Find the real solns for (10/x) - (12/x - 3) + 4 = 0

Guitarman

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Aug 22, 2006
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I'm having some trouble finding the real solutions for this problem:

. . .(10/x) - (12/x - 3) + 4 = 0

The "10/x" and "12/x-3" are fractions.

Looking at it, the answer appears to be 5, but I'm having a hard time getting it. Could someone point me in the right direction? Thank you!
 
Multiply by your LCD, x(x-3):

\(\displaystyle \L\\\sout{x}(x-3)\frac{10}{\sout{x}}-x\sout{(x-3)}\frac{12}{\sout{x-3}}+4x(x-3)=0\)
 
Guitarman said:
The "10/x" and "12/x-3" are fractions.
So you mean the equation to be as follows?

. . . . .10/x - 12/(x - 3) + 4 = 0

Guitarman said:
Looking at it, the answer appears to be 5....
I'm sorry, but what do you mean "looking at it"? Do you mean that you're looking at the equation and trying to solve it in your head, without pencil or paper...?

A good first step for solving rational equations is to multiply through by the common denominator. Then solve the resulting equation. (In this case, it will be a simple quadratic equation. So there could be two solutions.)

Eliz.
 
well yea...lol. I sometimes try to plug in numbers before I work the problem out. In this case, plugging 5 in for x.

sorry.

thanks for the help.
 
Guitarman said:
ok after I did all that I get x=3 or x=3.5. seems right to me
You can check the solution to any "solving" problem by plugging your answer back into the original question. In this case:

. . .x = 3:

. . . . .10/[3] - 12/([3] - 3) ...

No; 3 - 3 = 0, and division by zero is not defined. So "x = 3" cannot be a solution. Continuing:

. . .x = 3.5:

. . . . .10/[3.5] - 12/([3.5] - 3) + 4
. . . . .= 10/(35/10) - 12/0.5 + 4
. . . . .= (10/1)(10/35) - (12/1)(2/1) + 4
. . . . .= 100/35 - 24 + 4
. . . . .= 20/7 - 20
. . . . .= 20/7 - 140/7
. . . . .= -120/7

So "x = 3.5" cannot be a solution.

Please reply showing all of your steps. Thank you.

Eliz.
 
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