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Math Help - Factorisation/Algebra

  1. #1
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    Factorisation/Algebra

    Gah, i'm beat i give up... i lose, maths wins

    Can someone help me solve these?

    1.) (X-1)/2 - (x-2)/3 = 1

    2.) 2x/3x+1 = 0.85

    3.) (2x - 1)/x = 1

    4.) (x-1)(x+3) = 5

    5.) -10x^2 +15

    6.) 1/4 (x+1/2)^2 - 1/4

    Could someone please leave working out and explanations particularly with the factorisation ones. Thanks, appreciate it.
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  2. #2
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    Quote Originally Posted by andrew2322 View Post
    Can someone help me solve these?

    1.) (X-1)/2 - (x-2)/3 = 1

    2.) 2x/3x+1 = 0.85

    3.) (2x - 1)/x = 1

    4.) (x-1)(x+3) = 5

    5.) -10x^2 +15

    6.) 1/4 (x+1/2)^2 - 1/4
    to #1)

    multiply the equation by 6:
    3(x-1) - 2(x-2) = 6 ...... Expand the brackets, collect like terms, separate x at the LHS, constants to the RHS, I've got x = 5

    to #2)

    I assume that you mean
    \frac{2x}{3x+1}=0.85 ...... Muliply by (3x+1), expand the bracket at the RHS, I've got x = -\frac{17}{11}

    to #3)
    multiply by x, collect the x at the RHS, divide by the coefficient of x (by the way, it's (-1)), I've got x = 1

    to #4)

    expand the brackets, collect like terms, solve the quadratic for x:

    x^2+2x-8=0~\iff~(x-2)(x+4)=0...... A product equals zero if one factor equals zero. Thus:
    x-2=0~\vee~x+4=0~\iff~x=2~\vee~x=-4

    to #5

    there is nothing to solve because this is only a term. (I assume that there is a typo)

    to #6)

    \frac14 \cdot \left(x+\frac12\right)^2-\frac14 = \frac14 \cdot \left(\left(x+\frac12\right)^2-1\right) = \frac14 \cdot \left(\left(x+\frac12+1\right)\left(x+\frac12-1\right)\right)  = \frac14 \cdot \left(\left(x+\frac32\right)\left(x-\frac12\right)\right) =
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  3. #3
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    Hey thanks for helping out

    Question 5 is not a typo, Could you please explain on question 6, im still lost
    Last edited by andrew2322; February 7th 2008 at 01:10 AM.
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  4. #4
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    Sorry i have another quick question

    hey sorry to bother you, but i really want to learn.

    consider: 0.01(2y+0.6)^2 - 0.04

    I took out the common factor here as 0.01 hence

    0.01 (2y+0.6) ^ 2 - 4
    0.01 (2y+0.6)^2 -(2)^2
    0.01 (2y + 0.6-2)(2y+0.6+2)
    0.01(2y-1.4)(2y+2.6)

    i checked the answer and it appeared as 0.04(y-0.7)(y+1.3)

    I realised it could possibly due to the simplification of the two 2y's but on other problems when i removed 1 '2y' the other side got the 2 chopped off as well which leads me to another answer 0.02 (y-0.07)(y+1.3) could you please explain how it works? thanks.
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  5. #5
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    Quote Originally Posted by andrew2322 View Post
    ...
    0.01(2y-1.4)(2y+2.6)

    i checked the answer and it appeared as 0.04(y-0.7)(y+1.3)

    ....
    You've done a great job. Now continue:

    0.01(2y-1.4)(2y+2.6)= 0.01 \cdot \left(2(x-0.7)\right) \left(2(x+1.3) \right) = 4 \cdot 0.01 (x-0.7)(x+1.3)

    which will give you the result you found obviously at the back of your book.
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  6. #6
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    Sorry to bother again

    Consider: (2x+4)(x+6)

    If i were to remove the 2 as the common factor.

    2(x+2)(x+3)

    Why is the second term affected by me removing two as the common factor for the first term?
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