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Math Help - Basic Random algebra Q's

  1. #1
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    Basic Random algebra Q's

    Firstly is (2x-5)(x+3) equal to 2x^2+x-15? Just wanna make sure im getting this right.


    Simultaneous Equations
    Ive forgotten how to do them and the sites ive looked at arent too helpful. Could you explain how to do it with algebra instead of using graphs like ive seen

    x+3y=5
    3x-y=5

    I know x is 2 and y is 1 but how would you work it out.

    Also 2x=3y-2
    3y=1+4x

    What can y^2-1
    ------ be simplfied to?
    y^2-y-2

    What are the solutions to 2x^2-11x+15=0 and 3x^2-2x+4=0?


    Last one! :P

    Simplify (Sqrt)12x(Sqrt)8x(Sqrt)98

    and

    Mulitply (2-(Sqrt)3)(1+3(Sqrt)3)



    If you wouldnt mind explaining how you did them that would save me asking next time
    Thanks a lot.
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  2. #2
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    Quote Originally Posted by Yppolitia View Post
    Firstly is (2x-5)(x+3) equal to 2x^2+x-15? Just wanna make sure im getting this right.
    yes.


    Quote Originally Posted by Yppolitia View Post
    Simultaneous Equations...
    x+3y=5
    3x-y=5
    I know x is 2 and y is 1 but how would you work it out.
    There are a lot of different method to solve such a system of equations. I'll show you the substitution method:

    From the first equation you get: x = 5-3y
    Plug in this value into the 2nd equation: 3(5-3y)-y=5
    Expand to 15-9y-y=5
    you get 15 -10y = 5. Thus -10y = -10. Therefore y = 1
    Re-substitute x = 5-3(1) = 2


    Quote Originally Posted by Yppolitia View Post
    Also 2x=3y-2
    3y=1+4x
    If you use that 4x = 2*2x then your 2nd equation becomes:
    3y = 1+2*(3y-2). Expand
    3y = 1+6y-4
    -3y = -3. therefore y = 1. Re-substitute 2x = 3-2. So x = 1/2




    Quote Originally Posted by Yppolitia View Post
    What can y^2-1
    ------ be simplfied to?
    y^2-y-2
    \frac{y^2-1}{y^2-y-2}=\frac{(y+1)(y-1)}{(y+1)(y-2)}=\frac{y-1}{y-2}


    uff, I need a break. Next explanations will come soon!

    Promised
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Yppolitia View Post
    What are the solutions to 2x^2-11x+15=0 and 3x^2-2x+4=0?
    2x^2-11x+15=0

    You can try factoring, but if you can't figure out how to factor it, there's always the quadratic formula:
    If ax^2 + bx + c = 0 then x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

    So, a = 2, b = -11, c = 15. Thus:
    x = \frac{-(-11) \pm \sqrt{(-11)^2 - 4 \cdot 2 \cdot 15}}{2 \cdot 2} = \frac{11 \pm \sqrt{1}}{4}

    So x = 3 or x = 5/2

    By the way, this means that the quadratic DID factor: (x - 3)(2x - 5) = 2x^2 - 11x +15 (Check!)

    Now for
    3x^2-2x+4=0

    Again, I'll use the quadratic formula:
    x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 3 \cdot 4}}{2 \cdot 3} = \frac{2 \pm \sqrt{-44}}{6} = \frac{2 \pm \sqrt{-11 \cdot 4}}{6}

    x = \frac{2 \pm 2 \sqrt{-11}}{6} = \frac{1 \pm  \sqrt{-11}}{3}

    -Dan
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Yppolitia View Post
    Simplify (Sqrt)12x(Sqrt)8x(Sqrt)98
    I presume this is \sqrt{12} \sqrt{8} \sqrt{98}?

    \sqrt{12} \sqrt{8} \sqrt{98} = \sqrt{3 \cdot 4} \sqrt{2 \cdot 4} \sqrt{2 \cdot 49} = 2 \sqrt{3} \cdot 2 \sqrt{2} \cdot 7 \sqrt{2}

    = (2 \cdot 2 \cdot 7) \sqrt{3} \sqrt{2} \sqrt{2} = 28 \sqrt{3} \sqrt{2} \sqrt{2} <-- Note: \sqrt{2} \cdot \sqrt{2} = 2

    = 28 \cdot 2 \sqrt{3} = 56 \sqrt{3}

    Quote Originally Posted by Yppolitia View Post
    Mulitply (2-(Sqrt)3)(1+3(Sqrt)3)
    (2-\sqrt{3})(1+3 \sqrt{3}) = 2(1+3 \sqrt{3}) - \sqrt{3}(1+3 \sqrt{3})

    = 2 + 6 \sqrt{3} - \sqrt{3} - 3 \sqrt{3} \sqrt{3} = 2 + 6 \sqrt{3} - \sqrt{3} - 3 \cdot 3 = 2 + 6 \sqrt{3} - \sqrt{3} - 9

    = -7 + 5 \sqrt{3}

    -Dan
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