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Thread: If this = 0 then that = ?

  1. #1
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    If this = 0 then that = ?

    Hello.

    I'll get right to the point:

    IF $\displaystyle (2a-b-3)^2+(3a+b-7)^2=0$ THEN $\displaystyle 3a-7b=?$

    The solution is $\displaystyle -1$, but I need to know how they got that so I can tackle problems like this in the future by myself.
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  2. #2
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    Quote Originally Posted by ikislav View Post
    Hello.

    I'll get right to the point:

    IF $\displaystyle (2a-b-3)^2+(3a+b-7)^2=0$ THEN $\displaystyle 3a-7b=?$

    The solution is $\displaystyle -1$, but I need to know how they got that so I can tackle problems like this in the future by myself.
    note that $\displaystyle (anything)^2 \ge 0$

    so ... the only way that equation will equal zero is if

    $\displaystyle 2a - b - 3 = 0$ and $\displaystyle 3a + b - 7 = 0$

    ... take it from here?
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  3. #3
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    Take $\displaystyle (2a-b-3)^2$ from both sides, square root both sides and then isolate $\displaystyle 3a-7b$ to one side.
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  4. #4
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    $\displaystyle 2a-b-3=0$

    $\displaystyle -b=-2a+3$

    $\displaystyle b=2a-3$


    $\displaystyle 3a+b-7=0$

    $\displaystyle 3a+2a-3-7=0$

    $\displaystyle a=\frac{10}{5}$

    $\displaystyle a=2$


    $\displaystyle 2a-b-3=0$

    $\displaystyle 4-b-3=0$

    $\displaystyle -b=-4+3$

    $\displaystyle -b=-1$

    $\displaystyle b=1$


    $\displaystyle a=2$

    $\displaystyle b=1$


    $\displaystyle 3a-7b=?$

    $\displaystyle 3*2-7*1=?$

    $\displaystyle 6-7=-1$

    $\displaystyle 3a-7b=-1$

    Yes, that is the correct answer. Is that the right path for solving this after the hint you gave me?
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  5. #5
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    solving the system using substitution is fine ... elimination would also work

    $\displaystyle 2a-b-3 = 0$

    $\displaystyle 3a+b-7 = 0$
    ------------------
    $\displaystyle 5a - 10 = 0$

    $\displaystyle a = 2$ , and so on
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  6. #6
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    Quote Originally Posted by pickslides View Post
    Take $\displaystyle (2a-b-3)^2$ from both sides, square root both sides and then isolate $\displaystyle 3a-7b$ to one side.
    Can you show me this method please? I'm very interrested.
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  7. #7
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    Quote Originally Posted by ikislav View Post
    Can you show me this method please? I'm very interrested.
    You do not need to that.
    Look carefully at replies #2 & #5. Follow those.
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  8. #8
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    Yes, I already solved it, but I am stil interested in method mentioned by mr. pickslides.
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  9. #9
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    Quote Originally Posted by pickslides View Post
    Take $\displaystyle (2a-b-3)^2$ from both sides, square root both sides and then isolate $\displaystyle 3a-7b$ to one side.
    $\displaystyle \displaystyle (2a - b - 3)^2 + (3a + b - 7)^2 = 0$

    $\displaystyle \displaystyle (3a + b - 7)^2 = -(2a - b - 3)^2$
    And you can't take the square root. (Pesky little negative sign!)

    -Dan
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