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Math Help - Solving Systems

  1. #16
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    Hi\ Vuong,

    You\ may\ consider\ simple\ substitutions\ also\ as\ follows..

    3x-2y=10,\ hence\ 3x=2y+10,\ x=\frac{2y+10}{3}

    This\ x\ may\ now\ be\ used\ to\ solve\ for\ y\ in\ the\ 2nd\ equation

    5x+3y=15,\ hence\ \frac{10y+50}{3}+3y=15

    10y+50+9y=45,\ 19y=45-50=-5,\ y=-\frac{5}{19}

    Now\ use\ this\ y\ to\ find\ x.

    You\ may\ use\ cancellation\ or\ substitution\ methods.

    The\ second\ set\ of\ equations\ is\ just\ a\ single\ equation.
    Last edited by Archie Meade; December 30th 2009 at 01:21 AM.
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  2. #17
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    Now how about the problem

    3x-5y+10=0
    -9x+15y=-30
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  3. #18
    Super Member Bacterius's Avatar
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    Quote Originally Posted by Vuong View Post
    Now how about the problem

    3x-5y+10=0
    -9x+15y=-30
    Can't you solve it after all the different ways we showed you ?

    Here is my hint : in the second equation, try to express 9x in terms of y, and substitute into the first equation to find x. Then substitute the value of x you found into any of the two equations to find y.
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  4. #19
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    Hi\ Vuong,

    Pairs\ of\ linear\ simultaneous\ equations\ in\ two\ variables\ are\ line\ equations.

    This\ time\ your\ two\ lines\ are\ parallel.

    Parallel\ lines\ never\ intersect.

    The\ solution\ of\ a\ pair\ of\ such\ equations\ is\ the\ point\ of\ intersection.

    There\ is\ no\ point\ of\ intersection\ for\ these\ two\ lines.

    Your\ equations\ may\ be\ re-written

    9x-15y+30=0

    9x-15y-30=0

    or,

    y=\frac{3}{5}x+2

    y=\frac{3}{5}x-2

    The\ lines\ have\ the\ same\ slope\ but\ cross\ the\ y-axis\ at\ 2\ and\ -2.

    Cancellation\ and\ substitution\ yield\ no\ solutions.
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