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Solve-for-x-a-ax-1-b-bx-1-a-b-x-1-a-1-b-

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Question Number 179284 by Agnibhoo98 last updated on 27/Oct/22
Solve for x (a/(ax − 1)) + (b/(bx − 1)) = a + b [x ≠ (1/a), (1/b)]
Solveforxaax−1+bbx−1=a+b[x≠1a,1b]
Answered by FelipeLz last updated on 27/Oct/22
((a(bx−1)+b(ax−1))/((ax−1)(bx−1))) = a+b ((2abx−(a+b))/(abx^2 −(a+b)x+1)) = a+b { ((a+b = s)),((ab = p)) :} ((2px−s)/(px^2 −sx+1)) = s psx^2 −(2p+s^2 )x+2s = 0 ⇒ x = ((2p+s^2 ±(√([−(2p+s^2 )]^2 −8ps^2 )))/(2ps)) x = ((2p+s^2 ±(√(4p^2 −4ps^2 +s^4 )))/(2ps)) x = ((2p+s^2 ±(2p−s^2 ))/(2ps)) { ((x_1 = ((2p+s^2 +2p−s^2 )/(2ps)) = ((4p)/(2ps)) = (2/s) = (2/(a+b)))),((x_2 = ((2p+s^2 −2p+s^2 )/(2ps)) = ((2s^2 )/(2ps)) = (s/p) = ((a+b)/(ab)))) :}
a(bx−1)+b(ax−1)(ax−1)(bx−1)=a+b2abx−(a+b)abx2−(a+b)x+1=a+b{a+b=sab=p2px−spx2−sx+1=spsx2−(2p+s2)x+2s=0⇒x=2p+s2±[−(2p+s2)]2−8ps22psx=2p+s2±4p2−4ps2+s42psx=2p+s2±(2p−s2)2ps{x1=2p+s2+2p−s22ps=4p2ps=2s=2a+bx2=2p+s2−2p+s22ps=2s22ps=sp=a+bab
Answered by som(math1967) last updated on 28/Oct/22
(a/(ax−1)) −b=a−(b/(bx−1)) ⇒((a+b−abx)/(ax−1))=((abx−a−b)/(bx−1)) ⇒(((a+b−abx))/(ax−1)) +((a+b−abx)/(bx−1))=0 ⇒(a+b−abx)((1/(ax−1))+(1/(bx−1)))=0 either (a+b−abx)=0 ∴ x=(1/a)+(1/b) or (1/(ax−1))+(1/(bx−1))=0 ⇒ bx−1+ax−1=0 ∴x=(2/(a+b))
aax−1−b=a−bbx−1⇒a+b−abxax−1=abx−a−bbx−1⇒(a+b−abx)ax−1+a+b−abxbx−1=0⇒(a+b−abx)(1ax−1+1bx−1)=0either(a+b−abx)=0∴x=1a+1bor1ax−1+1bx−1=0⇒bx−1+ax−1=0∴x=2a+b

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