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Question Number 156244 by Eric002 last updated on 09/Oct/21
solve the pairs of simultaneous equations  ax−2y=2  x+3y=3  qy−px=(q^2 −p^2 )/pq  py+qx=2
solvethepairsofsimultaneousequationsax2y=2x+3y=3qypx=(q2p2)/pqpy+qx=2
Answered by MJS_new last updated on 10/Oct/21
(1) x=((12)/(3a+2))∧y=((3a−2)/(3a+2))∧a≠−(2/3)  (2) x=((p^2 +2pq−q^2 )/(2pq^2 ))∧y=−((p^2 −2pq−q^2 )/(2p^2 q))∧p≠0∧q≠0
(1)x=123a+2y=3a23a+2a23(2)x=p2+2pqq22pq2y=p22pqq22p2qp0q0

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