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Thread: equation

  1. #1
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    Unhappy equation

    hey i have yet another maths question... i tried looking for the solution but the site for the exam papers is down. the question is solve:

    for x and y
    x + y = 7

    x^2+y^2=29

    the answer is (2,5) and (5,2) but i got (10,0) and (0,10)


    i have ALOT of difficulty with these sums so i would be really grateful for the solution to do it , thanks




    then it asks in part (2) " which one of the values of y in the above satisfies the inequality 6-2y<0
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  2. #2
    A riddle wrapped in an enigma
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    Quote Originally Posted by cliste09 View Post
    hey i have yet another maths question... i tried looking for the solution but the site for the exam papers is down. the question is solve:

    for x and y
    x + y = 7

    x^2+y^2=29

    the answer is (2,5) and (5,2) but i got (10,0) and (0,10)


    i have ALOT of difficulty with these sums so i would be really grateful for the solution to do it , thanks




    then it asks in part (2) " which one of the values of y in the above satisfies the inequality 6-2y<0
    Hi cliste09,

    (1) $\displaystyle x+y=7$
    (2) $\displaystyle x^2+y^2=29$

    Solve the (1) equation for x
    (1)$\displaystyle x=7-y$

    Substitute into (2)

    $\displaystyle (7-y)^2+y^2=29$
    $\displaystyle 49-14y+y^2+y^2-29=0$
    $\displaystyle 2y^2-14y+20=0$
    $\displaystyle y^2-7y+10=0$
    $\displaystyle (y-5)(y-2)=0$
    $\displaystyle y=5 \ \ or \ \ y=2$

    Substitute each value of y back into (1) to retrieve your x-values. Your points of intersection are (2,5) and (5, 2).

    As far as part 2 goes, use your y-values of 5 and 2 and see which one makes the inequality true.
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  3. #3
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    Thumbs up

    thank you soooo much!!!!! your solution is very easy to understand , phew these q's are one of my worst areas in maths ... dont even get me started on trigonometry lol
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