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Math Help - A lot of algebra problems, a little time to do them!

  1. #1
    Newbie Ana-Maria's Avatar
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    Exclamation A lot of algebra problems, a little time to do them!

    Hi everybody!

    I have few problems that needs to be solve by to tomorrow!!! Some of the problems are quite easy, but I just don't seem to get the right answers.

    Here are the problems:

    1) (x-y)(x+y)-(x-y)^2-(x+y)^2

    2) 2/5x^2 - 5/2x(x-2)-3x

    3) (x+4)^2 / x^2-16

    4) 3a/4 + 1/4b(b-a/b) + 1/2b^2

    5) x^2+2x-35 < 0

    6) x-2y+3z=4
    3x+2y-4z=10
    x+y+z=9

    7) x+y= -1
    x*y= -6

    8) Let Y be the number of units of a product and let p be the price per unit of the same product. One of the equations below describes the correspondence between the demand for the product and the price per unit. The other equation describes the correspondence between the supply and the price per unit.

    Y=900-100p
    Y=300+50p

    a) Which of the equations describes the demand situation and which describes the supply situation? Explain why.

    b) Draw the graph of the two equations in the same coordinate system.

    c) Solve the equation system both graphically and algebraically.



    I know theres a lot of question, but I'll appreciate if you could help!

    Thank you!!
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  2. #2
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    Quote Originally Posted by Ana-Maria View Post
    Hi everybody!

    1) (x-y)(x+y)-(x-y)^2-(x+y)^2
    Multiply it out, but keep everything in parantheses!
    (x^2-xy+xy-y^2)-(x^2-xy-xy+y^2)-(x^2+xy+xy+y^2)
    Combine what you can still in parantheses,
    (x^2-y^2)-(x^2-2xy+y^2)-(x^2+2xy+y^2)
    Open parantheses, careful with signs
    x^2-y^2-x^2+2xy-y^2-x^2-2xy-y^2
    Combine (some will die),
    -x^2-y^2=-(x^2+y^2)
    2) 2/5x^2 - 5/2x(x-2)-3x
    Common denominator is: 10x^2(x-2)
    The first fraction becomes,
    4(x-2)/10x^2(x-2)
    The second fraction becomes,
    25x/10x^2(x-2)
    The third fraction becomes,
    3x(10x^2)(x-2)/10x^2(x-2)
    Find the denominators are all the same combine the numerator:
    4(x-2)-25x-3x(10x^2)(x-2)
    Work with third term,
    4(x-2)-25x-30x^3(x-2)
    Open parantheses,
    4x-8-25x-30x^4+60x^3
    Combine,
    -30x^4+60x^3-21x-8
    The answer is,
    -30x^4+60x^3-21x-8/10x^2(x-2)
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  3. #3
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    Quote Originally Posted by Ana-Maria View Post

    3) (x+4)^2 / x^2-16
    Factor the numerator,
    (x+4)(x+4)
    Factor the denominator,
    (x-4)(x+4)
    Kill what is the same,
    (x+4)/(x-4)

    5) x^2+2x-35 < 0
    Factor,
    (x+7)(x-5)<0
    Thus,
    Case 1 (x+7)>0 and (x-5)<0

    OR

    Case 2 (x+7)<0 and (x-5)>0

    Work with the frist case,
    x>-7 and x<5 thus, -7<x<5

    Work with the second case,
    x>-7 and x<5 impossible thus no such numbers.

    The only thing that we are left with is,
    x=(-7,5)

    6) x-2y+3z=4
    3x+2y-4z=10
    x+y+z=9
    1)Add first two equations,
    4x-z=14

    2)Mutliply third equation by 2:
    2x+2y+2z=18

    3)Add first equation with one in 2):
    3x+5z=22

    4)Thus,
    4x-z=14
    3x+5z=22

    5)Multiply first equation in 4) by 5 and add,
    23x=92
    Thus,
    x=4

    6)Substitute x=4 into first equation in 4) and solve for z:
    16-z=14 thus, z=2

    7)Substite x=4 and z=2 into third equation:
    x+y+z=9 thus, 4+y+2=9 thus, y=3

    8)Solution (if exists) must be,
    (x,y,z)=(4,3,2)

    9)Check to confirm it is a solution.

    This is my 29th post!!!
    Last edited by ThePerfectHacker; October 19th 2006 at 06:11 PM.
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  4. #4
    Newbie Ana-Maria's Avatar
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    Thank you so much!!! I just have few questions.

    Quote Originally Posted by Ana-Maria View Post

    2) 2/5x^2 - 5/2x(x-2)-3x
    The second question is that x(x-2)-3x in in the middle of the 5/2.
    so the x(x-2)-3x in no denumerator, what should I do then?

    Quote Originally Posted by Ana-Maria View Post

    5) x^2+2x-35 < 0
    The answer that you gave me in the question 5, does it mean that you have two ways to do it ? or is there just one answer?


    There is just two more questions


    Thanks again!
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  5. #5
    Senior Member OReilly's Avatar
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    Quote Originally Posted by Ana-Maria View Post
    The answer that you gave me in the question 5, does it mean that you have two ways to do it ? or is there just one answer?
    Thanks again!

    ThePerfectHacker answered you one answer.

    (x+7)(x-5)<0 can be negative (less then 0) only if one multiple is negative and other positive.

    So you have two cases as ThePerfectHacker told you when result can be negative.

    Case 1 (x+7)>0 and (x-5)<0

    OR

    Case 2 (x+7)<0 and (x-5)>0


    Simplest way to explain is:
    a*b < 0 if and only if
    a>0 and b<0
    or
    a<0 and b>0

    Substitute a=(x+7) and b=(x-5).
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  6. #6
    Senior Member OReilly's Avatar
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    Quote Originally Posted by Ana-Maria View Post
    7) x+y= -1
    x*y= -6

    From first equation we have x=-1-y

    Substite x in second equation:
    (-1-y)*y=-6
    -y^2-y=-6
    -y^2-y+6=0 {*-1}
    y^2+y-6=0

    (y-2)(y+3)=0

    y=2 and y=-3

    Substitue both y values in first equation to get x.
    x+2=-1 --> x=-3

    x-3=-1 --> x=2


    So you have solutions
    x=-3,y=2
    and
    x=2, y=-3
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  7. #7
    Newbie Ana-Maria's Avatar
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    Thank you for your help!!
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  8. #8
    Senior Member OReilly's Avatar
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    Quote Originally Posted by Ana-Maria View Post

    8) Let Y be the number of units of a product and let p be the price per unit of the same product. One of the equations below describes the correspondence between the demand for the product and the price per unit. The other equation describes the correspondence between the supply and the price per unit.

    Y=900-100p
    Y=300+50p

    a) Which of the equations describes the demand situation and which describes the supply situation? Explain why.

    b) Draw the graph of the two equations in the same coordinate system.

    c) Solve the equation system both graphically and algebraically.

    b) and c)
    Algebraically, system solution is:
    900-100p=300+50p
    600=150p
    p=4

    Then Y = 500.

    Graphically system is shown in attached picture.

    Blue line represents Y=900-100p and red line represents Y=300+50p.

    a) Y=900-100p (blue line) represents demand because when you increase price p you see that Y is getting less which means that higher price means lower selling (lower demand). Y=300+50p (red line) represents supply.
    Attached Thumbnails Attached Thumbnails A lot of algebra problems, a little time to do them!-image1.gif  
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  9. #9
    Newbie Ana-Maria's Avatar
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    Quote Originally Posted by OReilly View Post
    b) and c)
    Algebraically, system solution is:
    900-100p=300+50p
    600=150p
    p=4

    Then Y = 500.
    I understand the first part, but how did you get Y=500?
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  10. #10
    Senior Member OReilly's Avatar
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    Quote Originally Posted by Ana-Maria View Post
    I understand the first part, but how did you get Y=500?
    Substite p in one of equations with 4 and you will get Y=500.
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  11. #11
    Newbie Ana-Maria's Avatar
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    I got it !

    Do you know answer to the question 2 & 4 ?


    Thank you so much !
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  12. #12
    Senior Member OReilly's Avatar
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    Quote Originally Posted by Ana-Maria View Post
    The second question is that x(x-2)-3x in in the middle of the 5/2.
    so the x(x-2)-3x in no denumerator, what should I do then?
    I am not sure what is the second question.

    Since LaTex is working now can you type it again?

    Is maybe this question 2)?

    \frac{2}{5}x^2  - \frac{5}{2}x(x - 2) - 3x
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  13. #13
    Senior Member OReilly's Avatar
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    Quote Originally Posted by Ana-Maria View Post
    I got it !

    Do you know answer to the question 2 & 4 ?


    Thank you so much !
    Also, for question 4) please use parenthesses or even better use LaTex to show it.
    I don't see exactly what is what.
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  14. #14
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    Quote Originally Posted by Ana-Maria View Post
    I understand the first part, but how did you get Y=500?
    Hi,

    I've modified OReilly's diagram.

    You can get the values of p and Y directly from the graph:
    Attached Thumbnails Attached Thumbnails A lot of algebra problems, a little time to do them!-graph_loes.gif  
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