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Question Number 147944 by mathdanisur last updated on 24/Jul/21

if x + y + z = 13 find x, yz^(−) + y, zx^(−) + z, xy^(−) = ?

$${if}\:\:\:{x}\:+\:{y}\:+\:{z}\:=\:\mathrm{13} \\ $$ $${find}\:\:\:\overline {{x},\:{yz}}\:+\:\overline {{y},\:{zx}}\:+\:\overline {{z},\:{xy}}\:=\:? \\ $$

Answered by Rasheed.Sindhi last updated on 25/Jul/21

x+y+z=13 ; x, yz^(−) + y, zx^(−) + z, xy^(−) =? ⌣⌢⌣⌢⌣⌢⌣⌢⌣⌢⌣⌢⌣⌢⌣ (x+(y/(10))+(z/(100)))+(y+(z/(10))+(x/(100)))+(z+(x/(10))+(y/(100))) =((100x+10y+z)/(100))+((100y+10z+x)/(100))+((100z+10x+y)/(100)) =((111x+111y+111z)/(100)) z=13−x−y : =((111x+111y+111(13−x−y))/(100)) =((111x+111y+1443−111x−111y)/(100)) =1443/100=14.43

$$\:{x}+{y}+{z}=\mathrm{13}\:;\:\:\overline {{x},\:{yz}}\:+\:\overline {{y},\:{zx}}\:+\:\overline {{z},\:{xy}}=? \\ $$ $$\smile\frown\smile\frown\smile\frown\smile\frown\smile\frown\smile\frown\smile\frown\smile \\ $$ $$\left({x}+\frac{{y}}{\mathrm{10}}+\frac{{z}}{\mathrm{100}}\right)+\left({y}+\frac{{z}}{\mathrm{10}}+\frac{{x}}{\mathrm{100}}\right)+\left({z}+\frac{{x}}{\mathrm{10}}+\frac{{y}}{\mathrm{100}}\right) \\ $$ $$=\frac{\mathrm{100}{x}+\mathrm{10}{y}+{z}}{\mathrm{100}}+\frac{\mathrm{100}{y}+\mathrm{10}{z}+{x}}{\mathrm{100}}+\frac{\mathrm{100}{z}+\mathrm{10}{x}+{y}}{\mathrm{100}} \\ $$ $$=\frac{\mathrm{111}{x}+\mathrm{111}{y}+\mathrm{111}{z}}{\mathrm{100}} \\ $$ $${z}=\mathrm{13}−{x}−{y}\:: \\ $$ $$=\frac{\mathrm{111}{x}+\mathrm{111}{y}+\mathrm{111}\left(\mathrm{13}−{x}−{y}\right)}{\mathrm{100}} \\ $$ $$=\frac{\cancel{\mathrm{111}{x}}+\cancel{\mathrm{111}{y}}+\mathrm{1443}−\cancel{\mathrm{111}{x}}−\cancel{\mathrm{111}{y}}}{\mathrm{100}} \\ $$ $$=\mathrm{1443}/\mathrm{100}=\mathrm{14}.\mathrm{43} \\ $$

Commented bymathdanisur last updated on 24/Jul/21

Thank you Ser

$${Thank}\:{you}\:{Ser} \\ $$

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