QUADRATIC EQUATION Questions in English

(D) For equation $I$: $x^{2}-5x+6=0$
Factorizing the quadratic equation: $x^{2}-3x-2x+6=0$
$x(x-3)-2(x-3)=0$
$(x-3)(x-2)=0$
Thus,the roots are $x = 3$ and $x = 2$.
For equation $II$: $2y^{2}-15y+27=0$
Factorizing the quadratic equation: $2y^{2}-6y-9y+27=0$
$2y(y-3)-9(y-3)=0$
$(2y-9)(y-3)=0$
Thus,the roots are $y = \frac{9}{2} = 4.5$ and $y = 3$.
Comparing the values:
When $x = 3$,$y$ can be $4.5$ or $3$. Here $x \le y$.
When $x = 2$,$y$ can be $4.5$ or $3$. Here $x < y$.
Combining these,we get $x \le y$.

(D) For equation $I$: $2x^{2} + x - 1 = 0$
Using the quadratic formula or factoring: $2x^{2} + 2x - x - 1 = 0 \Rightarrow 2x(x + 1) - 1(x + 1) = 0 \Rightarrow (2x - 1)(x + 1) = 0$.
Thus,the roots are $x = \frac{1}{2}$ and $x = -1$.
For equation $II$: $2y^{2} - 3y + 1 = 0$
Factoring: $2y^{2} - 2y - y + 1 = 0 \Rightarrow 2y(y - 1) - 1(y - 1) = 0 \Rightarrow (2y - 1)(y - 1) = 0$.
Thus,the roots are $y = \frac{1}{2}$ and $y = 1$.
Comparing the values:
If $x = \frac{1}{2}$,then $y$ can be $\frac{1}{2}$ (equal) or $1$ $(x < y)$.
If $x = -1$,then $y$ can be $\frac{1}{2}$ or $1$ $(x < y)$.
In all cases,$x \le y$.

(D) Given equations:
$8x + 7y = 135$ ---$(1)$
$5x + 6y = 99$ ---$(2)$
To solve,multiply eq$(1)$ by $5$ and eq$(2)$ by $8$:
$40x + 35y = 675$ ---$(3)$
$40x + 48y = 792$ ---$(4)$
Subtract eq$(3)$ from eq$(4)$:
$(40x + 48y) - (40x + 35y) = 792 - 675$
$13y = 117$
$y = 9$
Substitute $y = 9$ into eq$(1)$:
$8x + 7(9) = 135$
$8x + 63 = 135$
$8x = 72$
$x = 9$
Since $x = 9$ and $y = 9$,therefore $x = y$.

(D) Step $1$: Solve for $x$ from equation $I$.
$x^{2} = 64 \Rightarrow x = \pm 8$.
So,$x = 8$ or $x = -8$.
Step $2$: Solve for $y$ from equation $II$.
$2y^{2} + 25y + 72 = 0$.
We need to factorize the quadratic equation: $2y^{2} + 16y + 9y + 72 = 0$.
$2y(y + 8) + 9(y + 8) = 0$.
$(2y + 9)(y + 8) = 0$.
So,$y = -8$ or $y = -4.5$.
Step $3$: Compare $x$ and $y$ values.
If $x = 8$,then $x > y$ (since $8 > -8$ and $8 > -4.5$).
If $x = -8$,then $x = y$ (when $y = -8$) and $x < y$ (when $y = -4.5$).
Since the relationship varies depending on the values chosen,the relationship between $x$ and $y$ cannot be established.

(A) Given equations are:
$7x + 3y = 26$ --- $(1)$
$2x + 17y = -41$ --- $(2)$
To eliminate $x$,multiply equation $(1)$ by $2$ and equation $(2)$ by $7$:
$(7x + 3y = 26) \times 2 \Rightarrow 14x + 6y = 52$ --- $(3)$
$(2x + 17y = -41) \times 7 \Rightarrow 14x + 119y = -287$ --- $(4)$
Subtract equation $(4)$ from equation $(3)$:
$(14x + 6y) - (14x + 119y) = 52 - (-287)$
$-113y = 339$
$y = -3$
Substitute $y = -3$ into equation $(1)$:
$7x + 3(-3) = 26$
$7x - 9 = 26$
$7x = 35$
$x = 5$
Comparing the values,$x = 5$ and $y = -3$. Since $5 > -3$,we have $x > y$.

(C) Step $1$: Solve for $x$.
$x = \sqrt[3]{2197} = 13$.
Step $2$: Solve for $y$.
$y^2 = 169 \Rightarrow y = \pm 13$.
Step $3$: Compare $x$ and $y$.
Since $x = 13$ and $y$ can be $13$ or $-13$,we have $x = 13$ and $y = 13$ (so $x = y$) or $x = 13$ and $y = -13$ (so $x > y$).
Combining these,we get $x \ge y$.

(C) For equation $I$: $x = \sqrt{1764} = 42$.
For equation $II$: $y^2 = 1764$,which implies $y = \pm \sqrt{1764} = \pm 42$.
Comparing the values:
If $y = 42$,then $x = y$.
If $y = -42$,then $x > y$.
Combining these results,we get $x \ge y$.

(B) For equation $I: 3x^2 + 5x - 2 = 0$
Factorizing: $3x^2 + 6x - x - 2 = 0$
$3x(x + 2) - 1(x + 2) = 0$
$(3x - 1)(x + 2) = 0$
So,$x = 1/3$ or $x = -2$.
For equation $II: 2y^2 - 7y + 5 = 0$
Factorizing: $2y^2 - 2y - 5y + 5 = 0$
$2y(y - 1) - 5(y - 1) = 0$
$(2y - 5)(y - 1) = 0$
So,$y = 5/2 = 2.5$ or $y = 1$.
Comparing the values:
$x_1 = 0.33, x_2 = -2$
$y_1 = 2.5, y_2 = 1$
In all cases,$x < y$.

(A) For equation $I$: $5x^{2} + 2x - 3 = 0$
Using the splitting the middle term method: $5x^{2} + 5x - 3x - 3 = 0$
$5x(x + 1) - 3(x + 1) = 0$
$(5x - 3)(x + 1) = 0$
So,$x = 0.6$ or $x = -1$.
For equation $II$: $2y^{2} + 7y + 6 = 0$
Using the splitting the middle term method: $2y^{2} + 4y + 3y + 6 = 0$
$2y(y + 2) + 3(y + 2) = 0$
$(2y + 3)(y + 2) = 0$
So,$y = -1.5$ or $y = -2$.
Comparing the values:
If $x = 0.6$,then $x > y$ (since $0.6 > -1.5$ and $0.6 > -2$).
If $x = -1$,then $x > y$ (since $-1 > -1.5$ and $-1 > -2$).
In all cases,$x > y$.

(B) To solve the system of linear equations:
$7x + 4y = 3$ --- $(1)$
$5x + 3y = 3$ --- $(2)$
Multiply equation $(1)$ by $3$ and equation $(2)$ by $4$ to eliminate $y$:
$21x + 12y = 9$ --- $(3)$
$20x + 12y = 12$ --- $(4)$
Subtract equation $(4)$ from equation $(3)$:
$(21x - 20x) + (12y - 12y) = 9 - 12$
$x = -3$
Substitute $x = -3$ into equation $(1)$:
$7(-3) + 4y = 3$
$-21 + 4y = 3$
$4y = 24$
$y = 6$
Comparing the values,we have $x = -3$ and $y = 6$.
Since $-3 < 6$,it follows that $x < y$.

(B) For equation $I$: $x^{2}+5x-6=0$
Factorizing: $x^{2}+6x-x-6=0$
$x(x+6)-1(x+6)=0$
$(x-1)(x+6)=0$
So,$x = 1$ or $x = -6$.
For equation $II$: $2y^{2}-11y+15=0$
Factorizing: $2y^{2}-6y-5y+15=0$
$2y(y-3)-5(y-3)=0$
$(2y-5)(y-3)=0$
So,$y = 2.5$ or $y = 3$.
Comparing the values:
For $x=1$: $1 < 2.5$ and $1 < 3$ (i.e.,$x < y$).
For $x=-6$: $-6 < 2.5$ and $-6 < 3$ (i.e.,$x < y$).
In all cases,$x < y$.

(A) First,solve for $y$ using equation $II$:
$y = \sqrt[3]{17576} = \sqrt[3]{26^3} = 26$
Next,substitute the value of $y$ into equation $I$ to find $x$:
$7x = 4(26) + 85$
$7x = 104 + 85$
$7x = 189$
$x = 189 / 7 = 27$
Comparing the values,we have $x = 27$ and $y = 26$.
Since $27 > 26$,it follows that $x > y$.

(D) For equation $I$: $x^{2}-25=0 \Rightarrow x^{2}=25 \Rightarrow x = 5, -5$.
For equation $II$: $4y^{2}-24y+35=0$.
Using the quadratic formula or factorization: $4y^{2}-14y-10y+35=0$.
$2y(2y-7)-5(2y-7)=0 \Rightarrow (2y-5)(2y-7)=0$.
Thus,$y = 2.5$ or $y = 3.5$.
Comparing the values:
If $x = 5$,then $x > 2.5$ and $x > 3.5$ $(x > y)$.
If $x = -5$,then $x < 2.5$ and $x < 3.5$ $(x < y)$.
Since we get different relationships ($x > y$ and $x < y$),the relationship between $x$ and $y$ cannot be established.

(A) For equation $I$: $4x^2 - 43x + 105 = 0$
We factorize the quadratic equation: $4x^2 - 28x - 15x + 105 = 0$
$4x(x - 7) - 15(x - 7) = 0$
$(4x - 15)(x - 7) = 0$
So,$x_1 = \frac{15}{4} = 3.75$ and $x_2 = 7$.
For equation $II$: $7y^2 - 29y + 30 = 0$
We factorize the quadratic equation: $7y^2 - 14y - 15y + 30 = 0$
$7y(y - 2) - 15(y - 2) = 0$
$(7y - 15)(y - 2) = 0$
So,$y_1 = \frac{15}{7} \approx 2.14$ and $y_2 = 2$.
Comparing the values:
$x_1 = 3.75$ and $x_2 = 7$
$y_1 \approx 2.14$ and $y_2 = 2$
Since all values of $x$ are greater than all values of $y$ $(3.75 > 2.14, 3.75 > 2, 7 > 2.14, 7 > 2)$,we conclude that $x > y$.

(D) For equation $I$: $x^{2}+13x+40=0$
Factorizing the quadratic equation: $x^{2}+8x+5x+40=0$
$x(x+8)+5(x+8)=0$
$(x+5)(x+8)=0$
Therefore,the roots are $x = -5$ and $x = -8$.
For equation $II$: $y^{2}+7y+10=0$
Factorizing the quadratic equation: $y^{2}+5y+2y+10=0$
$y(y+5)+2(y+5)=0$
$(y+2)(y+5)=0$
Therefore,the roots are $y = -2$ and $y = -5$.
Comparing the values:
When $x = -5$,$y$ can be $-2$ or $-5$. Here $x \le y$.
When $x = -8$,$y$ can be $-2$ or $-5$. Here $x < y$.
Combining these,we get $x \le y$.

(D) Step $1$: Solve for $x$.
$x = \sqrt[3]{2197} = 13$.
Step $2$: Solve the quadratic equation for $y$.
$2y^2 - 54y + 364 = 0$.
Divide the entire equation by $2$:
$y^2 - 27y + 182 = 0$.
We need two numbers that multiply to $182$ and add to $-27$. These numbers are $-13$ and $-14$.
$(y - 13)(y - 14) = 0$.
Therefore,$y = 13$ or $y = 14$.
Step $3$: Compare $x$ and $y$.
$x = 13$.
$y = 13$ or $14$.
Comparing the values,$x = 13$ and $y = 13$ or $14$. In both cases,$x \le y$ is satisfied.
Thus,the correct option is $D$.

(B) Given equations:
$13x - 8y = -81$ ---$(1)$
$15x - 5y = -65$ ---$(2)$
Multiply equation $(1)$ by $5$ and equation $(2)$ by $8$ to eliminate $y$:
$(13x - 8y) \times 5 = -81 \times 5 \Rightarrow 65x - 40y = -405$
$(15x - 5y) \times 8 = -65 \times 8 \Rightarrow 120x - 40y = -520$
Subtract the first result from the second:
$(120x - 40y) - (65x - 40y) = -520 - (-405)$
$55x = -115$
$x = -115 / 55 \approx -2.09$
Substitute $x$ into equation $(2)$:
$15(-2.09) - 5y = -65$
$-31.35 - 5y = -65$
$-5y = -65 + 31.35$
$-5y = -33.65$
$y = 6.73$
Comparing $x$ and $y$:
$x \approx -2.09$ and $y \approx 6.73$
Since $-2.09 < 6.73$,we have $x < y$.

(B) Step $1$: Solve for $x$.
$x = \sqrt{172}$. Since $13^2 = 169$ and $14^2 = 196$,the value of $x$ is approximately $13.11$.
Step $2$: Solve the quadratic equation for $y$.
$y^{2} - 29y + 210 = 0$.
We need two numbers that multiply to $210$ and add up to $29$. These numbers are $14$ and $15$ ($14 \times 15 = 210$ and $14 + 15 = 29$).
So,$(y - 14)(y - 15) = 0$.
This gives $y = 14$ or $y = 15$.
Step $3$: Compare $x$ and $y$.
Since $x \approx 13.11$ and $y$ is either $14$ or $15$,it is clear that $x < y$ in both cases.
Therefore,the correct option is $B$.

(B) For equation $I$: $676 x^{2} - 1 = 0$
$676 x^{2} = 1$
$x^{2} = \frac{1}{676}$
$x = \pm \frac{1}{26}$
So,$x = \frac{1}{26} \approx 0.0384$ or $x = -\frac{1}{26} \approx -0.0384$.
For equation $II$: $y = \frac{1}{\sqrt[3]{13824}}$
Since $24^{3} = 13824$,we have $\sqrt[3]{13824} = 24$.
Thus,$y = \frac{1}{24} \approx 0.0416$.
Comparing the values:
$x = 0.0384$ or $-0.0384$ and $y = 0.0416$.
In both cases,$x < y$.

(A) For equation $I$: $x^{2}+12x+36=0$
This can be written as $(x+6)^{2}=0$
Therefore,$x = -6, -6$.
For equation $II$: $y^{2}+15y+56=0$
Factoring the quadratic: $y^{2}+8y+7y+56=0$
$y(y+8)+7(y+8)=0$
$(y+7)(y+8)=0$
Therefore,$y = -7, -8$.
Comparing the values:
Since $x = -6$ and $y$ values are $-7$ and $-8$,we observe that $-6 > -7$ and $-6 > -8$.
Thus,$x > y$.

(A) For equation $I$: $x^{2}+12x+36=0$
This can be written as $(x+6)^{2}=0$,so $x = -6, -6$.
For equation $II$: $y^{2}+15y+56=0$
Factoring the quadratic: $y^{2}+7y+8y+56=0$
$y(y+7)+8(y+7)=0$
$(y+7)(y+8)=0$,so $y = -7, -8$.
Comparing the values:
$x = -6$
$y = -7$ or $-8$
Since $-6 > -7$ and $-6 > -8$,it follows that $x > y$.

(D) For equation $I$: $2x^2 - 3x - 35 = 0$
Using the quadratic formula or factoring: $2x^2 - 10x + 7x - 35 = 0$
$2x(x - 5) + 7(x - 5) = 0$
$(2x + 7)(x - 5) = 0$
So,$x = 5$ or $x = -3.5$.
For equation $II$: $y^2 - 7y + 6 = 0$
Factoring: $y^2 - 6y - y + 6 = 0$
$y(y - 6) - 1(y - 6) = 0$
$(y - 1)(y - 6) = 0$
So,$y = 1$ or $y = 6$.
Comparing the values:
If $x = 5$,then $x > y$ (when $y = 1$) and $x < y$ (when $y = 6$).
Since the relationship changes depending on the values,the relationship between $x$ and $y$ cannot be established.

(B) For equation $I$: $12x^2 - 47x + 40 = 0$
Splitting the middle term: $12x^2 - 32x - 15x + 40 = 0$
$4x(3x - 8) - 5(3x - 8) = 0$
$(4x - 5)(3x - 8) = 0$
So,$x = \frac{5}{4} = 1.25$ and $x = \frac{8}{3} \approx 2.67$.
For equation $II$: $4y^2 + 3y - 10 = 0$
Splitting the middle term: $4y^2 + 8y - 5y - 10 = 0$
$4y(y + 2) - 5(y + 2) = 0$
$(4y - 5)(y + 2) = 0$
So,$y = \frac{5}{4} = 1.25$ and $y = -2$.
Comparing the values:
$x_1 = 1.25, x_2 = 2.67$
$y_1 = 1.25, y_2 = -2$
Since $x_1 = y_1$,$x_2 > y_1$,and $x_1, x_2 > y_2$,we conclude that $x \ge y$.

(C) Step $1$: Solve for $x$.
$x = \frac{\sqrt{256}}{\sqrt{576}} = \frac{16}{24} = \frac{2}{3} \approx 0.666$.
Step $2$: Solve the quadratic equation $3y^2 + y - 2 = 0$.
Using the factorization method: $3y^2 + 3y - 2y - 2 = 0$.
$3y(y + 1) - 2(y + 1) = 0$.
$(3y - 2)(y + 1) = 0$.
So,$y = \frac{2}{3}$ or $y = -1$.
Step $3$: Compare $x$ and $y$.
$x = \frac{2}{3}$.
The values of $y$ are $\frac{2}{3}$ and $-1$.
Since $x = \frac{2}{3}$ and $y$ can be $\frac{2}{3}$ or $-1$,we observe that $x \ge y$ (because $x$ is equal to one value of $y$ and greater than the other value of $y$).

(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 $x \ge y$.

(C) Given equations:
$I. 42x - 17y = -67$
$II. 7x - 12y = -26$
Multiply equation $(II)$ by $6$ to eliminate $x$:
$6 \times (7x - 12y) = 6 \times (-26)$
$42x - 72y = -156$ $(III)$
Subtract equation $(III)$ from equation $(I)$:
$(42x - 17y) - (42x - 72y) = -67 - (-156)$
$42x - 17y - 42x + 72y = -67 + 156$
$55y = 89$
$y = \frac{89}{55} \approx 1.618$
Substitute $y = \frac{89}{55}$ into equation $(II)$:
$7x - 12(\frac{89}{55}) = -26$
$7x - \frac{1068}{55} = -26$
$7x = \frac{1068}{55} - 26$
$7x = \frac{1068 - 1430}{55}$
$7x = -\frac{362}{55}$
$x = -\frac{362}{385} \approx -0.94$
Comparing the values,$x \approx -0.94$ and $y \approx 1.618$.
Since $-0.94 < 1.618$,we have $x < y$.

(A) Step $1$: Solve equation $I$ for $x$.
$2.3x - 20.01 = 0$
$2.3x = 20.01$
$x = \frac{20.01}{2.3} = 8.7$
Step $2$: Solve equation $II$ for $y$ using the value of $x$.
$2.9y - x = 0$
$2.9y = x$
$2.9y = 8.7$
$y = \frac{8.7}{2.9} = 3$
Step $3$: Compare the values of $x$ and $y$.
Since $x = 8.7$ and $y = 3$,it is clear that $x > y$.

(D) For equation $I$: $x^{2}-26x+168=0$
We need two numbers whose sum is $26$ and product is $168$. These numbers are $12$ and $14$.
$x^{2}-12x-14x+168=0$
$x(x-12)-14(x-12)=0$
$(x-12)(x-14)=0$
Therefore,$x = 12$ or $x = 14$.
For equation $II$: $y^{2}-25y+156=0$
We need two numbers whose sum is $25$ and product is $156$. These numbers are $12$ and $13$.
$y^{2}-12y-13y+156=0$
$y(y-12)-13(y-12)=0$
$(y-12)(y-13)=0$
Therefore,$y = 12$ or $y = 13$.
Comparing the values:
If $x=12$ and $y=12$,then $x=y$.
If $x=12$ and $y=13$,then $xIf $x=14$ and $y=12$,then $x>y$.
Since we get different results ($x=y$,$xy$),the relationship between $x$ and $y$ cannot be established.

(B) For equation $I$: $2x^2 + 13x - 7 = 0$
Using the quadratic formula or factorization: $2x^2 + 14x - x - 7 = 0$
$2x(x + 7) - 1(x + 7) = 0$
$(2x - 1)(x + 7) = 0$
So,$x = 0.5$ or $x = -7$.
For equation $II$: $2y^2 - 5y + 3 = 0$
Using the quadratic formula or factorization: $2y^2 - 2y - 3y + 3 = 0$
$2y(y - 1) - 3(y - 1) = 0$
$(2y - 3)(y - 1) = 0$
So,$y = 1.5$ or $y = 1$.
Comparing the values:
When $x = 0.5$,$x < y$ (since $0.5 < 1$ and $0.5 < 1.5$).
When $x = -7$,$x < y$ (since $-7 < 1$ and $-7 < 1.5$).
In all cases,$x < y$.

(D) For equation $I: x^{2} + 12x + 32 = 0$
Factorizing the quadratic equation: $x^{2} + 8x + 4x + 32 = 0$
$x(x + 8) + 4(x + 8) = 0$
$(x + 4)(x + 8) = 0$
Thus,$x = -4$ or $x = -8$.
For equation $II: 2y^{2} + 15y + 27 = 0$
Factorizing the quadratic equation: $2y^{2} + 6y + 9y + 27 = 0$
$2y(y + 3) + 9(y + 3) = 0$
$(2y + 9)(y + 3) = 0$
Thus,$y = -4.5$ or $y = -3$.
Comparing the values:
When $x = -4$,$x > y$ (since $-4 > -4.5$) and $x < y$ (since $-4 < -3$).
When $x = -8$,$x < y$ (since $-8 < -4.5$ and $-8 < -3$).
Since the relationship changes depending on the values chosen,the relationship between $x$ and $y$ cannot be established.

(D) For equation $I$: $x^{2}-82x+781=0$.
We need two numbers whose product is $781$ and sum is $82$. These numbers are $71$ and $11$.
So,$(x-71)(x-11)=0$,which gives $x = 71$ or $x = 11$.
For equation $II$: $y^{2}=5041$.
Taking the square root on both sides,$y = \pm \sqrt{5041} = \pm 71$.
So,$y = 71$ or $y = -71$.
Comparing the values:
If $x = 71$ and $y = 71$,then $x = y$.
If $x = 71$ and $y = -71$,then $x > y$.
If $x = 11$ and $y = 71$,then $x < y$.
Since we get different relationships ($x = y$,$x > y$,and $x < y$) depending on the values chosen,the relationship between $x$ and $y$ cannot be established.

(D) For equation $I$: $3x^2 - 7x - 20 = 0$
Using the splitting the middle term method: $3x^2 - 12x + 5x - 20 = 0$
$3x(x - 4) + 5(x - 4) = 0$
$(3x + 5)(x - 4) = 0$
So,$x = 4$ or $x = -5/3 \approx -1.67$.
For equation $II$: $y^2 - 8y + 16 = 0$
This is a perfect square: $(y - 4)^2 = 0$
So,$y = 4$.
Comparing the values:
When $x = 4$,$x = y$.
When $x = -1.67$,$x < y$.
Combining these,we get $x \le y$.

(A) For equation $I$: $9x^{2} - 114x + 361 = 0$.
This is a perfect square trinomial: $(3x - 19)^{2} = 0$.
Solving for $x$: $3x = 19$,so $x = \frac{19}{3} \approx 6.33$.
For equation $II$: $y^{2} = 36$.
Taking the square root: $y = \pm 6$,so $y = 6$ or $y = -6$.
Comparing the values:
If $y = 6$,then $x = 6.33 > 6$.
If $y = -6$,then $x = 6.33 > -6$.
In both cases,$x > y$.

(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 $x < y$.

(D) For equation $I: 3x^2 - 4x - 32 = 0$
Using the quadratic formula or factorization: $3x^2 - 12x + 8x - 32 = 0$
$3x(x - 4) + 8(x - 4) = 0$
$(3x + 8)(x - 4) = 0$
So,$x = 4$ or $x = -8/3 \approx -2.67$.
For equation $II: 2y^2 - 17y + 36 = 0$
Factorizing: $2y^2 - 9y - 8y + 36 = 0$
$y(2y - 9) - 4(2y - 9) = 0$
$(y - 4)(2y - 9) = 0$
So,$y = 4$ or $y = 9/2 = 4.5$.
Comparing the values:
$x = \{4, -2.67\}$
$y = \{4, 4.5\}$
Since $4 \le 4$,$4 \le 4.5$,$-2.67 \le 4$,and $-2.67 \le 4.5$,we conclude that $x \le y$.

(D) For equation $I$: $x^{2}+14x+49=0$
This can be written as $(x+7)^{2}=0$.
Therefore,$x = -7$.
For equation $II$: $y^{2}+9y=0$
Factoring out $y$,we get $y(y+9)=0$.
Therefore,$y = 0$ or $y = -9$.
Comparing the values:
If $x = -7$ and $y = 0$,then $x < y$.
If $x = -7$ and $y = -9$,then $x > y$.
Since the relationship changes depending on the value of $y$,the relationship between $x$ and $y$ cannot be established.

(C) For equation $I$: $35x^{2} - 53x + 20 = 0$
We factorize the quadratic equation: $35x^{2} - 28x - 25x + 20 = 0$
$7x(5x - 4) - 5(5x - 4) = 0$
$(7x - 5)(5x - 4) = 0$
Thus,$x = \frac{5}{7} \approx 0.714$ or $x = \frac{4}{5} = 0.8$.
For equation $II$: $56y^{2} - 97y + 42 = 0$
We factorize the quadratic equation: $56y^{2} - 49y - 48y + 42 = 0$
$7y(8y - 7) - 6(8y - 7) = 0$
$(7y - 6)(8y - 7) = 0$
Thus,$y = \frac{6}{7} \approx 0.857$ or $y = \frac{7}{8} = 0.875$.
Comparing the values:
$x$ values are ${0.714, 0.8}$ and $y$ values are ${0.857, 0.875}$.
Since all values of $x$ are less than all values of $y$,we conclude that $x < y$.

(A) Step $1$: Solve for $x$ in equation $I$.
$x = \sqrt[3]{4913} = 17$.
Step $2$: Substitute the value of $x$ into equation $II$ to solve for $y$.
$13y + 3(17) = 246$
$13y + 51 = 246$
$13y = 246 - 51$
$13y = 195$
$y = \frac{195}{13} = 15$.
Step $3$: Compare the values of $x$ and $y$.
Since $x = 17$ and $y = 15$,it is clear that $x > y$.

(D) From equation $I$: $x^{2} = 3481 \Rightarrow x = \pm 59$.
From equation $II$: $3y^{2} = \sqrt[3]{216000}$.
Since $60^{3} = 216000$,we have $3y^{2} = 60 \Rightarrow y^{2} = 20 \Rightarrow y = \pm \sqrt{20} \approx \pm 4.47$.
Comparing the values:
If $x = 59$,then $x > y$ (since $59 > 4.47$ and $59 > -4.47$).
If $x = -59$,then $x < y$ (since $-59 < 4.47$ and $-59 < -4.47$).
Since we get different results for different values of $x$,the relationship between $x$ and $y$ cannot be established.

(A) For equation $I: 20 x^{2}-67 x+56=0$
We factorize the quadratic equation: $20 x^{2}-35 x-32 x+56=0$
$5 x(4 x-7)-8(4 x-7)=0$
$(5 x-8)(4 x-7)=0$
So,$x = \frac{8}{5} = 1.6$ or $x = \frac{7}{4} = 1.75$
For equation $II: 56 y^{2}-67 y+20=0$
We factorize the quadratic equation: $56 y^{2}-32 y-35 y+20=0$
$8 y(7 y-4)-5(7 y-4)=0$
$(8 y-5)(7 y-4)=0$
So,$y = \frac{5}{8} = 0.625$ or $y = \frac{4}{7} \approx 0.57$
Comparing the values: $x_1 = 1.6, x_2 = 1.75$ and $y_1 = 0.625, y_2 = 0.57$
Since both values of $x$ are greater than both values of $y$,we conclude $x > y$.

(D) From equation $I$,$x^{2} = 14641$,which implies $x = \pm \sqrt{14641} = \pm 121$.
So,$x$ can be $121$ or $-121$.
From equation $II$,$y = \sqrt{14641} = 121$.
Comparing the values:
If $x = 121$,then $x = y$.
If $x = -121$,then $x < y$.
Combining these two cases,we get $x \le y$.

(B) For equation $I$: $x^{2} - 13x + 42 = 0$.
Splitting the middle term: $x^{2} - 7x - 6x + 42 = 0$.
$x(x - 7) - 6(x - 7) = 0$.
$(x - 6)(x - 7) = 0$.
Thus,$x = 6$ or $x = 7$.
For equation $II$: $y = \sqrt[4]{1296}$.
Since $6^{4} = 6 \times 6 \times 6 \times 6 = 1296$,we have $y = 6$.
Comparing the values:
When $x = 6$,$x = y$.
When $x = 7$,$x > y$.
Combining these,we get $x \ge y$.

(C) For equation $I: 15x^2 - 46x + 35 = 0$
We factorize the quadratic equation: $15x^2 - 25x - 21x + 35 = 0$
$5x(3x - 5) - 7(3x - 5) = 0$
$(5x - 7)(3x - 5) = 0$
So,$x = 7/5 = 1.4$ or $x = 5/3 \approx 1.67$.
For equation $II: 4y^2 - 15y + 14 = 0$
We factorize the quadratic equation: $4y^2 - 8y - 7y + 14 = 0$
$4y(y - 2) - 7(y - 2) = 0$
$(4y - 7)(y - 2) = 0$
So,$y = 7/4 = 1.75$ or $y = 2$.
Comparing the values:
$x = 1.4, 1.67$
$y = 1.75, 2$
In all cases,$x < y$.

(A) Given equations:
$(i) 7x + 6y + 4z = 122$
$(ii) 4x + 5y + 3z = 88$
$(iii) 9x + 2y + z = 78$
Step $1$: Eliminate $z$ using equations $(ii)$ and $(iii)$.
Multiply $(iii)$ by $3$: $27x + 6y + 3z = 234$
Subtract $(ii)$ from this: $(27x + 6y + 3z) - (4x + 5y + 3z) = 234 - 88$
$23x + y = 146 \dots (iv)$
Step $2$: Eliminate $z$ using equations $(i)$ and $(iii)$.
Multiply $(iii)$ by $4$: $36x + 8y + 4z = 312$
Subtract $(i)$ from this: $(36x + 8y + 4z) - (7x + 6y + 4z) = 312 - 122$
$29x + 2y = 190 \dots (v)$
Step $3$: Solve for $x$ and $y$ using $(iv)$ and $(v)$.
Multiply $(iv)$ by $2$: $46x + 2y = 292$
Subtract $(v)$ from this: $(46x + 2y) - (29x + 2y) = 292 - 190$
$17x = 102 \Rightarrow x = 6$
Substitute $x = 6$ into $(iv)$: $23(6) + y = 146 \Rightarrow 138 + y = 146 \Rightarrow y = 8$
Step $4$: Find $z$ using $(iii)$.
$9(6) + 2(8) + z = 78 \Rightarrow 54 + 16 + z = 78 \Rightarrow 70 + z = 78 \Rightarrow z = 8$
Thus,$x = 6, y = 8, z = 8$. Therefore,$x < y = z$.

(C) Multiply equation $(II)$ by $2$ and subtract equation $(I)$ from it:
$2 \times (4x + 3y = 59) \Rightarrow 8x + 6y = 118$
$(8x + 6y = 118) - (7x + 6y = 110) \Rightarrow x = 8$
Substitute $x = 8$ into equation $(I)$:
$7(8) + 6y = 110$
$56 + 6y = 110$
$6y = 110 - 56 = 54$
$y = 9$
Substitute $x = 8$ into equation $(III)$:
$8 + z = 15$
$z = 7$
Comparing the values: $x = 8, y = 9, z = 7$. Thus,$x < y$ and $y > z$,which means $x < y > z$.

(D) Step $1$: Solve for $x$.
$x = \sqrt{(36)^{1/2} \times (1296)^{1/4}} = \sqrt{6 \times 6} = \sqrt{36} = 6$.
Step $2$: Solve the system of linear equations for $y$.
Given equations:
$(ii) \quad 2y + 3z = 33$
$(iii) \quad 6y + 5z = 71$
Multiply equation $(ii)$ by $3$:
$6y + 9z = 99$
Subtract equation $(iii)$ from this result:
$(6y + 9z) - (6y + 5z) = 99 - 71$
$4z = 28 \Rightarrow z = 7$.
Substitute $z = 7$ into equation $(ii)$:
$2y + 3(7) = 33$
$2y + 21 = 33$
$2y = 12 \Rightarrow y = 6$.
Step $3$: Compare $x$ and $y$.
Since $x = 6$ and $y = 6$,we have $x = y$.

(D) Multiply equation $(I)$ by $5$ and equation $(II)$ by $8$:
$40x + 35y = 675$
$40x + 48y = 792$
Subtracting the first from the second:
$(40x + 48y) - (40x + 35y) = 792 - 675$
$13y = 117 \Rightarrow y = 9$
Substitute $y = 9$ into equation $(I)$:
$8x + 7(9) = 135 \Rightarrow 8x + 63 = 135 \Rightarrow 8x = 72 \Rightarrow x = 9$
Substitute $y = 9$ into equation $(III)$:
$9(9) + 8z = 121 \Rightarrow 81 + 8z = 121 \Rightarrow 8z = 40 \Rightarrow z = 5$
Comparing the values: $x = 9, y = 9, z = 5$.
Therefore,$x = y > z$.

(D) From equation $I$: $(x+y)^{3} = 1331 \implies x+y = \sqrt[3]{1331} = 11$.
So,$y = 11-x$.
Substitute this into equation $III$: $x(11-x) = 28 \implies 11x - x^{2} = 28$.
Rearranging gives the quadratic equation: $x^{2} - 11x + 28 = 0$.
Factoring the quadratic: $x^{2} - 7x - 4x + 28 = 0 \implies x(x-7) - 4(x-7) = 0 \implies (x-7)(x-4) = 0$.
Thus,$x = 7$ or $x = 4$.
If $x = 7$,then $y = 11-7 = 4$.
If $x = 4$,then $y = 11-4 = 7$.
Now,using equation $II$: $x-y+z = 0 \implies z = y-x$.
Case $1$: If $x = 7, y = 4$,then $z = 4-7 = -3$.
Case $2$: If $x = 4, y = 7$,then $z = 7-4 = 3$.
Since $x$ can be greater than $y$ or less than $y$,no fixed relationship between $x$ and $y$ can be established.

(C) Given equations are:
$I. x + 3y + 4z = 96$
$II. 2x + 8z = 80$
$III. 2x + 6y = 120$
From equation $(III)$,$2x = 120 - 6y$,so $x = 60 - 3y$.
Substitute $x$ into equation $(II)$:
$2(60 - 3y) + 8z = 80$
$120 - 6y + 8z = 80$
$8z - 6y = -40 \Rightarrow 4z - 3y = -20 \Rightarrow 3y = 4z + 20 \Rightarrow y = \frac{4z + 20}{3}$.
Substitute $x$ and $y$ into equation $(I)$:
$(60 - 3y) + 3y + 4z = 96$
$60 + 4z = 96$
$4z = 36 \Rightarrow z = 9$.
Now,find $y$:
$y = \frac{4(9) + 20}{3} = \frac{36 + 20}{3} = \frac{56}{3} \approx 18.67$.
Now,find $x$:
$x = 60 - 3(18.67) = 60 - 56 = 4$.
Comparing the values: $x = 4, y = 18.67, z = 9$.
Thus,$x < z < y$,which implies $x < y$ and $y > z$. Therefore,$x < y > z$.

(B) For equation $I$: Let $k = \frac{3x}{3x+7}$. Then $k - \frac{1}{k} = 14 \implies k^2 - 14k - 1 = 0$.
Using the quadratic formula $k = \frac{14 \pm \sqrt{196 + 4}}{2} = 7 \pm 5\sqrt{2}$.
Since $k = \frac{3x}{3x+7}$,we solve for $x$. Both roots for $x$ will be negative because the constant term in the resulting quadratic $126x^2 + 336x + 49 = 0$ is positive and the middle term is positive.
For equation $II$: Let $m = \frac{y}{18y-5}$. Then $m - \frac{1}{m} = 2 \implies m^2 - 2m - 1 = 0$.
Using the quadratic formula $m = \frac{2 \pm \sqrt{4 + 4}}{2} = 1 \pm \sqrt{2}$.
Solving for $y$,we find that the roots for $y$ are positive because the resulting quadratic $359y^2 - 190y + 25 = 0$ has a positive constant term and a negative middle term.
Since all values of $x$ are negative and all values of $y$ are positive,it follows that $x < y$.