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Thread: Solving Systems

  1. #16
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    $\displaystyle Hi\ Vuong,$

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

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

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

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

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

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

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

    $\displaystyle The\ second\ set\ of\ equations\ is\ just\ a\ single\ equation.$
    Last edited by Archie Meade; Dec 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 $\displaystyle 9x$ in terms of $\displaystyle y$, and substitute into the first equation to find $\displaystyle x$. Then substitute the value of $\displaystyle x$ you found into any of the two equations to find $\displaystyle y$.
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  4. #19
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    $\displaystyle Hi\ Vuong,$

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

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

    $\displaystyle Parallel\ lines\ never\ intersect.$

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

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

    $\displaystyle Your\ equations\ may\ be\ re-written$

    $\displaystyle 9x-15y+30=0$

    $\displaystyle 9x-15y-30=0$

    $\displaystyle or,$

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

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

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

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