How do you solve #1/x-1/(2x)=2x# ?
I know the solution to
#1/x-1/(2x)=2x#
is
#x=+-0.5#
At least it is the solution in my math book.
But I don't understand how you solve it.
When I tried i got #x=+-0,25# I think.
So I typed the problem into Cymath. And it tells me to do the following:
One. Simplify #1/x-1/(2x)# to #1/(2x)#
#1/(2x)=2x#
Two. Multiply both sides by #2x#
#1=2x*2x#
Three. Simplify #2x*2x# to #4x^2#
#1=4x^2#
Four. Divide both sides by #4#
#1/4=x^2#
Five. Take the square root of both sides.
#+-sqrt(1/4)=x#
So the answer is #x=+-0.5#
But the thing I don't understand is the first step. Why does #1/x-1/(2x)# equals to #1/(2x)# ?
It just seems so illogical to me. Is there another way to solve the equation? Or can someone please explain why this is?
I know the solution to
is
At least it is the solution in my math book.
But I don't understand how you solve it.
When I tried i got
So I typed the problem into Cymath. And it tells me to do the following:
One. Simplify
Two. Multiply both sides by
Three. Simplify
Four. Divide both sides by
Five. Take the square root of both sides.
So the answer is
But the thing I don't understand is the first step. Why does
It just seems so illogical to me. Is there another way to solve the equation? Or can someone please explain why this is?
2 Answers
See the explanation please.
Explanation:
Let's continue:
Multiply both sides by
2nd method:
To get rid of denominators multiply both sides of equation by the Lowest Common Denominator (LCD) in this case
Explanation:
Given:
Considering the left hand side only for a moment
We need to make the bottom values (denominators) the same. Multiply by 1 and you do not change the value. However, 1 comes in many forms.
Multiply both sides by
Divide both sides by 2
Square root both sides
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Set
Set