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anonymous

  • 5 years ago

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

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  1. shadowfiend
    • 5 years ago
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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?

  2. anonymous
    • 5 years ago
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    what do you mean multiply through by x?

  3. shadowfiend
    • 5 years ago
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    We multiply both sides by x, so that we can get x out of the denominator in 1 / x.

  4. anonymous
    • 5 years ago
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    arent you looking for a common denominator of 2x though?

  5. shadowfiend
    • 5 years ago
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    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.

  6. anonymous
    • 5 years ago
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    i dont understand what your saying

  7. shadowfiend
    • 5 years ago
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    The easiest way to solve an equation like this is to make sure there is no x in any denominators.

  8. anonymous
    • 5 years ago
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    ok so multiply everything by x/x?

  9. anonymous
    • 5 years ago
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    i dont really understand i thought you would multiply everything by 2x to get a common denominator and then add the left side?

  10. anonymous
    • 5 years ago
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    there are multiple ways to approach this problem. finding common denominator for LHS of equation and cross multiplying is also another way.

  11. anonymous
    • 5 years ago
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    ok can you show me steps on how to solve this equation please

  12. shadowfiend
    • 5 years ago
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    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.)

  13. anonymous
    • 5 years ago
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    you can multiply both sides of equations by x, 2x or 4x. All of which should give you the same answer

  14. anonymous
    • 5 years ago
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    thank you i got the answer..can you take a look at my other problems?

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