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stapel · #3 April 19th, 2009, 01:28 PM

Quote:

Originally Posted by strunz

$\frac{8}{y-2}-\frac{13}{2}=\frac{3}{2y-4}$

LCD must be 2y-4

$(\frac{8}{y-2})(\frac{2y-4}{2y-4})-(\frac{13}{2})(\frac{2y-4}{2y-4})=(\frac{3}{2y-4})(\frac{2y-4}{2y-4})$

...Someplace I've must made a mistake as my book says y=3.

If the LCD is 2(y - 2) (and you're right: it is), then why are you converting to the various different denominators you did...?

Instead, try multiplying through (since this is an equation, so "multiplying through" is allowed) by what that common denominator would be, were you converting to it.

. . . . . $\frac{8}{y\, -\, 2}\, -\, \frac{13}{2}\, =\, \frac{3}{2(y\, -\, 2)}$

. . . . . $\left(\frac{8}{y\, -\, 2}\right)\left(\frac{2(y\, -\, 2)}{1}\right)\, -\, \left(\frac{13}{2}\right)\left(\frac{2(y\, -\, 2)}{1}\right)\, =\, \left(\frac{3}{2(y\, -\, 2)}\right)\left(\frac{2(y\, -\, 2)}{1}\right)$

. . . . . $(8)(2)\, -\, (13)(y\, -\, 2)\, =\, 3$

. . . . . $16\, -\, 13y\, +\, 26\, =\, 3$

. . . . . $16\, +\, 26\, -\, 3\, =\, 13y$

. . . . . $39\, =\, 13y$

...and so forth.