# LCD, first grade equation, solving help

• Apr 19th 2009, 10:18 AM
strunz
LCD, first grade equation, solving help
$\displaystyle \frac{8}{y-2}-\frac{13}{2}=\frac{3}{2y-4}$

LCD must be 2y-4

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

then

(16y-32)-(26y-52)=6y-12
16y-32-26y+52=6y-12
-16y=-32
y=2

Someplace I've must made a mistake as my book says y=3. Somebody?
Thanks
• Apr 19th 2009, 10:49 AM
skeeter
Quote:

Originally Posted by strunz
$\displaystyle \frac{8}{y-2}-\frac{13}{2}=\frac{3}{2y-4}$

LCD must be 2y-4

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

then

(16y-32)-(26y-52)=6y-12
16y-32-26y+52=6y-12
-16y=-32
y=2

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

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

try again.
• Apr 19th 2009, 11:28 AM
stapel
Quote:

Originally Posted by strunz
$\displaystyle \frac{8}{y-2}-\frac{13}{2}=\frac{3}{2y-4}$

LCD must be 2y-4

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

...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.

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

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

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

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

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

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

...and so forth. (Wink)
• Apr 22nd 2009, 07:57 AM
strunz
Quote:

Originally Posted by skeeter
$\displaystyle \frac{2 \cdot 8}{2(y-2)}-\frac{13(y-2)}{2(y-2)}=\frac{3}{2y-4}$

try again.

(Punch) I don't get it.

16 -13y-26 = 3

y= 1
y is supposed to be 3

Some site I can find that could be of help here?
• Apr 22nd 2009, 09:23 AM
Grep
You didn't solve the equation properly. Go over what stapel posted carefully until it's clear. Notice you got a sign wrong in front of the 26. A negative times a negative is *positive*. It's -13 times -2 = 26, not 13 times -2 or whatever. Very very important to make sure you get those right.

From:

16 - 13y + 26 = 3

Move -13y to other side of equal sign:
16 + 26 = 3 + 13y

Move the 3 from the right to the left side and combine the numbers on the left side:
16 + 26 - 3 = 13y
39 = 13y

Which gives:
y = 39/13
y = 3

Anyways, go through stapel's math very carefully. You've got to get fractions and such things down well or the rest of math will be very difficult for you. Build a good foundation and it'll be much easier. And mind those negative signs. :)

As always: practice, practice, practice. It'll seem easy in no time, don't worry. Good luck!
• Apr 22nd 2009, 01:57 PM
skeeter
Quote:

Originally Posted by strunz
(Punch) I don't get it.

16 -13y-26 = 3 ... should be 16 - 13y + 26 = 3

y= 1
y is supposed to be 3

.