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(C) For equation $I$: $8x^2 + 6x - 5 = 0$Factoring the quadratic equation: $8x^2 + 10x - 4x - 5 = 0$$2x(4x + 5) - 1(4x + 5) = 0$$(4x + 5)(2x - 1) = 0$

Vedclass · Accepted Answer

$x \leq y$

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(C) For equation $I$: $8x^2 + 6x - 5 = 0$Factoring the quadratic equation: $8x^2 + 10x - 4x - 5 = 0$$2x(4x + 5) - 1(4x + 5) = 0$$(4x + 5)(2x - 1) = 0$Therefore,$x = -\frac{5}{4}$ or $x = \frac{1}{2}$.For equation $II$: $12y^2 - 22y + 8 = 0$Dividing by $2$: $6y^2 - 11y + 4 = 0$Factoring the quadratic equation: $6y^2 - 8y - 3y + 4 = 0$$2y(3y - 4) - 1(3y - 4) = 0$$(2y - 1)(3y - 4) = 0$Therefore,$y = \frac{1}{2}$ or $y = \frac{4}{3}$.Comparing the values:If $x = -1.25$,then $x If $x = 0.5$,then $x \leq y$ (since $y = 0.5$ or $1.33$).Combining these,we get $x \leq y$.

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.$ $12y^2 - 22y + 8 = 0$