algebra

solve the elimination method 7r-3s=0

3r+7s=58

what is the solution of the system

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  1. From the first equation, we know that r and s are in the ratio 3:7.
    Trial substitution of 3 and 7 into the second equation gives 58, which means that 3 and 7 are the solutions for r and s respectively.

    By elimination:
    7r-3s=0...(1)
    3r+7s=58...(2)

    Use linear combination to eliminate r: 7(2)-3(1)
    21r+49s-(21r-9s)=58*7
    58s=58*7
    s=7
    Substitute s=3 in (1)
    7r-3(7)=0
    r=3(7)/7=3
    this gives
    r=3, s=7.

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  2. -rs^3-r^3s=

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