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Math Help - Solve two algebra problems

  1. #1
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    Solve two algebra problems

    Please help me to solve these equations:

    2 8
    X = -
    15

    and:
    2
    X x 3x = -X + 4


    Thank you !
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  2. #2
    MHF Contributor
    Grandad's Avatar
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    Explain, please

    Hello Shandy
    Quote Originally Posted by Shandy View Post
    Please help me to solve these equations:

    2 8
    X = -
    15

    and:
    2
    X x 3x = -X + 4


    Thank you !
    I don't understand what you mean. Can you write these more clearly? Use x^2 to mean x^2 if you like, and write the word 'times' to mean multiply, so that we don't think it's a letter x.

    Grandad

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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Shandy View Post
    Please help me to solve these equations:

    2 8
    X = -
    15

    and:
    2
    X x 3x = -X + 4


    Thank you !
    please fix the format of your question. just use / to indicate fractions. example, type 1/x to mean \frac 1x and 1/(x + 1) to mean \frac 1{x + 1} etc, also, use ^ to indicate powers. type x^2 to mean x^2 for instance. you can use * to mean multiply. what you posted is confusing
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  4. #4
    MHF Contributor
    Grandad's Avatar
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    Algebra

    Hello Shandy

    I'm just guessing that your first equation might be

    x^2 = \frac{8}{15}

    If it is, the answers are

    x = \pm \sqrt{\frac{8}{15}}

    So get your calculator working! Divide 8 by 15. Then find the square root. Then put a \pm sign in front, because there will be two answers.

    The second one, I think, might be:

    x \times 3x = -x^2 + 4

    If so, then you solve it like this:

    x \times 3x = -x^2 + 4

    \Rightarrow 3x^2 = -x^2 + 4

    \Rightarrow 4x^2 = 4

    \Rightarrow x^2 = 1

    \Rightarrow x = \pm 1

    How am I doing? Did I guess right?

    Grandad
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  5. #5
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    Quote Originally Posted by Shandy View Post
    Please help me to solve these equations:

    2 8
    X = -
    15

    and:
    2
    X x 3x = -X + 4
    Looking at your post inside "Quote" tags, some of the original formatting is displayed while in message-entry mode. Assuming you are using "X" and "x" to actually be the same variable, your equations appear to be:

    . . . . .x^2 = 8/15

    . . . . .(x)(3x) = -x^2 + 4

    If so, then you can solve the first equation by taking the square root of each side of the equation, remembering the "plus-minus" on the right-hand side. You'll probably need to "rationalize the denominator" to get the answer in the "right" format.

    For the second one, simplify the left-hand side to get 3x^2. Get the variable term together, and then divide 4x^2 = 4 on both sides by 4. Then you can solve by taking the square root of either side.

    Have fun!
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