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(B) For equation $I$: $2x^2 + 3x + 1 = 0$$2x^2 + 2x + x + 1 = 0$$2x(x + 1) + 1(x + 1) = 0$$(2x + 1)(x + 1) = 0$So,$x = -0.5$ or $x = -1$.For equation

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if $x < y$

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(B) For equation $I$: $2x^2 + 3x + 1 = 0$$2x^2 + 2x + x + 1 = 0$$2x(x + 1) + 1(x + 1) = 0$$(2x + 1)(x + 1) = 0$So,$x = -0.5$ or $x = -1$.For equation $II$: $12y^2 + 7y + 1 = 0$$12y^2 + 4y + 3y + 1 = 0$$4y(3y + 1) + 1(3y + 1) = 0$$(4y + 1)(3y + 1) = 0$So,$y = -0.25$ or $y = -0.33$.Comparing the values:$x$ values are $\{-1, -0.5\}$ and $y$ values are $\{-0.33, -0.25\}$.Since both values of $x$ are less than both values of $y$,we conclude that $x < y$.

Solve the given two equations and select the correct option.
$I.$ $2x^2 + 3x + 1 = 0$
$II.$ $12y^2 + 7y + 1 = 0$

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Solve the given two equations and select the correct option.$I.$ $2x^2 + 3x + 1 = 0$$II.$ $12y^2 + 7y + 1 = 0$