subtracting done, I get:
-2x[sup:2czybzko]2[/sup:2czybzko] + 57x -21
But, substituting into the quadratic formula yields two totally different answers. I guess my math teacher hasnt been doing her job :?

Thnaks in advance for any help

You may want to recheck "the adding and subtracting done".....

(3x[sup:2czybzko]2[/sup:2czybzko] + 25x) - (5x[sup:2czybzko]2[/sup:2czybzko] + 35x) = 3x + 21

3x[sup:2czybzko]2[/sup:2czybzko] + 25x - 5x[sup:2czybzko]2[/sup:2czybzko] - 35x = 3x + 21

-2x[sup:2czybzko]2[/sup:2czybzko] - 10x = 3x + 21

Now continue, and see if the quadratic formula gives you "better" results.

 

[Bladesofhalo]

Argh, again always those simple mistakes. But I was proceeding in the right direction, so thats good. Thank you very much mrspi and eliz for the help again.

 

[kasie-tutor]

Dear Bladesofhalo,

Based on context I am asuming you are studying how to solve rational equations. This involves "clearing the fractions" by multiplying both sides of the equation by the LCD (which you did when you multiplied both sides by \(\displaystyle x(x+7)\)).

Note: Do you see how your math syntax is ambiguous (like stapel said)? If you hadn't mentioned anything about x(x + 1) I wouldn't have known what you were asking.

You should check how you added your like terms (i.e., the terms that are like to \(\displaystyle -25x\)). That is where you made at least one error.

You are right that the quadratic formula will come in handy, once you get to the form \(\displaystyle ax^2+bx+c=0\).