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 the fixed charge of a taxi be x and the charge per km be Rs. y
According to the given conditions,
\(\therefore\) x + 10y = 75 ---------- (1)
x + 15y = 110 ---------- (2)
(1) \(\Rightarrow\) x+10y = 75
(2) \(\Rightarrow\) \(\cfrac { x+15y=110 }{ -5y=-35 } \)
y = 7
Substitute y = 7 in (1)
x + 10(7) = 75
x+ 70 = 75
x = 75 -70 = 5
\(\therefore\) Amount that a person will have to pay for travelling a distance of 25 km
x + 25y = 5 + 25 (7)
= 5 + 175
= Rs. 180
\(\therefore\)  Amount that a person will have to pay for travelling a distance of 25 km is Rs. 180.

\(\cfrac { x }{ y } =\cfrac { 5 }{ 6 } \)
\(\Rightarrow\) 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 two digit number be x y
Its place value = 10x +y
After interchanging the digits the number will be y x
Its place value = 10y + x
Their sum = 10x +y + 10y + x = 110
11x + 11y = 110
x +y = 10 ----- (1)
If 10 is subtracted from the first number, the new number is 10x +y - 10
The sums of the digits of the first number is x +y.
Its 4 more than 5 times is = 5(x +y) + 4
10x + y -10 = 5x + 5y + 4
10x + y - 5x - 5y = 4 + 10
5x - 4y = 14    ....(2)
(10x 5 \(\Rightarrow\) 5x + 5y = 50
(2)\(\Rightarrow\) \(\cfrac { 5x-4y=14 }{ 9y=36 } \)

Substitute y = 4 in (1)
x + 4 = 10
x = 10 - 4
x = 6
\(\therefore\) The first number is 64

Let the man's present age = x
Five years ago his age is = x - 5
Let his son's age be = y
5 years ago his son's age = y - 5
\(\therefore\) x - 5 = 7(y- 5)
x - 5 = 7y-35
x - 7y = -35 + 5
x - 7y = -30    ....(1)
After 5 years, man's age will be = x + 5
His son's age will be = y + 5
\(\therefore\) x + 5 = 4(y + 5)
x+ 5 = 4y+ 20
x-4y = 20-5
\(\Rightarrow\) x-4y = 15
(1) \(\Rightarrow\) x-7y = -30
(2) \(\Rightarrow\) \(\cfrac { x-4y=15 }{ 3y\quad =\quad 45 } \)
y= 15
Substitute y = 15 in (1)
x -7 (15) = -30
x-105 =- 30
x = -30 + 105
x = 75
\(\therefore\) Man's Age = 75, His son's Age = 15

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 2 hrs.
    ...(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

(i) \(\cfrac { 2x+2 }{ 3 } =\cfrac { 3x-6 }{ 5 } \)
5(2x+2) = 3(3x - 6) [By cross multiplication]
10x + 10 = 9x - 18
10x = 9x - 18 - 10
10x = 9x-28
10x- 9x = -28
x = -28
(ii)

2 = 4 (x + 1)-x
2 = 4x+4-x
3x + 4 = 2'
3x+4-4 = 2-4
3x =-2

\(3\left( \cfrac { 1 }{ 4 } +1 \right) =3\left( 5-\cfrac { 1 }{ 4 } \right) -2\left( 5+\cfrac { 1 }{ 4 } \right) \)

\(3\left( \cfrac { 5 }{ 4 } \right) =3\left( \cfrac { 20-1 }{ 4 } \right) -2\left( \cfrac { 20+1 }{ 4 } \right) \)
\(\cfrac { 15 }{ 4 } =\cfrac { 57 }{ 4 } -\cfrac { 42 }{ 4 } \)
\(\cfrac { 15 }{ 4 } =\cfrac { 15 }{ 4 } \)
Yes \(\cfrac { 1 }{ 4 } \) is a solution of the given equation

(i) 8x- 3y = 12      ...(1)
5x-2y = 7      ..(2)
8x- 3y-12 = 0
5x-2y- 7 = 0
For cross multiplication method, we write the co-efficients as

\(\cfrac { x }{ (-3)(-7)(-2)(-12) } =\cfrac { y }{ (-12)(5)-(-7)(8) } =\cfrac { 1 }{ (8)(-2)-(5)(-3) } \)
\(\cfrac { x }{ 21-24 } =\cfrac { y }{ -60+56 } =\cfrac { 1 }{ -16+15 } \)
\(\cfrac { x }{ -3 } =\cfrac { y }{ -60+56 } =\cfrac { 1 }{ -16+15 } \)
\(\therefore \ \cfrac { x }{ 3 } =\cfrac { 1 }{ -1 } \ \cfrac { y }{ -4 } =\cfrac { 1 }{ -1 } \)
x = 3 , y = 4
\(\therefore\)  Solutions: x = 3; y = 4
(ii) 6x+ 7y-11 = 0
5x+ 2y-13 = 0
For cross multiplication method, we write the co-efficients as
\(\cfrac { x }{ -91-(-22) } =\cfrac { y }{ -55-(-68) } =\cfrac { 1 }{ 12-35 } \)
\(\cfrac { x }{ -91+22 } =\cfrac { y }{ -55+78 } =\cfrac { 1 }{ -23 } \)
\(\cfrac { x }{ -69 } =\cfrac { y }{ 23 } =\cfrac { 1 }{ -23 } \)
\(\cfrac { x }{ -69 } =\cfrac { 1 }{ -23 } \quad \cfrac { y }{ 23 } =\cfrac { 1 }{ -23 } \)

x = 3, y= -1
\(\therefore\)  x = 3; y= -1
(iii) 

\(\cfrac { 2 }{ x } +\cfrac { 3 }{ y } -5=0\)
\(\cfrac { 3 }{ x } -\cfrac { 1 }{ y } +9=0\)
In (1), (2) Put \(\cfrac { 1 }{ x } =a,\cfrac { 1 }{ y } =b\)
(1) \(\Rightarrow\) 2a + 3b - 5 = 0
(2) \(\Rightarrow\) 3a - b + 9 = 0
For cross multiplication method, we write the co-efficients as
\(\cfrac { a }{ (3)(9)-(-1)(-5) } =\cfrac { b }{ (-5)(3)-(9)(2) } =\cfrac { 1 }{ (2)(-1)-(3)(3) } \)
\(\cfrac { a }{ 27-5 } =\cfrac { b }{ -15-18 } =\cfrac { 1 }{ -2-9 } \)
\(\cfrac { a }{ 22 } =\cfrac { b }{ -33 } =\cfrac { 1 }{ -11 } \)
\(\therefore \ \cfrac { a }{ 22 } =\cfrac { 1 }{ -11 } \cfrac { b }{ -33 } =\cfrac { 1 }{ -11 } \)

a = -2    b = 3
\(a=\cfrac { 1 }{ x } =-2\quad b=\cfrac { 1 }{ y } =3\)
\(\therefore \ x=\cfrac { -1 }{ 2 } \ \ y=\cfrac { 1 }{ 3 } \)
solution \(\ x=\cfrac { -1 }{ 2 } \ \ y=\cfrac { 1 }{ 3 } \)

5b = 75  which gives b = 15
Put b = 15 in (2):
a + 2(15) = 35 which simplifies to a = 5
Thus the solution is a = 5, b = 15.
Verification :
4a+3b = 65 ...(1)
4(5)+3(15) = 65
20 + 45 = 65
65 = 65 True
a + 2b = 35 ...(2)
5 + 2(15) = 35
5+30 = 35
35 = 35 True

The given equation is 2y – 3x = 12
\(\Rightarrow\) 2y = +3x + 12
\(\Rightarrow \ \frac { 2y }{ 2 } =\frac { 3x+12 }{ 2 } \)
\(\Rightarrow \ y=\frac { 3x }{ 2 } +\frac { 12 }{ 2 } \)
\(\Rightarrow \ y=\frac { 3 }{ 2 } x+6\)
compare with, y=mx + c
Slope m = \(\frac { 3 }{ 2 } \), y-intercept c = 6

The given system of equations can be written as
2x + 7y − 5 = 0
−3x + 8y +11 = 0
For the cross multiplication method, we write the coefficients as

\(\frac { x }{ (7)(11)-(8)(-5) } =\frac { y }{ (-5)(-3)-(11)(2) } =\frac { 1 }{ (2)(8)-(-3)(7) } \)
\(\frac { x }{ 77+40 } =\frac { y }{ 15-22 } =\frac { 1 }{ 16+21 } \)
\(\frac { x }{ 117 } =\frac { y }{ -7 } =\frac { 1 }{ 37 } \)
\(\frac { x }{ 117 } =\frac { 1 }{ 37 } ,\frac { y }{ -7 } =\frac { 1 }{ 37 } \)
Hence the solution is \(\left( \frac { 117 }{ 37 } ,\frac { -7 }{ 37 } \right) \)
Verification:
2x + 7y -5 = 0 .....(1)
\(2\left( \frac { 117 }{ 37 } \right) +7\left( \frac { -7 }{ 37 } \right) -5=0\)
\(\frac { 234 }{ 57 } -\frac { 49 }{ 37 } -5=0\)
\(\frac { 185 }{ 37 } -5=0\)
5 - 5 =0 True
3x+8y+11 = 0 ...(2)
\(3\left( \frac { 117 }{ 37 } \right) +8\left( \frac { -7 }{ 37 } \right) +11=0\)
\(\frac { -351 }{ 37 } -\frac { 56 }{ 37 } +11=0\)
\(\frac { -407 }{ 37 } +11=0\)
-11 + 11 = 0 True

The given system of equations are 8x − 3y = 5xy ...(1)
6x − 5y = −2xy ...(2)
Observe that the given system is not linear because of the occurrence of xy term. Also note that if x =0, then y =0 and vice versa. So, (0,0) is a solution for the system and any other solution would have both x \(\neq \) 0 and y \(\neq \) 0
Let us take up the case where x \(\neq \) 0 and y \(\neq \) 0
Dividing both sides of each equation by xy,
\(\frac { 8x }{ xy } -\frac { 3y }{ xy } =\frac { 5xy }{ xy } \)
\(\frac { 6x }{ xy } -\frac { 5y }{ xy } =\frac { -2xy }{ xy } \)
Let \(a=\frac { 1 }{ x } ,b=\frac { 1 }{ y } \)
We get, \(\frac { 8 }{ y } -\frac { 3 }{ x } =5\)  .....(3)
\(\frac { 6 }{ y } -\frac { 5 }{ x } =-2\)  .......(4)
(3)&(4) respectively become, 8b − 3a = 5 ...(5)
b − 5a = −2 ...(6)
which are linear equations in a and b.
To eliminate a, we have,  (5) \(\times\) 5\(\Rightarrow\) 40b −15a = 25 .....(7)
(6) × 3\(\Rightarrow\) 18b −15a = −6 .....(8)
Now proceed as in the previous example to get the solution\(\left( \frac { 11 }{ 23 } ,\frac { 22 }{ 31 } \right) \).
Thus, the system have two solutions \(\left( \frac { 11 }{ 23 } ,\frac { 22 }{ 31 } \right) \) and (0,0).

Let x be the digit at ten’s place and y be the digit at unit place.
Given that x + y = 5 …… (1)

Given, Original number − reversing number = 27
(10x + y) − (10y + x) = 27
10x − x + y −10y = 27
9x − 9y = 27
\(\Rightarrow\) x − y = 3 ... (2)
Also from (1), y = 5 – x ... (3)
Substitute (3) in (2) to get x − (5 − x) = 3
x − 5 + x = 3
2x = 8
x = 4
Substituting x = 4 in (3), we get y = 5 − x = 5 − 4
y = 1
Thus, 10x + y = 10 × 4 +1 = 40 +1 = 41.
Therefore, the given two-digit number is 41.
Verification :
sum of the digits = 5
x + y = 5
4 + 1 = 5
5 = 5 true
Original number – reversed number = 27
41 - 14 = 27
27 = 27 true

Given x + y = 5 ...(1)
2x – y = 4 ...(2)
To draw the graph (1) is very easy. We can find the x and y intercepts and thus two of the points on the line (1).
When x = 0, (1) gives y = 5.
Thus A(0,5) is a point on the line.
When y = 0, (1) gives x = 5.
Thus B(5,0) is another point on the line.
Plot A and B; join them to produce the line (1).
To draw the graph of (2), we can adopt the same procedure.
When x = 0, (2) gives y = −4.
Thus P(0,−4) is a point on the line.
When y = 0, (2) gives x = 2.
Thus Q(2,0) is another point on the line.
Plot P and Q; join them to produce the line (2).
The point of intersection (3, 2) of lines (1) and (2) is a solution.
The solution is the point that is common to both the lines. Here we find it to be (3,2). We can give the solution as x = 3 and y = 2.