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Math Help - system of equations

  1. #1
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    system of equations

    im not sure how to solve problem 1, need a lil help.
    solve the following:
    1)

    1/x + 1/y = 1/19
    x/y = 12

    2)solve the variable v

    3 = 25x + 5v
    xv = v + x

    im not sure if im doing 2 right, but i times the 1st equation by (v) and 2nd equation by (25). so got:
    -25xv - 5v^2 +3x = 0
    25xv -25v -25x = 0
    so:
    -5v^2 -22v -25x=0, if thats right do i use the quadratic equation to solve for v? if not please explain.
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  2. #2
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    Hello,
    Quote Originally Posted by viet View Post
    im not sure how to solve problem 1, need a lil help.
    solve the following:
    1)

    1/x + 1/y = 1/19
    x/y = 12
    In the second equation, write x=12y and substitute in the first

    2)solve the variable v

    3 = 25x + 5v
    xv = v + x

    im not sure if im doing 2 right, but i times the 1st equation by (v) and 2nd equation by (25). so got:
    -25xv - 5v^2 +3x = 0
    25xv -25v -25x = 0
    so:
    -5v^2 -22v -25x=0, if thats right do i use the quadratic equation to solve for v? if not please explain.
    Not very correct... The aim of it is to leave only v or x in the equation.

    The second equation looks more "difficult" to isolate x or v than the first one. I'll show if you take the 1st equation, you'll do the second one

    3=25x+5v ----> 5v=3-25x ----> v=(3-25x)/5

    In the first question, you will just have to add 2 fractions. In the second question, you will have to use the quadratic formula.
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  3. #3
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    im working on the first question.
    after plugging  x = 12y into the first equation, i got:
    1/12y + 1/y = 1/19
    =
    1/13y = 1/19
    at this point do i solve for y?
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  4. #4
    Moo
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    Quote Originally Posted by viet View Post
    im working on the first question.
    after plugging  x = 12y into the first equation, i got:
    1/12y + 1/y = 1/19
    =
    1/13y = 1/19
    at this point do i solve for y?


    \frac{1}{12y}=\frac1{12} \cdot \frac 1y+\frac 1y

    Be careful ! \frac 1a+\frac 1b \neq \frac{1}{a+b}

    And after that, you can solve for y by taking the inverse
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  5. #5
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    what happen to the \frac{1}{19} ?
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