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    Solve simultaneously
    xy=1
    3x+2y=5

    i've done 5xy=3x+2y so they both equal 5
    don't know what to do now
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    (Original post by AsianG99)
    Solve simultaneously
    xy=1
    3x+2y=5
    You need to get an equation in one variable in order to solve it. What methods have you been taught to do that?
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    (Original post by Plagioclase)
    You need to get an equation in one variable in order to solve it. What methods have you been taught to do that?
    just updated it
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    (Original post by AsianG99)
    just updated it
    Okay, so what you've done is technically correct but it's not really helping you because you've still got an equation in two unknowns, which you can't solve. You need to get down to an equation in one unknown, which you can solve.

    I'll give you a clue: try to find an expression for y in terms of x, and substitute it into the other equation.
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    (Original post by Plagioclase)
    You need to get an equation in one variable in order to solve it. What methods have you been taught to do that?
    just solved it
    got x=1 and y=1
    i did x=0 y=5/2
    y=0 x= 5/3

    3/5x+2/5y=xy
    3/5*5/3= 1 for x
    5/2*2/5= 1 for y
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    (Original post by Plagioclase)
    Okay, so what you've done is technically correct but it's not really helping you because you've still got an equation in two unknowns, which you can't solve. You need to get down to an equation in one unknown, which you can solve.

    I'll give you a clue: try to find an expression for y in terms of x, and substitute it into the other equation.
    Find set of values for k where x^2-kx+k has no real roots
    I've done
    -k^2 - 4*1*k < 0
    so -k^2 -4k < 0

    4k < k^2
    2k < k

    i don't get this question at all
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    (Original post by AsianG99)
    just solved it
    got x=1 and y=1
    i did x=0 y=5/2
    y=0 x= 5/3

    3/5x+2/5y=xy
    3/5*5/3= 1 for x
    5/2*2/5= 1 for y
    That's one set of answers, but there are actually two sets of (x,y) that fulfil this equation. Again, I recommend that you find an expression for y in terms of x, substitute it into the other equation, and solve it. I'll do the first couple of steps.

     y = \frac{1}{x}
    3x + 2 \left( \frac{1}{x} \right) = 5
    3x + \frac{2}{x} = 5
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    (Original post by AsianG99)
    Find set of values for k where x^2-kx+k has no real roots
    I've done
    -k^2 - 4*1*k < 0
    so -k^2 -4k < 0

    4k < k^2
    2k < k

    i don't get this question at all
    Firstly, it should be k^2 -4k &lt; 0 because (-k)^2 = k^2. To get further, factorise that quadratic.
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    (Original post by Plagioclase)
    That's one set of answers, but there are actually two sets of (x,y) that fulfil this equation. Again, I recommend that you find an expression for y in terms of x, substitute it into the other equation, and solve it. I'll do the first couple of steps.

     y = \frac{1}{x}
    3x + 2 \left( \frac{1}{x} \right) = 5
    3x + \frac{2}{x} = 5
    converted it into a quadratic and got an x value of 2/3
    y=1
    so they both equal 5
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    (Original post by Plagioclase)
    Firstly, it should be k^2 -4k &lt; 0 because (-k)^2 = k^2. To get further, factorise that quadratic.
    k< +-2
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    (Original post by AsianG99)
    converted it into a quadratic and got an x value of 2/3
    y=1
    so they both equal 5
    If x = 2/3, y is not equal to 1!
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    (Original post by AsianG99)
    k< +-2
    I'm not really sure how you got there... please try factorising the expression. It is the only way you are going to get to an answer.
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    (Original post by Plagioclase)
    If x = 2/3, y is not equal to 1!
    3x+2y=5
    3x=2
    2y=3
    y=3/2
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    (Original post by Plagioclase)
    I'm not really sure how you got there... please try factorising the expression. It is the only way you are going to get to an answer.
    crapppp i did d.o.t.s by accident
    k(k-4)<0
    k<0 or k<4
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    (Original post by AsianG99)
    3x+2y=5
    3x=2
    2y=3
    y=3/2
    Yep exactly, or even easier just use the fact that y = \frac{1}{x} \rightarrow \frac{1}{\frac{2}{3}} = \frac{3}{2}
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    (Original post by Plagioclase)
    Yep exactly, or even easier just use the fact that y = \frac{1}{x} \rightarrow \frac{1}{\frac{2}{3}} = \frac{3}{2}
    Thank You!!!!!!
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    (Original post by AsianG99)
    crapppp i did d.o.t.s by accident
    k(k-4)<0
    k<0 or k<4
    Nearly. If the question were k(k-4) = 0 then you'd be right, k = 0, 4. However, we've got the inequality. According to your answer, -1 should fulfil the equation but if you plug -1 into your expression for k, you will see that it's greater than 0 so therefore the constraint k<0 is wrong.

    Instead, try sketching the graph of k(k-4) (using the roots you've already found) and find where that is less than 0.
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    (Original post by Plagioclase)
    Nearly. If the question were k(k-4) = 0 then you'd be right, k = 0, 4. However, we've got the inequality. According to your answer, -1 should fulfil the equation but if you plug -1 into your expression for k, you will see that it's greater than 0 so therefore the constraint k<0 is wrong.

    Instead, try sketching the graph of k(k-4) (using the roots you've already found) and find where that is less than 0.
    would it have to be inbetween
    0<k<4
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    (Original post by AsianG99)
    would it have to be inbetween
    0<k<4
    Yep! Well done.
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    (Original post by Plagioclase)
    Yep! Well done.
    i have 1 last question
 
 
 
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