The Purplemath Forums

by **Merz** on Mon Feb 23, 2009 4:29 pm

I forgot some of my algebra and can't recall what to do with x+7/x-9=11...

nearly done.

**Sponsor**

by **DAiv** on Mon Feb 23, 2009 7:28 pm

Hi Merz, welcome to the forums. :wave:

As written, the equation you gave was $x \,+\, \frac7x \,-\, 9 \,=\, 11$ , which I presume you meant rather than $\frac{x \,+\, 7}{x \,-\, 9} \,=\, 11$ , (x+7) / (x-9) = 11. In either case, you don't need logs to solve these.

So, assuming you did mean the first version, I'll take you part way and give you a pointer for the rest. (If you didn't mean that version, let me know and I'll post a reply for the other one.)

Solve for $x$ :

$x \,+\, \frac7x \,-\, 9 \,=\, 11$

Add 9 to both sides:

$x \,+\, \frac7x \,-\, 9 \,+\, 9 \,=\, 11 \,+\, 9$

$x \,+\, \frac7x \,=\, 20$

Get rid of the fraction on the left hand side by multiplying both sides by $x$ :

$x \, (x \,+\, \frac7x) \,=\, 20x$

... and multiplying out the parentheses:

$x^2 \,+\, 7 \,=\, 20x$

Bring the $20x$ over and rearrange:

$x^2 \,-\, 20x \,+\, 7 \,=\, 0$

And now you're left with a quadratic equation.

Can you continue from there?

DAiv

The Purplemath Forums

Solving Log functions... x + 7 / x - 9 = 11

Solving Log functions... x + 7 / x - 9 = 11

Re: Solving Log functions...