Another Math Wiz question. Come on in. 2 questions

BigJ

Lifer
Nov 18, 2001
21,335
1
81
For the first one, find the LCD and solve. For the second one, get all the numbers divided by y-4 on the same side then solve. Not very hard questions.
 

BigPoppa

Golden Member
Oct 9, 1999
1,930
0
0
First one:
Multiply 1/x by (x-1)/(x-1) and 1/(x-1) by x/x. Easy to solve from there.

Second one:
Not solvable? I'm getting 1 = 5/3.
 
Dec 1, 2003
145
0
0
Originally posted by: BigPoppa
First one:
Multiply 1/x by (x-1)/(x-1) and 1/(x-1) by x/x. Easy to solve from there.

Second one:
Not solvable? I'm getting 1 = 5/3.
It's solvable...I got an integer answer. Not really that hard (then again, I have a bachelor's in math).
 

minendo

Elite Member
Aug 31, 2001
35,553
12
81
Originally posted by: GrouchyLadybug
Originally posted by: BigPoppa
First one:
Multiply 1/x by (x-1)/(x-1) and 1/(x-1) by x/x. Easy to solve from there.

Second one:
Not solvable? I'm getting 1 = 5/3.
It's solvable...I got an integer answer. Not really that hard (then again, I have a bachelor's in math).
Is it just -4?
 

waffel

Member
Mar 16, 2004
78
0
0
I can't get the second one, I don't think it has a solution
y/(y-4) - 5/3 = 4/(y-4)
(y-4)/(y-4) = 5/3
1 = 5/3?
 

pray4mojo

Diamond Member
Mar 8, 2003
3,647
0
0
Originally posted by: waffel
I can't get the second one, I don't think it has a solution
y/(y-4) - 5/3 = 4/(y-4)
(y-4)/(y-4) = 5/3
1 = 5/3?

Woops I bombed it. Got confused with the way the equation is written.
 

DXM

Senior member
Jul 26, 2003
264
0
0
Originally posted by: minendo
Originally posted by: DXM
I got y = +4 for the second one.
That gives you a divide by 0 error on the right side of the equation.
You're right minendo. Funny thing is, I just worked out the problem twice and got +4 twice. So I either need to be sent back to Algebra I or I'm missing a negative sign somewhere. ;)
 

waffel

Member
Mar 16, 2004
78
0
0
Originally posted by: pray4mojo
Originally posted by: waffel
I can't get the second one, I don't think it has a solution
y/(y-4) - 5/3 = 4/(y-4)
(y-4)/(y-4) = 5/3
1 = 5/3?
y = -6
I don't know how you got from y/(y-4) - 5/3 = 4/(y-4) to (y-4)/(y-4) = 5/3.
y/(y-4) - 5/3 = 4/(y-4)
add 5/3 to both sides, subtract 4/(y-4) from both sides
y/(y-4) - 4/(y-4) = 5/3
combine terms
(y-4)/(y-4) = 5/3
 
Dec 1, 2003
145
0
0
Originally posted by: waffel
I can't get the second one, I don't think it has a solution
y/(y-4) - 5/3 = 4/(y-4)
(y-4)/(y-4) = 5/3
1 = 5/3?
Okay, my husband the math teacher agrees with you.

y/(y-4) - (5/3) = 4/(y-4)
y - (5/3)(y-4) = 4
y - (5/3)y + (20/3) = 4
(1 - 5/3)y = 4 - (20/3)
-(2/3)y = -(8/3)
y = -(8/3)*-(3/2)
y = 4

Since the (possible) solution y = 4 is undefined (can't divide by zero), there is no solution.
 

rgwalt

Diamond Member
Apr 22, 2000
7,393
0
0
Originally posted by: waffel
Originally posted by: pray4mojo
Originally posted by: waffel
I can't get the second one, I don't think it has a solution
y/(y-4) - 5/3 = 4/(y-4)
(y-4)/(y-4) = 5/3
1 = 5/3?
y = -6
I don't know how you got from y/(y-4) - 5/3 = 4/(y-4) to (y-4)/(y-4) = 5/3.
y/(y-4) - 5/3 = 4/(y-4)
add 5/3 to both sides, subtract 4/(y-4) from both sides
y/(y-4) - 4/(y-4) = 5/3
combine terms
(y-4)/(y-4) = 5/3
Yep, the equation has no solution.

R

 

ASK THE COMMUNITY