Question-83373

Question Number 83373 by Power last updated on 01/Mar/20

Commented by jagoll last updated on 02/Mar/20

(ax+by)(x+y) = 5(x+y) ax^2 +by^2 +axy+bxy=5(x+y) 10 + xy (a+b)= 5(x+y) (i) (ax^2 +by^2 )(x+y) = 10(x+y) ax^3 + by^3 +ax^2 y+bxy^2 = 10(x+y) 50 + xy(ax+by) = 10(x+y) 50 + 5xy = 10(x+y) ⇒ 10+xy = x+y (ii) (ax^3 +by^3 )(x+y)=50(x+y) ax^4 +by^4 +ax^3 y+bxy^3 = 50(x+y) 130 + xy(ax^2 +by^2 )=50(x+y) 130 + 10xy = 50(x+y) 13 + xy = 5(x+y) (iii) so we get the result

$$\left(\mathrm{ax}+\mathrm{by}\right)\left(\mathrm{x}+\mathrm{y}\right)\:=\:\mathrm{5}\left(\mathrm{x}+\mathrm{y}\right) \\ $$$$\mathrm{ax}^{\mathrm{2}} +\mathrm{by}^{\mathrm{2}} +\mathrm{axy}+\mathrm{bxy}=\mathrm{5}\left(\mathrm{x}+\mathrm{y}\right) \\ $$$$\mathrm{10}\:+\:\mathrm{xy}\:\left(\mathrm{a}+\mathrm{b}\right)=\:\mathrm{5}\left(\mathrm{x}+\mathrm{y}\right)\:\left(\mathrm{i}\right) \\ $$$$\left(\mathrm{ax}^{\mathrm{2}} +\mathrm{by}^{\mathrm{2}} \right)\left(\mathrm{x}+\mathrm{y}\right)\:=\:\mathrm{10}\left(\mathrm{x}+\mathrm{y}\right) \\ $$$$\mathrm{ax}^{\mathrm{3}} \:+\:\mathrm{by}^{\mathrm{3}} +\mathrm{ax}^{\mathrm{2}} \mathrm{y}+\mathrm{bxy}^{\mathrm{2}} \:=\:\mathrm{10}\left(\mathrm{x}+\mathrm{y}\right) \\ $$$$\mathrm{50}\:+\:\mathrm{xy}\left(\mathrm{ax}+\mathrm{by}\right)\:=\:\mathrm{10}\left(\mathrm{x}+\mathrm{y}\right) \\ $$$$\mathrm{50}\:+\:\mathrm{5xy}\:=\:\mathrm{10}\left(\mathrm{x}+\mathrm{y}\right)\:\Rightarrow\:\mathrm{10}+\mathrm{xy}\:=\:\mathrm{x}+\mathrm{y}\:\left(\mathrm{ii}\right) \\ $$$$\left(\mathrm{ax}^{\mathrm{3}} +\mathrm{by}^{\mathrm{3}} \right)\left(\mathrm{x}+\mathrm{y}\right)=\mathrm{50}\left(\mathrm{x}+\mathrm{y}\right) \\ $$$$\mathrm{ax}^{\mathrm{4}} +\mathrm{by}^{\mathrm{4}} +\mathrm{ax}^{\mathrm{3}} \mathrm{y}+\mathrm{bxy}^{\mathrm{3}} \:=\:\mathrm{50}\left(\mathrm{x}+\mathrm{y}\right) \\ $$$$\mathrm{130}\:+\:\mathrm{xy}\left(\mathrm{ax}^{\mathrm{2}} +\mathrm{by}^{\mathrm{2}} \right)=\mathrm{50}\left(\mathrm{x}+\mathrm{y}\right) \\ $$$$\mathrm{130}\:+\:\mathrm{10xy}\:=\:\mathrm{50}\left(\mathrm{x}+\mathrm{y}\right) \\ $$$$\mathrm{13}\:+\:\mathrm{xy}\:=\:\mathrm{5}\left(\mathrm{x}+\mathrm{y}\right)\:\left(\mathrm{iii}\right) \\ $$$$\mathrm{so}\:\mathrm{we}\:\mathrm{get}\:\mathrm{the}\:\mathrm{result} \\ $$

Commented by jagoll last updated on 02/Mar/20

(ii)−(iii) ⇒ 3 = 4(x+y) , x+y = (3/4) ⇒xy = (3/4)−10 = ((−37)/4) ⇒ 10+(−((37)/4))(a+b) = 5×(3/4) ⇒ 40−37(a+b) = 15 ⇒ a +b = ((25)/(37))

$$\left(\mathrm{ii}\right)−\left(\mathrm{iii}\right)\:\Rightarrow\:\mathrm{3}\:=\:\mathrm{4}\left(\mathrm{x}+\mathrm{y}\right)\:,\:\mathrm{x}+\mathrm{y}\:=\:\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\Rightarrow\mathrm{xy}\:=\:\frac{\mathrm{3}}{\mathrm{4}}−\mathrm{10}\:=\:\frac{−\mathrm{37}}{\mathrm{4}} \\ $$$$\Rightarrow\:\mathrm{10}+\left(−\frac{\mathrm{37}}{\mathrm{4}}\right)\left(\mathrm{a}+\mathrm{b}\right)\:=\:\mathrm{5}×\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\Rightarrow\:\mathrm{40}−\mathrm{37}\left(\mathrm{a}+\mathrm{b}\right)\:=\:\mathrm{15} \\ $$$$\Rightarrow\:\mathrm{a}\:+\mathrm{b}\:=\:\frac{\mathrm{25}}{\mathrm{37}} \\ $$

Commented by Power last updated on 02/Mar/20

$$\mathrm{thank}\:\mathrm{you} \\ $$

Commented by mr W last updated on 02/Mar/20

$${nice}\:{solution}\:! \\ $$

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