1/8-x=3/5 i dont get how to solve this
you can solve equations by applying the same operation on both sides this is ALWAYS possible in EVERY equation
you want x seperately, right? (the description for the unknown)
but what operation?
you can pick any operation, such as: addition of some number or constant, subtraction of some number or constant, multiplication of some number or constant, division of some number or constant. of course, one wants to make operations that help - however, any operation is allowed AS LONG AS it is applied to BOTH SIDES
and yes about the x
\[\frac{ 3 }{ 5 } - \frac{ 1 }{ 8 } = x \]
yes we want the x separately. what is still bothering us?
1/8-x=3/5
i just dont get how to solve it can u give a example
yes
so we have this: 1/8-x=3/5
yes
two things are wrong: 1. there are numbers on the side of x.... 2. the x is negative
okay
just take the -x to the opposite side and you will have an x,
the number is a positive fraction: 1/8 1/8-x=3/5 the following operation is allowed: subtracting a number we choose as number 1/8 this is what we do: 1/8-x -1/8=3/5 -1/8
okay
-1/8 was introduced to both sides. this is okay as long as we introduce the same amount, the same sign -- on both sides!
then, if we calculate that we get: 1/8-x -1/8=3/5 -1/8 gives -x = 3/5 - 1/8
okay
now the only issue is that x is still negative - that was problem 2)
to counter this problem, we can "MULTIPLY BY -1". all it does is change the signs of every term when we multiply with -1 it doesn't change the values...
ok, so operation two: multiply by -1 before, we had: -x = 3/5 - 1/8 now, we have: x = -3/5 + 1/8
but why -1
because the x was negative: -x = 3/5 - 1/8
we don't want -x = ... we want x = ,,,
and MULTIPLICATION with -1 just changes the sign 3 * -1 = -3 :)
oh
you could have made other steps, as well
do you want to see a second way to solve this (without -1) ?
yes please
ok original equation: 1/8-x=3/5
yes
now we can add x to both sides (Operation: addition of variable) 1/8-x+x=3/5 + x -x + x = 0, so it is the same as: 1/8=3/5 + x
with this we were able to also get a positive x :)
then 1/8=3/5 + x the 3/5 just needs to be subtracted away so x is alone
okay
\[\frac{ 1 }{ 8 } - x = \frac{ 3 }{ 5 }\] \[\frac{ 1 }{ 8 } = x + \frac{ 3 }{ 5 }\] \[-\frac{ 3 }{ 5 } + \frac{ 1 }{ 8 } = x\]
before: 1/8=3/5 + x subtraction of 3/5 from both sides: 1/8 - 3/5 = x
last part again: 1/8 -3/5 =3/5 + x - 3/5 1/8 -3/5 = 3/5 - 3/5 + x 1/8 -3/5 = 0 + x 1/8 - 3/5 = x
I chose to subtract 3/5 because it cancels the 3/5 on the x side that's why I subtract exactly that
okay so then you would find a common denominator and that would be 40
true
okay so it would be 5/40 - 24/40
yes
-19/40 ?
yes
yeah
that's the value of the uknown x in simple form
oh ok
the same value was already in the original equation, we just couldn't see it ;)
because everything was over the place and split up, on the wrong side...
all information was already there
remember with this equation you can verify your answer, if you unsure. substitute x = (your answer) and see if LHS = RHS .
yes that's a good tip from @LinuxMint
1 more question 3/4-5/2x=1/3
we want x alone, there's a fraction that shouldn't be there: 3/4
OK so: if there is a negative x, it is a good idea to add x to both sides
ok?
that's a new rule
k
ok, so we can add one x, or three x to both sides
however it only makes sense to add the same amount that we have in negative 3/4-5/2x=1/3
what is the amount of negative x here?
k
do you know what coefficient of a variable is?
no
oh
well it is the number that's with the variable 3x the 3 in this case
to solve equations with a NEGATIVE x, you must determine this number in front of x
3/4-5/2x=1/3
i think i got this now if i have any mor e ?s i will let you know
ok :)
in other terms, if there's -5/2 x then you have to add 5/2 x to both sides, if there's -3x you have to add 3x to both sides, if there's a -2/7 x you have to add 2/7 x to both sides.
the -2/7 x will disappear on one side and appear positive on the other
that's how to deal with negative x's
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