# 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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