LCD, first grade equation, solving help

strunz · #1 $Old$ April 19th, 2009, 12:18 PM

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

LCD must be 2y-4

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

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

skeeter · #2 $Old$ April 19th, 2009, 12:49 PM

Quote:

Originally Posted by strunz $View Post$

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

LCD must be 2y-4

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

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

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

try again.

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

Quote:

Originally Posted by strunz $View Post$

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

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.

strunz · #4 $Old$ April 22nd, 2009, 09:57 AM

Quote:

Originally Posted by skeeter $View Post$

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

try again.

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?

Grep · #5 $Old$ April 22nd, 2009, 11:23 AM

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!

skeeter · #6 $Old$ April 22nd, 2009, 03:57 PM

Quote:

Originally Posted by strunz $View Post$

I don't get it.

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

y= 1
y is supposed to be 3

.