Taking {tex} \frac { 1 } { x } = u{/tex} and {tex} \frac { 1 } { y } = v.{/tex}The given system of equations become
{tex} 2 u + \frac { 2 } { 3 } v = \frac { 1 } { 6 }{/tex}
Therefore, {tex} 12u+4v=1{/tex}............(i)
and, {tex}3u+2v=0{/tex}..........(ii)
Multiplying (ii) by 2 and subtracting from (i), we get
{tex} 6 u = 1 \Rightarrow u = \frac { 1 } { 6 }{/tex}
Putting {tex} u = \frac { 1 } { 6 }{/tex}in (i), we get
{tex} 2 + 4 v = 1 \Rightarrow v = - \frac { 1 } { 4 }{/tex}
Hence, {tex} x = \frac { 1 } { u } = 6{/tex} and {tex} y = \frac { 1 } { v } = - 4{/tex}
So. the solution of the given system of equations is {tex}x=6,y=-4{/tex} 
Putting x = 6, y = -4 in {tex}y=ax-4{/tex}, we get
{tex}-4=6a-4{/tex}
{tex} \Rightarrow a=0{/tex}

Let A → (–3, 10), B → (6, –8) and P → (–1, 6)
Let P divide AB in the ratio K: 1.

{tex}P \to \left\{ {\frac{{(K)(6) + (1)( - 3)}}{{K + 1}},\frac{{(K)( - 8) + (1)(10)}}{{K + 1}}} \right\}{/tex}
or {tex}P \to \left( {\frac{{6K - 3}}{{K + 1}},\frac{{ - 8K + 10}}{{K + 1}}} \right){/tex}
But P {tex}\rightarrow{/tex} (-1, 6)
{tex}\therefore \;\frac{{6K - 3}}{{K + 1}} = - 1{/tex}
{tex}\Rightarrow{/tex} 6K - 3 = -K - 1
{tex}\Rightarrow{/tex} 7K = 2 
{tex}\Rightarrow K = \frac{2}{7}{/tex}
and {tex}\frac{{ - 8K + 10}}{{K + 1}} = 6{/tex}
{tex}\Rightarrow{/tex} -8k + 10 = 6K + 6
{tex}\Rightarrow{/tex} 14K = 4
{tex}\Rightarrow K = \frac{4}{{14}} = \frac{2}{7}{/tex}

Let the polynomial is f(x) and zeros are α and β

then f(x)=x²-(α+β)x+ αβ

Given {tex}\alpha + \beta = \frac { 5 } { 2 } , \alpha \beta = 1{/tex}
 {tex}x ^ { 2 } - ( \alpha + \beta ) x + \alpha \beta{/tex}

f(x)={tex}= x ^ { 2 } - \frac { 5 } { 2 } x + 1 = \frac { 1 } { 2 } \left( 2 x ^ { 2 } - 5x + 2 \right){/tex}
The polynomial whose zero are {tex}\alpha , \beta \text { is } 2 x ^ { 2 } - 5 x + 2{/tex}
Further, {tex}f ( x ) = \frac { 1 } { 2 } \left( 2 x ^ { 2 } - 5 x + 2 \right) = \frac { 1 } { 2 } \left( 2 x ^ { 2 } - 4 x - x + 2 \right){/tex}
{tex}= \frac { 1 } { 2 } [ 2 x ( x - 2 ) - ( x - 2 ) ]{/tex}
{tex}= \frac { 1 } { 2 } ( x - 2 ) ( 2 x - 1 ){/tex}
f(x) = 0 {tex}\Rightarrow \frac { 1 } { 2 } ( x - 2 ) ( 2 x - 1 ) = 0{/tex}
{tex}\therefore{/tex} for that x - 2 = 0 or 2x - 1 = 0
i.e., Either x = 2 or {tex}x = \frac { 1 } { 2 }{/tex}
{tex}\therefore{/tex} Zeros of polynomial are 2 and {tex} \frac { 1 } { 2 }{/tex}.

Let the cubic polynomial be ax³ + bx² +cx + d
and its zeroes be {tex}\alpha ,\beta {/tex} and {tex}\gamma{/tex}.
Then, {tex}\alpha + \beta + \gamma= 2 = \frac { - b } { a }{/tex}
{tex}\alpha \beta + \beta \gamma + \gamma \alpha = - 7 = \frac { C } { a }{/tex}
and {tex}\alpha \beta \gamma = - 14 = \frac { - d } { a }{/tex}
If a = 1, then b = -2, c = -7, d = 14
So, one cubic polynomial which fits the given
conditions is x³ - 2x² - 7x + 14.

Given polynomial is  f(x) = x³ - 3x² + x + 1
Let {tex} \alpha{/tex} = (a - b), {tex} \beta{/tex} = a and {tex} \gamma{/tex} = (a + b)
Now, {tex} \alpha + \beta + \gamma{/tex} = {tex} - \frac { ( - 3 ) } { 1 }{/tex}
⇒ (a - b) + a + ( a + b ) = 3
⇒ a - b + a + a+ b = 3
⇒ a + a + a = 3
⇒ 3a = 3
⇒ a = 3/3
⇒ a = 1
Also, {tex} \alpha \beta + \beta y + \gamma \alpha = \frac { 1 } { 1 }{/tex}
⇒ (a - b)a + a (a + b) + (a + b)(a - b) = 1 
⇒ a² - ab + a² +ab + a² - b² = 1
⇒ 3a² - b² = 1 ( ∵ a = 1)
⇒ 3(1)² - b² = 1( ∵ a = 1)
⇒ 3 - b² = 1
⇒ b² = 2
⇒ b = {tex} \pm \sqrt{2}{/tex}
Hence, a = 1 and b = {tex} \pm \sqrt{2}{/tex}

Let A(a, 2a), B(-2, 6) and C(3, 1) be the given points.
{tex}\Delta{/tex}Area of {tex}\Delta{/tex}ABC
{tex} = \frac{1}{2}\left\{ {a\left( {6 - 1} \right) + \left( { - 2} \right)\left( {1 - 2a} \right) + 3\left( {2a - 6} \right)} \right\}{/tex}
{tex} = \frac{1}{2}\left\{ {5a - 2 + 4a + 6a - 18} \right\}{/tex}
{tex} = \frac{1}{2}\left\{ {15a - 20} \right\}{/tex}
Since, Area of {tex}\Delta{/tex}ABC = 10 (given)
{tex} \Rightarrow \frac{1}{2}\left\{ {15a - 20} \right\} = 10{/tex}
{tex}\Rightarrow{/tex} 15a - 20 = 20
{tex}\Rightarrow{/tex} 15a = 40
{tex} \Rightarrow a = \frac{{40}}{{15}}{/tex}
{tex} \Rightarrow a = \frac{8}{3}{/tex}