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Math Help - 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 ...

    \sqrt{x+2} = x

    x+2 = x^2

    0 = x^2 - x - 2

    0 = (x - 2)(x + 1)

    x = 2 , x = -1

    x = -1 is an extraneous solution caused by the resulting quadratic ... \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 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 ...

    \sqrt{x+2} = x

    x+2 = x^2

    0 = x^2 - x - 2

    0 = (x - 2)(x + 1)

    x = 2 , x = -1

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

    A simple algebra error . . .


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

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


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

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


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

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

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


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

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

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

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


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


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


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

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

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