Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

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

thank you

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

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

nice solution !

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

Terms of Service

Privacy Policy

Contact: info@tinkutara.com