hi i need help with elimination method including mixfraction

Avatar.Of.Woe

New member
Joined
Sep 2, 2006
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hi could any1 pls help solve this problem... fractions in this parts confuses me...

-4x = 25 -3y

6x = 5y -39


im supose to solve one of them, then input the answer either x is equal to or y is equal to... then plug it in to the next one and solve for the opposit variable, then plug the asnwer back into the first problem that was solve for

im new to this sight and putting on forums...

i'd apreciate for all the help you guys can give me
 

Denis

Senior Member
Joined
Feb 17, 2004
Messages
1,442
Re: hi i need help with elimination method including mixfrac

Avatar.Of.Woe said:
-4x = 25 -3y
6x = 5y -39
-4x + 3y = 25 [1]
6x - 5y = -39 [2]
Multiply [1] by 3, [2] by 2:
-12x + 9y = 75
12x - 10y = -78
Add them:
-y = -3
y = 3

Finish it.
 

Avatar.Of.Woe

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Sep 2, 2006
Messages
6
wow... that was good.. but how did you know that 3 would fit at[1] and 2 on [2] ?

what did you do to figure it out?

coz i use the method to put everything on one side and then devide them until i the answer...

could you pls show me your method?

or at list do it step by step.. i'd rather learn it than get the answer = )
 

Avatar.Of.Woe

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Joined
Sep 2, 2006
Messages
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i think i got it... -4x = 25 - 3y
and 6x = 5y -39

3(-4x = 25 -3y) becomes -12x+9y=75

and 2(6x = 5y - 39) becomes 12x -10y = -78


9y=12x+75 9y/9= 12x/9 + 75/9

y= 4x/3 + 25/3

then i plug it into the next problem...

12x - 10(4x/3 + 25/3) = -78

12x -40x/3 -250/3 = -78

so i multiplied them all with 3 to find the lcd

3(12x -40x/3 -250/3 = -78)

and come up with 36x -40x -250 = -234
-4x +250 +250

-4x = 16

-4x/4 = 16/-4 x = -4



then i plug in the asnwer to the next one

12x - 10y = -78 12(-4) - 10y = -78


-48 -10y= -78
+48 +48


-10y = -30
-10 -10


y= 3
 
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