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(C) Given equations are $8x^2 - 49x + 45 = 0$ and $8y^2 - y - 9 = 0$.For the first equation: $8x^2 - 40x - 9x + 45 = 0$$8x(x - 5) - 9(x - 5) = 0$$(8x

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if $x \ge y$

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(C) Given equations are $8x^2 - 49x + 45 = 0$ and $8y^2 - y - 9 = 0$.For the first equation: $8x^2 - 40x - 9x + 45 = 0$$8x(x - 5) - 9(x - 5) = 0$$(8x - 9)(x - 5) = 0$So,$x_1 = 5$ and $x_2 = 1.125$.For the second equation: $8y^2 - 9y + 8y - 9 = 0$$y(8y - 9) + 1(8y - 9) = 0$$(y + 1)(8y - 9) = 0$So,$y_1 = 1.125$ and $y_2 = -1$.Comparing the values: $x = \{5, 1.125\}$ and $y = \{1.125, -1\}$.Since $x_1 > y_1$,$x_1 > y_2$,$x_2 = y_1$,and $x_2 > y_2$,we conclude that $x \ge y$.

Solve the given two equations and select the correct option.

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