fractional equations{resolved}

[Dillinger4]

I started out fine with these but ive slipped up some where as i went along. :(

If i take a fractional equation such as x/2 + x/3 = 10 both sides have to be multiplied by the lcd which is 6. The same holds true if i take another equation such as 2/(y - 1) - 4/3y = 1/y2 - y with the lcd being 3y(y-1).

Now how would i apply this to an equation such as (a + 5)/2 = (a+2)/5 or (2r + 1)/3 -(r+1)/5 = (r+8)/6? For (2r + 1)/3 -(r+1)/5 = (r+8)/6? Using an lcd of 15 results in the right hand side of the equation to still be in the form of a fraction. :confused:

[TheManWhoCan]

A quicker way to think about it (without worrying about LCD's) is to just multiply everything by the denominator of each fraction in turn:

(2r + 1)/3 -(r + 1)/5 = (r + 8)/6

Multiply by 3:

(2r + 1) - 3(r + 1)/5 = (r + 8)/2

Multiply by 5:

5(2r + 1) - 3(r + 1) = 5(r + 8)/2

Multiply by 2:

10(2r + 1) - 6(r + 1) = 5(r + 8)

Now you can solve to get r = 4.

Note also that if the coefficients (10, -6 and 5 in this case) have no common factors, the product of the original denominators will be the LCD.

[prog_tom]

(a+5)/2 - (a+2)/5 = 0

5(a+5) - 2(a+2)/10 = 0

a= -7

Easy ****, rofl.

[Dillinger4]

Yeah i gotta stop working late. :rolleyes:


Posted by TheManWhoCan

Note also that if the coefficients (10, -6 and 5 in this case) have no common factors, the product of the original denominators will be the LCD.


Seems easy enough. That will come in handy when i start algebra b next semester. So the lcd for this equation shoud be 90.

(2r + 1)/3 -(r+1)/5 = (r+8)/6

30(2r + 1) -18(r + 1) = 15(r + 8)

60r + 30 - 18r - 18 = 15r + 120

42r + 12 = 15r + 120

27r/27 = 108/27

r = 4

Thanks for the help.

[Dillinger4]

Sorry. I ment 90 would be a common denominator not the lcd. I was using the process that TheManWhoCan pointed out. :)