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Mathematics

Using the properties of proportion, find x : y, given

x2+2x2x+4=y2+3y3y+9\dfrac{x^{2} + 2x}{2x + 4} = \dfrac{y^{2} + 3y}{3y + 9}.

Ratio Proportion

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Answer

Given,

⇒x2+2x2x+4=y2+3y3y+9\Rightarrow \dfrac{x^{2} + 2x}{2x + 4} = \dfrac{y^{2} + 3y}{3y + 9}

Applying componendo and dividendo, we get :

⇒x2+2x+2x+4x2+2x−(2x+4)=y2+3y+3y+9y2+3y−(3y+9)⇒x2+2x+2x+4x2+2x−2x−4=y2+3y+3y+9y2+3y−3y−9⇒x2+4x+4x2−4=y2+6y+9y2−9⇒(x+2)2(x−2)(x+2)=(y+3)2(y−3)(y+3)⇒x+2x−2=y+3y−3\Rightarrow \dfrac{x^{2} + 2x + 2x + 4}{x^{2} + 2x - (2x + 4)} = \dfrac{y^{2} + 3y + 3y + 9}{y^{2} + 3y - (3y + 9)} \\[1em] \Rightarrow \dfrac{x^{2} + 2x + 2x + 4}{x^{2} + 2x - 2x - 4} = \dfrac{y^{2} + 3y + 3y + 9}{y^{2} + 3y - 3y - 9} \\[1em] \Rightarrow \dfrac{x^{2} + 4x + 4}{x^{2} - 4} = \dfrac{y^{2} + 6y + 9}{y^{2} - 9} \\[1em] \Rightarrow \dfrac{(x + 2)^2}{(x - 2)(x + 2)} = \dfrac{(y + 3)^2}{(y - 3)(y + 3)} \\[1em] \Rightarrow \dfrac{x + 2}{x - 2} = \dfrac{y + 3}{y - 3}

Applying componendo and dividendo again we get,

⇒x+2+x−2x+2−(x−2)=y+3+y−3y+3−(y−3)⇒x+2+x−2x+2−x+2=y+3+y−3y+3−y+3⇒2x4=2y6⇒x2=y3⇒xy=23⇒x:y=2:3.\Rightarrow \dfrac{x + 2 + x - 2}{x + 2 - (x - 2)} = \dfrac{y + 3 + y - 3}{y + 3 - (y - 3)} \\[1em] \Rightarrow \dfrac{x + 2 + x - 2}{x + 2 - x + 2} = \dfrac{y + 3 + y - 3}{y + 3 - y + 3} \\[1em] \Rightarrow \dfrac{2x}{4} = \dfrac{2y}{6} \\[1em] \Rightarrow \dfrac{x}{2} = \dfrac{y}{3} \\[1em] \Rightarrow \dfrac{x}{y} = \dfrac{2}{3} \\[1em] \Rightarrow x : y = 2 : 3.

Hence, x : y = 2 : 3.

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