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    hey guys!
    so i can do simultaneous equations fairly well and i always find when i get to the end both factors are often negative. when this happens can i just turn the both positive?? will it still work?
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    (Original post by gertygreengage)
    hey guys!
    so i can do simultaneous equations fairly well and i always find when i get to the end both factors are often negative. when this happens can i just turn the both positive?? will it still work?
    Perhaps an example would help you best.

    If you had (6-x)(3-2x) = 0, then you can take out a factor of -1 from each break to give (x-6) * -1 * -1 * (2x-3) so you also get that (x-6)(2x-3) = 0.

    You can also take out -1 in things like (5-x)(x+3) = 0 to give -1 * (x-5)(x+3) = 0 , then divide both sides by -1 to give (x-5)(x+3) = 0.
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    (Original post by SeanFM)
    Perhaps an example would help you best.

    If you had (6-x)(3-2x) = 0, then you can take out a factor of -1 from each break to give (x-6) * -1 * -1 * (2x-3) so you also get that (x-6)(2x-3) = 0.

    You can also take out -1 in things like (5-x)(x+3) = 0 to give -1 * (x-5)(x+3) = 0 , then divide both sides by -1 to give (x-5)(x+3) = 0.
    oh right!! ok thanks!!
 
 
 
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