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Thread: Solving equations

  1. #1
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    Solving equations

    I would appreciate any assistance with the following problems.
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  2. #2
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    Hello, pashah!

    Solve each equation and check for extraneous solutions.

    $\displaystyle a)\;\sqrt{w - 3}\:=\:\sqrt{4w + 15}$

    Square: .$\displaystyle w - 3\:=\:4w + 15$

    Then: .$\displaystyle -3w\:=\:18\quad\Rightarrow\quad \boxed{w = -6}$


    But when we check, we get: .$\displaystyle \sqrt{-6 - 3}\:=\:\sqrt{-9}$ . . . an "imaginary" number.

    Therefore, this equation has no solutions.



    $\displaystyle b)\;\sqrt{x^2-6x=2} \:=\:\frac{x}{2}$

    Square: .$\displaystyle x^2-5x + 2\:=\:\frac{x^2}{4}$

    Multiply by 4: .$\displaystyle 4x^2 - 20x + 8\:=\:x^2\quad\Rightarrow\quad 3x^2 - 20x + 8 \:= \:0$

    Now use the Quadratic Formula (and check your answers).



    $\displaystyle \frac{5}{y-3} \;= \;1 + \frac{y+7}{2y-6}$

    We have: .$\displaystyle \frac{5}{y-3} \;= \;1 + \frac{y+7}{2(y-3)}$


    Multiply by the LCD: $\displaystyle 2(y-3)$

    . . $\displaystyle 2(y-3)\cdot\frac{5}{y-3} \;\;= \;\;2(y-3)\cdot1 \;+ \;2(y-3)\cdot\frac{y+7}{2(y-3)}$


    We have: .$\displaystyle 2\cdot5 \;= \;2(y-3) + (y + 7)\quad\Rightarrow\quad 10 \;= \;2y - 6 + y + 7$

    Then: .$\displaystyle 9\:=\:3y\quad\Rightarrow\quad \boxed{y = 3}$


    But when we check, it starts with: .$\displaystyle \frac{5}{3-3} \:= \:\frac{5}{0}$ . . . undefined!

    Therefore, this equation has no solutions.

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