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(C) $I.$ $8x^2 + 6x - 5 = 0$$8x^2 + 10x - 4x - 5 = 0$$2x(4x + 5) - 1(4x + 5) = 0$$(2x - 1)(4x + 5) = 0$$x = \frac{1}{2}, -\frac{5}{4}$$II.$ $12y^2 - 2

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if $x \leq y$

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(C) $I.$ $8x^2 + 6x - 5 = 0$$8x^2 + 10x - 4x - 5 = 0$$2x(4x + 5) - 1(4x + 5) = 0$$(2x - 1)(4x + 5) = 0$$x = \frac{1}{2}, -\frac{5}{4}$$II.$ $12y^2 - 22y + 8 = 0$$12y^2 - 16y - 6y + 8 = 0$$4y(3y - 4) - 2(3y - 4) = 0$$(4y - 2)(3y - 4) = 0$$y = \frac{1}{2}, \frac{4}{3}$Comparing the values:$x = 0.5, -1.25$$y = 0.5, 1.33$Since all values of $x$ are less than or equal to the values of $y$,we conclude $x \leq y$.

Solve the given two equations and select the correct answer from the given options.
$I.$ $8x^2 + 6x = 5$
$II.$ $12y^2 - 22y + 8 = 0$

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Solve the given two equations and select the correct answer from the given options.$I.$ $8x^2 + 6x = 5$$II.$ $12y^2 - 22y + 8 = 0$