Directions for Questions 31 - 35: In each question two equations numbered I and II are given. You have to solve both the equations and mark answer:
I. $9{{x}^{2}}-12x-5=0$
II. $3{{y}^{2}}-11y+10=0$
$x \lt y$
$x \gt y$
$x\le y$
$x\ge y$
$x=y$ or relationship between $x$ and $y$ cannot be established
Solution
I. $9{{x}^{2}}-12x-5=0$
$9{{x}^{2}}-\left( 15-3 \right)x-5=0$
$9{{x}^{2}}-15x+3x-5=0$
$3x\left( 3x-5 \right)+1\left( 3x-5 \right)=0$
(3x- 5)(3x + 1)=0
$x=~\frac{5}{3},~-\frac{1}{3}$
II. $3{{y}^{2}}-11y+10=0$
$3{{y}^{2}}-~\left( 6+5 \right)y+10=0$
$3{{y}^{2}}-6y-5y+10=0$
$3y\left( y-2 \right)-5\left( y-2 \right)=0$
$\left( y-2 \right)\left( 3y-5 \right)=~0$
$y=~\frac{5}{3},~2$
After comparison of both equations, the conclusion is
$x\le y$