solve: 3x/2+1/x=3x/4

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solve: 3x/2+1/x=3x/4

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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The first move would be to multiply through by x: 3x^2 / 2 + 1 = 3x^2 / 4 Then multiply through by 4: 6x^2 + 4 = 3x^2 Can you solve that?
what do you mean multiply through by x?
We multiply both sides by x, so that we can get x out of the denominator in 1 / x.

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arent you looking for a common denominator of 2x though?
Hm? No, you can just multiply both sides by x, and that'll cancel the x out in the denominator on the left, and make the other two x-es x^2s instead.
i dont understand what your saying
The easiest way to solve an equation like this is to make sure there is no x in any denominators.
ok so multiply everything by x/x?
i dont really understand i thought you would multiply everything by 2x to get a common denominator and then add the left side?
there are multiple ways to approach this problem. finding common denominator for LHS of equation and cross multiplying is also another way.
ok can you show me steps on how to solve this equation please
You don't have to multiply by x/x, you can just multiply by x. For example: 5 = 5 x*5 = x*5 5x = 5x So the equation is still valid. In this case: \[\frac{3x}{2}+\frac{1}{x} = \frac{3x}{4}\] \[x\left(\frac{3x}{2}+\frac{1}{x}\right) = x\frac{3x}{4}\] \[\frac{x\cdot 3x}{2} + \frac{x\cdot 1}{x} = \frac{x\cdot 3x}{4}\] \[\frac{3x^2}{2} + \frac{x}{x} = \frac{3x^2}{4}\] \[\frac{3x^2}{2} + 1 = \frac{3x^2}{4}\] (And Min-hee is correct about there being multiple approaches; just trying to explain mine. Have to go now though, sorry.)
you can multiply both sides of equations by x, 2x or 4x. All of which should give you the same answer
thank you i got the answer..can you take a look at my other problems?

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