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ax-by-7-ax-2-by-2-49-ax-3-by-3-133-ax-4-by-4-406-2014x-2014y-100a-100b-2014xy-




Question Number 157664 by cortano last updated on 26/Oct/21
  { ((ax+by=7)),((ax^2 +by^2 =49)),((ax^3 +by^3 =133)),((ax^4 +by^4 =406)) :}   ⇒2014x+2014y−100a−100b−2014xy=?
$$\:\begin{cases}{{ax}+{by}=\mathrm{7}}\\{{ax}^{\mathrm{2}} +{by}^{\mathrm{2}} =\mathrm{49}}\\{{ax}^{\mathrm{3}} +{by}^{\mathrm{3}} =\mathrm{133}}\\{{ax}^{\mathrm{4}} +{by}^{\mathrm{4}} =\mathrm{406}}\end{cases} \\ $$$$\:\Rightarrow\mathrm{2014}{x}+\mathrm{2014}{y}−\mathrm{100}{a}−\mathrm{100}{b}−\mathrm{2014}{xy}=? \\ $$
Commented by mr W last updated on 26/Oct/21
x+y=((7×406−49×133)/(7×133−49^2 ))=(5/2)  xy=((49×406−133^2 )/(7×133−49^2 ))=−(3/2)  a+b=((7×(5/2)−49)/(−(3/2)))=21     2014x+2014y−100a−100b−2014xy   =2014(x+y−xy)−100(a+b)   =2014((5/2)+(3/2))−100×21   =2014×4−100×21   =5956
$${x}+{y}=\frac{\mathrm{7}×\mathrm{406}−\mathrm{49}×\mathrm{133}}{\mathrm{7}×\mathrm{133}−\mathrm{49}^{\mathrm{2}} }=\frac{\mathrm{5}}{\mathrm{2}} \\ $$$${xy}=\frac{\mathrm{49}×\mathrm{406}−\mathrm{133}^{\mathrm{2}} }{\mathrm{7}×\mathrm{133}−\mathrm{49}^{\mathrm{2}} }=−\frac{\mathrm{3}}{\mathrm{2}} \\ $$$${a}+{b}=\frac{\mathrm{7}×\frac{\mathrm{5}}{\mathrm{2}}−\mathrm{49}}{−\frac{\mathrm{3}}{\mathrm{2}}}=\mathrm{21} \\ $$$$ \\ $$$$\:\mathrm{2014}{x}+\mathrm{2014}{y}−\mathrm{100}{a}−\mathrm{100}{b}−\mathrm{2014}{xy} \\ $$$$\:=\mathrm{2014}\left({x}+{y}−{xy}\right)−\mathrm{100}\left({a}+{b}\right) \\ $$$$\:=\mathrm{2014}\left(\frac{\mathrm{5}}{\mathrm{2}}+\frac{\mathrm{3}}{\mathrm{2}}\right)−\mathrm{100}×\mathrm{21} \\ $$$$\:=\mathrm{2014}×\mathrm{4}−\mathrm{100}×\mathrm{21} \\ $$$$\:=\mathrm{5956} \\ $$
Commented by mr W last updated on 26/Oct/21
for details see Q87533
$${for}\:{details}\:{see}\:{Q}\mathrm{87533} \\ $$
Commented by cortano last updated on 26/Oct/21
amazing
$${amazing}\: \\ $$

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