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Thread: 9 ?'s that i really need help on!!!

  1. #1
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    9 ?'s that i really need help on!!!

    questions are on attachment!
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  2. #2
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    #1:

    Let x=amount invested at 10%.

    Then 3500-x is the amount invested at 15%.

    The difference between the two interest rates is $85.

    $\displaystyle .15(3500-x)-.10x=85$

    Also, how about showing some of your attempts instead of just posting a littany of problems.
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  3. #3
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    Hello, Lane!

    Here are a few more . . .


    Solve: .$\displaystyle \begin{array}{cc}3x - 4y \,= \,7 \\ x + 2y \,=\,9\end{array}$

    With no method specified, I'll use Elimination.

    We have: .$\displaystyle \begin{array}{cc}3x - 4y \:= \:7 \\ x + 2y \:=\:9\end{array}\;\begin{array}{cc}(1)\\(2)\end{a rray}$

    Multiply (2) by 2: .$\displaystyle 2x + 4y \:=\:18$
    . . . . . . Add (1): . $\displaystyle 3x - 4y \:=\:7$

    And we have: .$\displaystyle 5x\,=\,25\quad\Rightarrow\quad \boxed{x = 5}$

    Substitute into (2): .$\displaystyle 5 + 2y\:=\:9\quad\Rightarrow\quad 2y \,=\,4\qiad\Rightarrow\quad \boxed{y = 2}$



    Solve with Cramer's Rule: .$\displaystyle \begin{array}{cc}6x - 4y \:=\:6 \\ 3x \:= \:2y - 5\end{array}$

    We have: .$\displaystyle \begin{array}{cc}6x-4y\:=\:7\\3x-2y\:=\:\text{-}5\end{array}$

    $\displaystyle D\:=\:\begin{vmatrix}6 & \text{-}4 \\ 3 & \text{-}2\end{vmatrix} \;= \;(6)(\text{-}2) - (\text{-}4)(3) \;= \;-12 + 12 \;= \;0$ . . . No solution!



    6. Solve: .$\displaystyle \begin{array}{cc}xy\:=\:10 \\ x^2 + 4y^2\:=\:36\end{array}\;\begin{array}{cc}(1)\\(2) \end{array}$

    Another trick question!

    From (1), we have: .$\displaystyle y = \frac{10}{x}$

    Substitute into (2): .$\displaystyle x^2 + 4\left(\frac{10}{x}\right)^2\;=\;36 \quad\Rightarrow\quad x^2 + \frac{400}{x^2}\:=\:36$

    Multiply by $\displaystyle x^2:\;\;x^4 + 400 \:=\:36x^2\quad\Rightarrow\quad x^4 - 36x^2 + 400 \:= \:0$

    Quadratic Formula: .$\displaystyle x^2\:=\:\frac{-(\text{-}36) \pm \sqrt{(\text{-}36)^2 - 4(1)(400)}}{2(1)}$

    . . . $\displaystyle x^2\;=\;\frac{36 \pm\sqrt{-304}}{2}$ . . . No real root


    Therefore, the system has no real solution.

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