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(A) For equation $I$: $17 x^{2}+48 x-9=0$$17 x^{2}+51 x-3 x-9=0$$17 x(x+3)-3(x+3)=0$$(17 x-3)(x+3)=0$$x = -3$ or $x = \frac{3}{17} \approx 0.176$For e

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if $x < y$

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(A) For equation $I$: $17 x^{2}+48 x-9=0$$17 x^{2}+51 x-3 x-9=0$$17 x(x+3)-3(x+3)=0$$(17 x-3)(x+3)=0$$x = -3$ or $x = \frac{3}{17} \approx 0.176$For equation $II$: $13 y^{2}-32 y+12=0$$13 y^{2}-26 y-6 y+12=0$$13 y(y-2)-6(y-2)=0$$(13 y-6)(y-2)=0$$y = \frac{6}{13} \approx 0.461$ or $y = 2$Comparing the values: Since all values of $x$ ($-3$ and $0.176$) are less than all values of $y$ ($0.461$ and $2$),we conclude that $x < y$.

Solve the given two equations and select the correct answer from the given options.
$I.$ $17 x^{2}+48 x=9$
$II.$ $13 y^{2}=32 y-12$

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Solve the given two equations and select the correct answer from the given options.$I.$ $17 x^{2}+48 x=9$$II.$ $13 y^{2}=32 y-12$