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Math Help - solving simultaneously...?

  1. #1
    Newbie suckatmaths's Avatar
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    solving simultaneously...?

    "Solve Simultaneously."

    "2x+3y=8
    3x+2y=7"

    "5x-4y=27
    4x-5y=18"


    I know I have to add/subtract somewhere...But not sure where...

    Can someone help me, please?
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  2. #2
    Like a stone-audioslave ADARSH's Avatar
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    Quote Originally Posted by suckatmaths View Post
    "Solve Simultaneously."

    "2x+3y=8
    3x+2y=7"

    "5x-4y=27
    4x-5y=18"


    I know I have to add/subtract somewhere...But not sure where...

    Can someone help me, please?
    -We are gonna get the value of x from first equation

    -We will put that in second equation

    -This gives an wequation in y

    -Get value of y

    -And thus the value of x

    1st one:

    2x+3y = 8

    2x = 8 - 3y

    Hence

    x= \frac{(8-3y)}{2}..................(1)

    3x+2y =7<br />

    3x = 7 -2y

    Put the value of x as in 1

    3 \frac{(8-3y)}{2}  = 7 -2y

    Hence
    3(8-3y) = 2(7-2y)

    24 - 9 y = 14 - 4y

    Hence

    -5y = - 10

    Thus
    y = \frac{10}{5}= ~2

    And
    x = \frac{8-3y}{2}   = \frac{ 8 -\frac{3\times 10}{5}}{2} =1





    2nd One ;

    5x - 4y = 27

    5x = 27 + 4y

    x =\frac{27+4y}{5} .................1


    4x - 5y = 18

    Put the value of x from 1 here

    \frac{4( 27 +4y)}{5} - 5y = 18

    4(27+4y) = 5(5y+ 18)

    108 + 16y = 25y +90

    Hence
    9y = 18

    Thus y= 2

     <br />
x = \frac{(27+4y)}{5} = 35/5=7    <br />
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