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Question Number 164496 by HongKing last updated on 18/Jan/22

{ ((x^2 + xy = 11)),((y^2 + xy = 24)) :} ⇒ ∣x+y∣=?

$$\begin{cases}{\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{xy}\:=\:\mathrm{11}}\\{\mathrm{y}^{\mathrm{2}} \:+\:\mathrm{xy}\:=\:\mathrm{24}}\end{cases}\:\:\Rightarrow\:\mid\mathrm{x}+\mathrm{y}\mid=? \\ $$

Answered by Rasheed.Sindhi last updated on 18/Jan/22

{ ((x^2 + xy = 11)),((y^2 + xy = 14)) :} ⇒ ∣x+y∣=? { ((x(x+y)=11)),((y(x + y = 14)) :}⇒(x/y)=((11)/(14))⇒x=((11y)/(14)) ((11y)/(14))(((11y)/(14))+y)=11 11y(25y)=11×14^2 y^2 =((14^2 )/(25)) y=±((14)/5) x=((11)/(14))(±((14)/5) )=±((11)/( 5)) ∣x+y∣=∣±((11)/( 5))±((14)/5) ∣=((25)/5)=5

$$\begin{cases}{\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{xy}\:=\:\mathrm{11}}\\{\mathrm{y}^{\mathrm{2}} \:+\:\mathrm{xy}\:=\:\mathrm{14}}\end{cases}\:\:\Rightarrow\:\mid\mathrm{x}+\mathrm{y}\mid=? \\ $$$$\begin{cases}{\mathrm{x}\left(\mathrm{x}+\mathrm{y}\right)=\mathrm{11}}\\{\mathrm{y}\left(\mathrm{x}\:+\:\mathrm{y}\:=\:\mathrm{14}\right.}\end{cases}\Rightarrow\frac{\mathrm{x}}{\mathrm{y}}=\frac{\mathrm{11}}{\mathrm{14}}\Rightarrow\mathrm{x}=\frac{\mathrm{11y}}{\mathrm{14}} \\ $$$$\frac{\mathrm{11y}}{\mathrm{14}}\left(\frac{\mathrm{11y}}{\mathrm{14}}+\mathrm{y}\right)=\mathrm{11} \\ $$$$\mathrm{11y}\left(\mathrm{25y}\right)=\mathrm{11}×\mathrm{14}^{\mathrm{2}} \\ $$$$\mathrm{y}^{\mathrm{2}} =\frac{\mathrm{14}^{\mathrm{2}} }{\mathrm{25}} \\ $$$$\mathrm{y}=\pm\frac{\mathrm{14}}{\mathrm{5}} \\ $$$$\mathrm{x}=\frac{\mathrm{11}}{\mathrm{14}}\left(\pm\frac{\mathrm{14}}{\mathrm{5}}\:\right)=\pm\frac{\mathrm{11}}{\:\mathrm{5}} \\ $$$$\mid\mathrm{x}+\mathrm{y}\mid=\mid\pm\frac{\mathrm{11}}{\:\mathrm{5}}\pm\frac{\mathrm{14}}{\mathrm{5}}\:\mid=\frac{\mathrm{25}}{\mathrm{5}}=\mathrm{5} \\ $$

Commented by HongKing last updated on 18/Jan/22

sorry dear Sir, y^2 +xy=14

$$\mathrm{sorry}\:\mathrm{dear}\:\mathrm{Sir},\:\mathrm{y}^{\mathrm{2}} +\mathrm{xy}=\mathrm{14} \\ $$

Answered by nurtani last updated on 18/Jan/22

x^2 +y^2 +2xy= 25 (x+y)^2 = 25 ⇒ x+y = ±5 ⇒∣x+y∣=5

$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{2}{xy}=\:\mathrm{25} \\ $$$$\left({x}+{y}\right)^{\mathrm{2}} \:=\:\mathrm{25}\:\Rightarrow\:{x}+{y}\:=\:\pm\mathrm{5}\:\Rightarrow\mid{x}+{y}\mid=\mathrm{5} \\ $$

Commented by Rasheed.Sindhi last updated on 18/Jan/22

Compact! ...∩i⊂∈!

$${Compact}!\:...\cap\mathrm{i}\subset\in! \\ $$

Commented by nurtani last updated on 18/Jan/22

thanks sir

$${thanks}\:{sir} \\ $$

Commented by HongKing last updated on 18/Jan/22

thank you dear Sir cool

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{dear}\:\mathrm{Sir}\:\mathrm{cool} \\ $$

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