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Math Help - solve by substitution

  1. #1
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    solve by substitution

    once again substitution! can anyone help?

    x/y-y/3=5/6
    x/5-y/4=-13/10

    solve by substitution
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  2. #2
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    Quote Originally Posted by mafai44 View Post
    once again substitution! can anyone help?

    x/y-y/3=5/6
    x/5-y/4=-13/10

    solve by substitution
    \displaystyle\huge\frac{x}{y}-\frac{y}{3}=\frac{5}{6}

    \displaystyle\huge\frac{x}{5}-\frac{y}{4}=-\frac{13}{10}


    \displaystyle\huge\frac{x}{y} is inconvenient, hence multiply all terms on both sides of the first equation by "y".

    \displaystyle\huge\ x-\frac{y^2}{3}=\frac{5y}{6}

    \displaystyle\huge\ x=\frac{y^2}{3}+\frac{5y}{6}

    That's one equation for x.

    \displaystyle\huge\frac{x}{5}=\frac{y}{4}-\frac{13}{10}

    \displaystyle\huge\ x=\frac{5y}{4}-\frac{65}{10}

    x=x, hence equate the two equations in y only.
    Subtract them and the answer is zero, so solve the resulting quadratic in y.
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  3. #3
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    so sorry lol i wrote the equation wrong x/2-y/3=5/6
    x/5-y/4=-13/10
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  4. #4
    Senior Member yeKciM's Avatar
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    Quote Originally Posted by mafai44 View Post
    so sorry lol i wrote the equation wrong x/2-y/3=5/6
    x/5-y/4=-13/10
    same thing as "Archi meade" wrote there, just now with different numbers

    multiply firs with 6 and second with 20

     3x-2y=5 \;\;\;\;\;\;\;\; 1
     4x-5y=-26 \;\;\;\;\;\;\;\; 2

    now i u can go as you wish

    first one multiply with 4 and second with (-3) (and then sum them )

     12x-8y=20 \;\;\;\;\;\;\;\;1
     -12x+15y=78 \;\;\;\;\;\;\;\;2

     7y=98 \Rightarrow y=98/7 = 14

    rest you know




    or u can :
     6x-2y=5 \Rightarrow 6x=5+2y \Rightarrow x = \frac {2y+5}{6} \;\;\;\;\;\;\;\;1
     4x-5y=-26 \;\;\;\;\;\;\;\; 2

    and put in second and solve for "y" and so on for x...

    P.S. you don't need to do that this way (that's up to you, and with practice u can decide better how and what would you do, so that you have a less work, and to keep it simple... don't complicate it to much )


    Edit: corrected that "6" and "20"... i really don't know from where and when did i type that
    Last edited by yeKciM; August 12th 2010 at 01:57 PM.
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  5. #5
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    Typo yeKciM !

    3x-2y=5
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  6. #6
    Senior Member yeKciM's Avatar
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    aaaaah
    thank you , i'll fix it now
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  7. #7
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    if anyone can do it once more i got the last one wrong lol i noticed the edit after i answered it

    x/2-y/3=5/6
    x/5-y/4=-71/10
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  8. #8
    Senior Member yeKciM's Avatar
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    Quote Originally Posted by mafai44 View Post
    if anyone can do it once more i got the last one wrong lol i noticed the edit after i answered it

    x/2-y/3=5/6
    x/5-y/4=-71/10

    here's using substitution :

     \displaystyle \frac {x}{2}- \frac {y}{3} = \frac {5}{6}
     \displaystyle \frac {x}{5}- \frac {y}{4} = -\frac {13}{10}

    first one multiply with (6) and second with (20) and you will get :

     3x-2y=5
     4x-5y=-26

    now from any ( i'll do first ) you can express either "x" or "y" (i'll do "x" )

    \displaystyle  3x-2y=5 \Rightarrow 3x=2y+5 \Rightarrow x= \frac {2y+5}{3}

    now that "x we put in second equation ...

     \displaystyle 4(\frac {2y+5}{3} ) -5y = -26

    now multiply with 3 :

     \displaystyle 4(2y+5) -15y = -78  \Rightarrow 8y+20-15y = -78 \Rightarrow 7y = 98 \Rightarrow y = 14


    now that "y" we will put in first equation

     3x-28 = 5 \Rightarrow 3x = 33 \Rightarrow x=11


    P.S. what ? now second equation is with -71/10 , where did go -13/10 ?
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