anonymous
  • anonymous
solve: 3x/2+1/x=3x/4
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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shadowfiend
  • shadowfiend
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?
anonymous
  • anonymous
what do you mean multiply through by x?
shadowfiend
  • shadowfiend
We multiply both sides by x, so that we can get x out of the denominator in 1 / x.

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More answers

anonymous
  • anonymous
arent you looking for a common denominator of 2x though?
shadowfiend
  • shadowfiend
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.
anonymous
  • anonymous
i dont understand what your saying
shadowfiend
  • shadowfiend
The easiest way to solve an equation like this is to make sure there is no x in any denominators.
anonymous
  • anonymous
ok so multiply everything by x/x?
anonymous
  • anonymous
i dont really understand i thought you would multiply everything by 2x to get a common denominator and then add the left side?
anonymous
  • anonymous
there are multiple ways to approach this problem. finding common denominator for LHS of equation and cross multiplying is also another way.
anonymous
  • anonymous
ok can you show me steps on how to solve this equation please
shadowfiend
  • shadowfiend
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.)
anonymous
  • anonymous
you can multiply both sides of equations by x, 2x or 4x. All of which should give you the same answer
anonymous
  • anonymous
thank you i got the answer..can you take a look at my other problems?

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