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Let-R-fixed-Solve-for-real-numbers-ax-by-2-1-ax-2-by-2-4-1-ax-3-by-3-8-1-ax-4-by-4-16-1-




Question Number 150646 by mathdanisur last updated on 14/Aug/21
Let 𝛌∈R fixed.Solve for real numbers:   { ((ax + by = 2λ + 1)),((ax^2  + by^2  = 4λ + 1)),((ax^3  + by^3  = 8λ + 1)),((ax^4  + by^4  = 16λ + 1)) :}
$$\boldsymbol{\mathrm{L}}\mathrm{et}\:\boldsymbol{\lambda}\in\mathbb{R}\:\mathrm{fixed}.\boldsymbol{\mathrm{S}}\mathrm{olve}\:\mathrm{for}\:\mathrm{real}\:\mathrm{numbers}: \\ $$$$\begin{cases}{\mathrm{ax}\:+\:\mathrm{by}\:=\:\mathrm{2}\lambda\:+\:\mathrm{1}}\\{\mathrm{ax}^{\mathrm{2}} \:+\:\mathrm{by}^{\mathrm{2}} \:=\:\mathrm{4}\lambda\:+\:\mathrm{1}}\\{\mathrm{ax}^{\mathrm{3}} \:+\:\mathrm{by}^{\mathrm{3}} \:=\:\mathrm{8}\lambda\:+\:\mathrm{1}}\\{\mathrm{ax}^{\mathrm{4}} \:+\:\mathrm{by}^{\mathrm{4}} \:=\:\mathrm{16}\lambda\:+\:\mathrm{1}}\end{cases} \\ $$
Answered by MJS_new last updated on 14/Aug/21
looks complicated but indeed is very simple  (1) solve the 1^(st)  for b  (2) solve the 2^(nd)  for a  (3) solve the 3^(rd)  for y (it′s linear!)  (4) solve the 4^(th)  for x       (x^2 −3x+2=0 !)  ⇒  a=b=1  x=1∧y=2 ∨ x=2∧y=1
$$\mathrm{looks}\:\mathrm{complicated}\:\mathrm{but}\:\mathrm{indeed}\:\mathrm{is}\:\mathrm{very}\:\mathrm{simple} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{solve}\:\mathrm{the}\:\mathrm{1}^{\mathrm{st}} \:\mathrm{for}\:{b} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{solve}\:\mathrm{the}\:\mathrm{2}^{\mathrm{nd}} \:\mathrm{for}\:{a} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{solve}\:\mathrm{the}\:\mathrm{3}^{\mathrm{rd}} \:\mathrm{for}\:{y}\:\left(\mathrm{it}'\mathrm{s}\:\mathrm{linear}!\right) \\ $$$$\left(\mathrm{4}\right)\:\mathrm{solve}\:\mathrm{the}\:\mathrm{4}^{\mathrm{th}} \:\mathrm{for}\:{x} \\ $$$$\:\:\:\:\:\left({x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{2}=\mathrm{0}\:!\right) \\ $$$$\Rightarrow \\ $$$${a}={b}=\mathrm{1} \\ $$$${x}=\mathrm{1}\wedge{y}=\mathrm{2}\:\vee\:{x}=\mathrm{2}\wedge{y}=\mathrm{1}\: \\ $$
Commented by mathdanisur last updated on 14/Aug/21
Thank you Ser
$$\mathrm{Thank}\:\mathrm{you}\:\boldsymbol{\mathrm{S}}\mathrm{er} \\ $$

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