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Thread: Simulatenous equations (Quadratic + linear)

  1. #1
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    Simulatenous equations (Quadratic + linear)

    If (2,1) is a solution fo the simultaneous equations,
    x^2 + xy + ay = b
    2ax + 3y = b

    A) Find the value of a and of b. [a=1, b =7]
    B) Find also the other solution [x=-7 and y = 7)

    I have solved both parts, but I face a problem.
    After part a), I substituted values of a and b into the equation to obtain
    x^2 + xy + y = 7 and 2x + 3y = 7

    Substituting x = (7-3y)/2 into the equation x^2 + xy + y = 7
    I obtain the quadratic expression y^2 - 9y + 14 = 0
    Hence, I get y = 7 and y =2 ; x = -7 and x = 1/2

    Why do I reject y =2 and x = 1/2 combination ?
    I know that substituting the above into 2x + 3y = 7 satisfies the equation
    But it doesn't satisfy x^2 + xy + y = 7. How would I know?

    Thanks
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  2. #2
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    Quote Originally Posted by Drdj View Post
    If (2,1) is a solution fo the simultaneous equations,
    x^2 + xy + ay = b
    2ax + 3y = b

    A) Find the value of a and of b. [a=1, b =7]
    B) Find also the other solution [x=-7 and y = 7)

    I have solved both parts, but I face a problem.
    After part a), I substituted values of a and b into the equation to obtain
    x^2 + xy + y = 7 and 2x + 3y = 7

    Substituting x = (7-3y)/2 into the equation x^2 + xy + y = 7
    I obtain the quadratic expression y^2 - 9y + 14 = 0
    Hence, I get y = 7 and y =2 ; x = -7 and x = 1/2

    Why do I reject y =2 and x = 1/2 combination ?
    I know that substituting the above into 2x + 3y = 7 satisfies the equation
    But it doesn't satisfy x^2 + xy + y = 7. How would I know?

    Thanks
    you don't ... extraneous solutions show up when you sub them back into the original equations

    for example ...

    $\displaystyle \sqrt{x+2} = x$

    $\displaystyle x+2 = x^2$

    $\displaystyle 0 = x^2 - x - 2$

    $\displaystyle 0 = (x - 2)(x + 1)$

    $\displaystyle x = 2$ , $\displaystyle x = -1$

    $\displaystyle x = -1$ is an extraneous solution caused by the resulting quadratic ... $\displaystyle \sqrt{whatever} \ge 0$
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Drdj View Post
    If (2,1) is a solution fo the simultaneous equations,
    x^2 + xy + ay = b
    2ax + 3y = b

    A) Find the value of a and of b. [a=1, b =7]
    B) Find also the other solution [x=-7 and y = 7)

    I have solved both parts, but I face a problem.
    After part a), I substituted values of a and b into the equation to obtain
    x^2 + xy + y = 7 and 2x + 3y = 7

    Substituting x = (7-3y)/2 into the equation x^2 + xy + y = 7
    I obtain the quadratic expression y^2 - 9y + 14 = 0
    Hence, I get y = 7 and y =2 ; x = -7 and x = 1/2

    Why do I reject y =2 and x = 1/2 combination ?
    I know that substituting the above into 2x + 3y = 7 satisfies the equation
    But it doesn't satisfy x^2 + xy + y = 7. How would I know?

    Thanks
    There are no extraneous solutions. Your quadratic is incorrect. I get $\displaystyle y^2 - 8y + 7 = 0$

    -Dan
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  4. #4
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    Thanks skeeter ! So, for all quadratic equations, I have to substitute the final answers back, in order to check if both sets of equations are valid! Many thanks for your quick reply !

    Quote Originally Posted by skeeter View Post
    you don't ... extraneous solutions show up when you sub them back into the original equations

    for example ...

    $\displaystyle \sqrt{x+2} = x$

    $\displaystyle x+2 = x^2$

    $\displaystyle 0 = x^2 - x - 2$

    $\displaystyle 0 = (x - 2)(x + 1)$

    $\displaystyle x = 2$ , $\displaystyle x = -1$

    $\displaystyle x = -1$ is an extraneous solution caused by the resulting quadratic ... $\displaystyle \sqrt{whatever} \ge 0$
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  5. #5
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    Hello, Drdj!

    A simple algebra error . . .


    $\displaystyle \text{If }(2,1)\text{ is a solution of the simultaneous equations:}$

    . . $\displaystyle \begin{array}{c} x^2 + xy + ay \;=\;b \\ 2ax + 3y \;=\;b \end{array}$


    $\displaystyle \text{(A) Find the values of }a\text{ and }b.\;\;[a=1,\:b =7]$

    $\displaystyle \text{(B) Find also the other solution. }\;\;[x=-7,\:y = 7)$


    $\displaystyle \text{I have solved both parts, but I face a problem. }$

    $\displaystyle \text{After part (A), I substituted values of }a\text{ and }b\text{ into the equations}$
    . . $\displaystyle \text{to obtain: }\:x^2 + xy + y \:=\: 7\:\text{ and }\:2x + 3y \:=\: 7 $ . Yes!

    $\displaystyle \text{Substituting }x \:=\:\frac{7-3y}{2}\text{ into the equation }x^2 + xy + y \:=\: 7$
    . . $\displaystyle \text{I obtain the quadratic expression: }\:y^2 - 9y + 14 \:=\: 0$ . No


    You should have had: .$\displaystyle \left(\dfrac{7-3y}{2}\right)^2 + \left(\dfrac{7-3y}{2}\right)\!y \,+\, y \;=\;7$

    . . . . . . . . . . . . . . . $\displaystyle \displaystyle \frac{49-42y + 9y^2}{4} + \frac{7y - 3y^2}{2} + y \;=\;7 $

    Multiply by 4: . $\displaystyle 49 - 42y + 9y^2 + 14y - 6y^2 + 4y \;=\;28$

    . . which simplifies to: .$\displaystyle 3y^2 - 24y + 21 \:=\:0$


    Divide by 3: .$\displaystyle y^2 - 8y + 7 \:=\:0$


    Factor: .$\displaystyle (y - 1)(y - 7) \:=\:0$


    Hence, we have: .$\displaystyle y \;=\;1,\:7$

    . . .which yields: .$\displaystyle x \;=\;2,\:\text{-}7$

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