2007-03-25 20:34:14 · 11 answers · asked by BingBang 1 in Science & Mathematics ➔ Mathematics
5(x-5)=5x+24
5x-25=5x+24
Both RHS & LHS sides are not equal.
x=0 will also not work.
2007-03-25 20:39:53 · answer #1 · answered by Anonymous · 2⤊ 0⤋
5(x-5)=5x+24
0=49
It is not a valid equation. The unknowns cancel out when like terms are collected. Zero cannot equal 49!
2007-03-26 03:59:53 · answer #2 · answered by Max 6 · 0⤊ 0⤋
5 ( x - 5) = 5x + 24
5x - 25 = 5x +24
5x - 5x = 24 + 25
0 = 49
2007-03-26 05:43:00 · answer #3 · answered by crestfallen 5 · 0⤊ 0⤋
When 'x' goes to infinity there is an answer.
Take the limit of both sides when 'x' tends to infinity & you;lll find both of them equal.
Divide both sides by x as x is not '0' (zero). Now take the limits.
LHS =
LIM (x --> 0 ) 5(1 - 5/x)
= LIM (x -->0) 5 ( 1- 0)
= 5
RHS =
LIM (x-->0) 5 + 24/x
= LIM (x-->0) 5 + 0
= 5
2007-03-26 04:52:25 · answer #4 · answered by CodeRed 3 · 0⤊ 0⤋
The answer is 49.
2007-03-26 05:09:36 · answer #5 · answered by Anonymous · 0⤊ 0⤋
what you have to do is distribute the five into the x-5, the from there add 25 to both sides...and im pretty sure you can figure it out from there...
2007-03-26 03:38:24 · answer #6 · answered by CruzA 1 · 0⤊ 1⤋
I think the only way it works is if x = 0
2007-03-26 03:36:36 · answer #7 · answered by marcusromus 1 · 0⤊ 1⤋
5x-25=5x+24
5x-25+25=5x+24+25
5x-5x=5x-5x+49
0x=49
undefined
2007-03-26 04:58:50 · answer #8 · answered by Dave aka Spider Monkey 7 · 0⤊ 0⤋
the statement is not ttue because once you move "x" it no longer exsit, you get 0x=-1 and that is not possiable.
2007-03-26 03:49:01 · answer #9 · answered by Brandon B 2 · 0⤊ 0⤋
Question is incorrect
2007-03-26 03:49:38 · answer #10 · answered by Como 7 · 0⤊ 0⤋