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Math Help - word problems

  1. #1
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    word problems

    Hi all, i'm after a bit of advice, problem is i'm not that great with word problems i am stuck on a couple and i am wodering if anyone has any tips for solving word problems and forming equations from them?? Here is one i'm a bit stuck on.

    The cost of an anorak rose by 6. As a result a shop could buy 5 fewer anoraks for 600. If the cost of the anorak was x before the rise, find expressions, in terms of x,for the number of anoraks which could be bought before and after the rise. Hence form an equation in x and show that it reduces to x^2+ 6x-720=0. solve this equation and state the original cost of the anorak.

    can somebody tell me how i should go about doing these??

    Thanks,
    Chris.
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  2. #2
    Senior Member DivideBy0's Avatar
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    We know x is the original cost per anorak.
    Let a be the amount of anoraks purchased.
    Let T = total cost

    From this we have

    ax=T

    Now, originally, you would have:

    ax=600

    But now, cost has risen by six, so amount must decrease by five. Remember that the new cost * new amount still = 600.

    (a-5)(x+6)=600

    But what is a? Well, we can put a in terms of x.

    ax=600

    a=\frac{600}{x}

    So we have the equation:
    \left(\frac{600}{x}-5 \right)(x+6)=600

    Hopefully you can take it from there
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  3. #3
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    Hello thanks, i see what your done. How would i go about solving the final equation at the bottom for the price of the anorak? I played around and got 24 i think thats right but it was trial and error. So as i say how would i solve the equation at the bottom properly?

    Thankyou for your help

    Chris.
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  4. #4
    Senior Member DivideBy0's Avatar
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    \left( \frac{600}{x}-5 \right)(x+6)=600

    600+\frac{3600}{x}-5x-30=600

    -5x-30+\frac{3600}{x}=0

    -5x^2-30x+3600=0

    x^2+6x-720=0

    You have to find two numbers with multiply to give -720 and add to give 6. Since -720 is negative, you will need a negative and a positive number. What you could do is list the factors of 720:
    (1,720), (2,360), (3,240), (4,180), (5,144), (6,120), (8,90), (9,80), (10,72), (12,60), (15,48), (16,45), (18,40), (20,36), (24,30)
    And test them to see if they fit the requirements. Here, 24 and 30 do.

    (x-24)(x+30)=0

    x-24=0 or x+30=0 (Null Factor Law)

    x=24 or x=-30


    Of course this is a very long and hard way to do the problem (especially for 720, as it has loads of factors), but there are easier ways.

    Another way is 'completing the square':

    x^2+6x-720=0

    x^2+6x+(\frac{6}{2})^2-720=(\frac{6}{2})^2 (add the square of half the coefficient of the x-term to both sides)

    x^2+6x+9=729 (notice the left hand side is a perfect square)

    (x+3)^2=729 (factorise the LHS)

    x+3=\pm 27

    x=-3 \pm 27 = -30,24

    But perhaps the easiest way for this problem, is to plug the values into the quadratic formula. For any equation ax^2+bx+c=0, we have

    x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}

    In this case we have a = 1, b = 6, c = -720

    x=\frac{-(6) \pm \sqrt{(6)^2-4(1)(-720)}}{2(1)}

    x=\frac{-6 \pm \sqrt{2916}}{2}

    x=\frac{-6 \pm 54}{2}

    x=-3 \pm 27=24,-30
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