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(C) Given equations are $9x^2 - 29x + 22 = 0$ and $y^2 - 7y + 12 = 0$.For the first equation: $9x^2 - 29x + 22 = 0$$9x^2 - 18x - 11x + 22 = 0$$9x(x -

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if $x < y$

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(C) Given equations are $9x^2 - 29x + 22 = 0$ and $y^2 - 7y + 12 = 0$.For the first equation: $9x^2 - 29x + 22 = 0$$9x^2 - 18x - 11x + 22 = 0$$9x(x - 2) - 11(x - 2) = 0$$(9x - 11)(x - 2) = 0$So,$x = 2$ or $x = 11/9 \approx 1.22$.For the second equation: $y^2 - 7y + 12 = 0$$y^2 - 4y - 3y + 12 = 0$$y(y - 4) - 3(y - 4) = 0$$(y - 3)(y - 4) = 0$So,$y = 3$ or $y = 4$.Comparing the values: $x$ values are ${1.22, 2}$ and $y$ values are ${3, 4}$.Since all values of $x$ are less than all values of $y$,we conclude $x < y$.

Solve the given two equations and select the correct option.

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