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Math Help - Identity or No Solution

  1. #1
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    Identity or No Solution

    And how to work it out?

    2+1/3t = 1+1/4t
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  2. #2
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    Re: Identity or No Solution

    Suggestion: Multiply both sides by 12t.
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  3. #3
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    Re: Identity or No Solution

    Quote Originally Posted by ingyaningya View Post
    And how to work it out?

    2+1/3t = 1+1/4t
    If we try t=1 we see that it doesn't work: 2 + \frac{1}{3} \neq 1 + \frac{1}{4} so it's not an identity
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  4. #4
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    Re: Identity or No Solution

    Original Problem:
    2+\frac1{3t} = 1+\frac1{4t}

    Find common denominators:

    \frac{2*(3t)+1}{3t} = \frac{4t+1}{4t}


    \frac{6t+1}{3t} = \frac{4t+1}{4t}

    Clear fractions:

    (3t)(4t)(\frac{6t+1}{3t} = \frac{4t+1}{4t})

    (4t)(6t+1) = (3t)(4t+1)

    Distribute:

    24t^2+4t=12t^2+3t

    Isolate your variable:

    12t^2-t=0

    Factor:
    t(12t-1) = 0

    solve for t:
    t = 0 or 12t-1=0

    t=0 or t=\frac{1}{12}
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  5. #5
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    Re: Identity or No Solution

    Quote Originally Posted by mdhafn View Post
    Original Problem:
    2+\frac1{3t} = 1+\frac1{4t}

    Find common denominators:

    \frac{2*(3t)+1}{3t} = \frac{4t+1}{4t}


    \frac{6t+1}{3t} = \frac{4t+1}{4t}

    Clear fractions:

    (3t)(4t)(\frac{6t+1}{3t} = \frac{4t+1}{4t})

    (4t)(6t+1) = (3t)(4t+1)

    Distribute:

    24t^2+4t=12t^2+3t

    Isolate your variable:

    12t^2-t=0

    Factor:
    t(12t-1) = 0

    solve for t:
    t = 0 or 12t-1=0

    t=0 or t=\frac{1}{12}
    0 is not a solution since you'd be dividing by 0 in the original equation.

    @OP: For the equation to be an identity all values of t must work. For no solution then there must be no value of t.
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  6. #6
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    Re: Identity or No Solution



    Isolate your variable:

    img.top { vertical-align: 15%; }

    Subtracting 3t leaves +t on the left.

    12t^2 + t = 0

    t(12t+1)=0

    t = 0 or -1/12

    Drop 0 because the original equation has t in the denominator.
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