# Thread: Having Trouble Solveing Equalities with Variables on Each Side

1. ## Having Trouble Solveing Equalities with Variables on Each Side

EDIT: One more problem I'm having with fuctions...

solve:
f(2/3) = (x^2+3)/(x-5)

having a hard time here

2. I'll do the second rational equation ... you try the first one after I show you how ...

$\displaystyle \frac{x+10}{x+3} = \frac{x+8}{x+5}$

cross multiply ...

$\displaystyle (x+10)(x+5) = (x+3)(x+8)$

expand (multiply out) both sides ...

$\displaystyle x^2 + 15x + 50 = x^2 + 11x + 24$

combine like terms ...

$\displaystyle 4x = -26$

$\displaystyle x = -\frac{13}{2}$

here's the first one ... I'll leave the easier one for you.

$\displaystyle \sqrt{x^2 + 16} = 3+x$

square both sides ...

$\displaystyle x^2 + 16 = 9 + 6x + x^2$

combine like terms ...

$\displaystyle 7 = 6x$

$\displaystyle \frac{7}{6} = x$

important! ... check the solution

$\displaystyle \sqrt{\frac{49}{36} + 16} = 3 + \frac{7}{6}$

$\displaystyle \sqrt{\frac{625}{36}} = \frac{25}{6}$

$\displaystyle \frac{25}{6} = \frac{25}{6}$

checks good.

3. Originally Posted by skeeter
I'll do the second rational equation ... you try the first one after I show you how ...

$\displaystyle \frac{x+10}{x+3} = \frac{x+8}{x+5}$

cross multiply ...

$\displaystyle (x+10)(x+5) = (x+3)(x+8)$

expand (multiply out) both sides ...

$\displaystyle x^2 + 15x + 50 = x^2 + 11x + 24$

combine like terms ...

$\displaystyle 4x = -26$

$\displaystyle x = -\frac{13}{2}$

here's the first one ... I'll leave the easier one for you.

$\displaystyle \sqrt{x^2 + 16} = 3+x$

square both sides ...

$\displaystyle x^2 + 16 = 9 + 6x + x^2$

combine like terms ...

$\displaystyle 7 = 6x$

$\displaystyle \frac{7}{6} = x$

important! ... check the solution

$\displaystyle \sqrt{\frac{49}{36} + 16} = 3 + \frac{7}{6}$

$\displaystyle \sqrt{\frac{625}{36}} = \frac{25}{6}$

$\displaystyle \frac{25}{6} = \frac{25}{6}$

checks good.

Thanks a bunch! I knew it would be somewhat easy once I got a little help.

Got one last problem I'm stuck on with my homework so I edited my post rather than make another thread.

4. Originally Posted by emximer
EDIT: One more problem I'm having with fuctions...

solve:
f(2/3) = (x^2+3)/(x-5)

having a hard time here
Is the question to find f(2/3) given that $\displaystyle f(x) = \frac{x^2 + 3}{x-5}$? Because if it is, all you have to do is substitute 2/3 into that expression for f(x).

Or is the question to solve the equation $\displaystyle \frac{2}{3} = \frac{x^2 + 3}{x-5}$?

5. Originally Posted by icemanfan
Is the question to find f(2/3) given that $\displaystyle f(x) = \frac{x^2 + 3}{x-5}$? Because if it is, all you have to do is substitute 2/3 into that expression for f(x).

Or is the question to solve the equation $\displaystyle \frac{2}{3} = \frac{x^2 + 3}{x-5}$?
you have to substitute 2/3 in and solve, for some reason I can't get the right answer though. These problems are for WebWorks, so you submit and answer and it tells you if you're right.

6. $\displaystyle \frac{(\frac{2}{3})^2 + 3}{\frac{2}{3}-5}$

$\displaystyle \frac{(\frac{4}{9}) + 3}{\frac{2}{3}-5}$

$\displaystyle \frac{\frac{4+27}{9}}{\frac{2-15}{3}}$

$\displaystyle \frac{\frac{31}{9}}{\frac{-13}{3}}$

$\displaystyle \frac{(3)(31)}{(9)(-13)}$

$\displaystyle -\frac{31}{39}$