Let for one Indians the' rate of working be \(\frac { 1 }{ x } \)
Let for one Chinese the rate of working be \(\frac { 1 }{ y } \)
\(\therefore \ \cfrac { 4 }{ x } +\cfrac { 4 }{ y } =\cfrac { 1 }{ 3 } \)
\(\cfrac { 2 }{ x } +\cfrac { 5 }{ y } =\cfrac { 1 }{ 4 } \)
put \(\cfrac { 1 }{ x } =a,\cfrac { 1 }{ y } =b\)
\(4a+4b=\cfrac { 1 }{ 3 } \)
\(4a+4b-\cfrac { 1 }{ 3 } =0\)
\(2a+5b=\cfrac { 1 }{ 4 } \)
\(2a+5b-\cfrac { 1 }{ 4 } =0\)
For cross multiplication method, we write the co-efficients as

\(\cfrac { a }{ 5 } =\cfrac { b }{ 2 } =\cfrac { 1 }{ 20-8 } \)
\(\cfrac { 3a }{ 2 } =3b={ \cfrac { 1 }{ 12 } }\)
\(\therefore \ \cfrac { 3a }{ 2 } =\cfrac { 1 }{ 12 } \quad 3b=\cfrac { 1 }{ 36 } \)
\(a=\cfrac { 1 }{ 12 } \times \cfrac { 2 }{ 3 } =\cfrac { 1 }{ 18 } \quad b=\cfrac { 1 }{ 36 } \)
\(\therefore\) 1boy can do the same piece of work = \(\cfrac { 1 }{ y } =b=\cfrac { 1 }{ 36 } =36days\)
1 Indians can do the piece of work = \(\cfrac { 1 }{ x } =a=\cfrac { 1 }{ 18 } =18days\)

Let basic fare = Rs. B and
Let reservation charge =Rs. R
\(\cfrac { B }{ 2 } +R=half\ ticket\)
B + R = full ticket
B + R = 216     ....(1)
\(\left( \cfrac { B }{ 2 } +R \right) +\left( B+R \right) =327\)
\(\cfrac { B }{ 2 } +R+216=327\)
\(\cfrac { B }{ 2 } +R=111\)
B+2R = 222
B+2R = 222                                ...(1)
(1) \(\Rightarrow\) B+R = 216
(2) \(\Rightarrow\) \(\cfrac { B+2R=222 }{ -R=-6 } \)
R = 6
Substitute R = 6 in (1)
B+6 = 216
B = 216 - 6
B = 210
\(\therefore\) The basic first class full fare = Rs. 210
The half fare = Rs. 105
The reservation charge = Rs. 6

\(\cfrac { x }{ y } =\cfrac { 5 }{ 6 } \)
\(\Rightarrow\) 6x = 5y
6x - 5y = 0 .......(1)
\(\cfrac { x-8 }{ y-8 } =\cfrac { 4 }{ 5 } \)
\(\Rightarrow\) 5(x - 8) = 4(y - 8)
5x-40 = 4y-32
5x -4y = 40 - 32
5x - 4y = 8 ------ (2)
(1) x 5 \(\Rightarrow\) 30x-25y = 0
(2) x 6 \(\Rightarrow\) \(\cfrac { 30x-24y=48 }{ -y=-48 } \)    ...(2)
Substitute y = 48 in (1)
6x- 5(48) = 0
6x-240 = 0
6x = 240
\(x=\cfrac { 240 }{ 6 } =40\)
y = 48
\(\cfrac { x }{ y } =\cfrac { 40 }{ 48 } =\cfrac { 5 }{ 6 } \)
\(\therefore\) The number are in the Ratio 5 : 6

Let the actual price of a T.V. = x
Let the actual price of a Fridge = y
\(\cfrac { 5 }{ 100 } x+\cfrac { 10 }{ 100 } y=2000\)
\(\cfrac { 5 }{ 100 } (x+2y)=2000\)
\(x+2y=2000\times \cfrac { 100 }{ 5 } \)
x+2y = 40000     ...(1)
\(\cfrac { 10 }{ 100 } x-\cfrac { 5 }{ 100 } y=1500\)
\(\cfrac { 5 }{ 100 } (2x-y)=1500\)
\(2x-y=1500\times \cfrac { 100 }{ 5 } \)
2x -y = 30000
(1) X 2 \(\Rightarrow\) 2x + 4y = 80000   
(2) \(\Rightarrow\) \(\cfrac { 2x-y=30000 }{ 5y=50000 } \)
y = 10000
Substitute y = 10000 in (1)
x + 2(10000) = 40000
x + 20000 = 40000
x = 40000 - 20000
x = 20000
\(\therefore\) Actual price of T.V = Rs. 20000
Actual price of Fridge = Rs. 10000

Let x, y be the speed of the two cars. If the two cars travel toward's each other they will meet in 1 hr. The distance between them d = 100; \(\frac { d }{ s } \) =t
i.e.,\(\cfrac { 100 }{ x+y } =1\Rightarrow x+y=100\)  .....(1)
If the two cars travel in the same direction they will meet in 2hrs.
    ...(2)
(1) \(\Rightarrow\) x + y = 100
y = -x + 100

(2) \(\Rightarrow\) x - y = 50
-y = -x + 50
y = x - 50

The points to be plotted
(-100, -50), (0, -50),
x + y =100 ...(1)
Put x = 0 in (1), then 0 + y = 100 \(\Rightarrow\)y = 100
A (0, 100) is a point on (1)
Put y = 0 in(1), then x + 0 = 100 \(\Rightarrow\) x = 100
B (100, 0) is another point on (1)
Plot A & B Join them to produce the line (1)
Similarly by x - y = 50
Put x = 0 in (2), then 0 - y = 50 
\(\Rightarrow\)y = 50
P (0, -50) is a point on (2)
Put y = 0 in (2), then x - 0 = 5
\(\Rightarrow\) x = 50
Q (50, 0) is another point on (2)
Plot P & Q Join them to produce the line (2)
The point of intersection (75, 25) of the two lines (1)& (2) is the solution.
\(\therefore\)The solution i.e., the speed of the two cars x and y is given by x = 75 km and y = 25 km

x⁵-16x = x[x⁴-16] = x[(x²)²-4²]
= x(x²+4)(x²-4)
= x(x²+4)(x+2)(x-2)

(a+b+c) (a²+b²+c²-ab-bc-ca) = a³+b³+c³-3abc
If a+b+c = 0, then a³+b³+c³ = 3abc
ஃ 7-10+3 = 0
⇒ 7³-10³+3³ = 3\(\times\)7\(\times\)-10\(\times\)3
= 9\(\times\)-70 = -630

(-p+2q+3r)² 
= (-p)²+(2q)²+(3r)²+2(-p)(2q)+2(2q)(3r)+2(-p)(3r) 
= p²+4q²+9r²-4pq+12qr- 6pr

(m + 5)(m - 8)   [we have (x + a)(x - b) = x²- (a - b)x - ab]
(m + 5)(m - 8) = m²+(5 - 8)m - (5)(8)   put [x = m, a = 5, b = 8]
= m²- 3m - 40

Number of zeros = 1 (The curve is intersecting the x-axis at one point)