37a + 29b - 45 = 0
29a + 37b - 21 = 0

\(\cfrac { a }{ 29(-21)-37(-45) } =\cfrac { b }{ (-45)(29)-(-21)37 } '=\cfrac { 1 }{ 37(37)-29(29) } \)
\(\cfrac { a }{ -609+1665 } =\cfrac { b }{ -1305+777 } =\cfrac { 1 }{ 1369-841 } \)
\(\cfrac { a }{ 1056 } =\cfrac { b }{ -528 } =\cfrac { 1 }{ 528 } \)
\(\cfrac { a }{ 1056 } =\cfrac { b }{ -528 } \Rightarrow a\cfrac { 1056 }{ 528 } =2\)
\(\cfrac { b }{ -528 } =\cfrac { 1 }{ 528 } \Rightarrow b=\cfrac { -528 }{ 528 } =-1\)

ka + 10b = 15
8a + 5b = 9
\(\cfrac { k }{ 8 } =\cfrac { 10 }{ 5 } \neq \cfrac { 15 }{ 9 } \)
\(\Rightarrow \cfrac { k }{ 8 } =\cfrac { 10 }{ 5 } \Rightarrow \cfrac { k }{ 8 } =2\Rightarrow k=16\)

5x - 2y - 10 = 0
3x +y-17 = 0

\(\cfrac { x }{ 34+10 } =\cfrac { y }{ -30+85 } =\cfrac { 1 }{ 5+6 } \)
\(\cfrac { x }{ 44 } =\cfrac { y }{ 55 } \Rightarrow x=\cfrac { 1 }{ 11 } =4\)
\(\cfrac { x }{ 44 } =\cfrac { 1 }{ 11 } \Rightarrow y=\cfrac { 55 }{ 11 } =5\)

2x +y = 10 ----------(1)
5x - y = 11 -----------(2)
(1)+(2)\(\Rightarrow\)
2x+y =10
\(\cfrac { 5x-y=11 }{ 7x=21 } \)
\(x=\cfrac { 21 }{ 7 } =3\)
(l) \(\Rightarrow\) 2(3) +y =10
6+y = 10
y = 10 - 6 = 4
\(\therefore\) Solution is x = 3, y = 4

17(2 -x) - 5(x + 12) = 8 (1-7x)
34 - 17x - 5x - 60 = 8 - 56x
56x - 17x - 5x = 60 + 8 - 34
56x-22x = 34
34x = 34
x = 1

\(\cfrac { 2x+1 }{ 3x-2 } =\cfrac { 9 }{ 10 } \)
\(\cfrac { 2x+1 }{ 3x-2 } =\cfrac { 9 }{ 10 } \)
10(2x+ 1) = 9 (3x - 2)
20x + 10 = 27x -18
18+10 = 27x-20x
28 = 7x
\(x=\cfrac { 28 }{ 7 } =4\)

Fixed charge = Rs. F and
Additional charge = Rs. A
For 6 days F+(6 - 2)A = 22
F + 4 A = 22           ...(1)
For 4 days
F+ (4 - 2)A = 22
F+ 2A = 16          ...(2)
(1) \(\Rightarrow\) F + 4A = 22
(2) \(\Rightarrow\) \(\cfrac { F+2A=16 }{ 2A=6 } \)
\(\therefore\) Substitute A = 3 in (1)
F + 4(3) = 22
F = 22 -12
F = 10
\(\therefore\) Fixed charge = Rs. 10
Additional charge = Rs. 3

Let the age of the father be x .
Let the sum of the age of his sons be y
At present x = 2y \(\Rightarrow\) x - 2y = 0
After 20 years x + 20 = y + 40
x-y = 40 -20
x-y = 20             ...(1)
(1) \(\Rightarrow\) x-2y = 0
(2) \(\Rightarrow\) \(\cfrac { x-y=20 }{ -y=-20 } \)
y = 20
Substitute y = 20 in (1)
x-2(20) = 0
x-40 = 0
x = 40
\(\therefore\) The age of the father is 40 years

\(\cfrac { x }{ y } =\cfrac { 5 }{ 6 } \)
\(\Rightarrow\) 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 numerator be x
Denominator be y
x +y = 12 ...(1)
\(\cfrac { x }{ y+3 } =\cfrac { 1 }{ 2 } \)
2x = y+3
2x-y = 3 ...(2)
(1) X 2 \(\Rightarrow\)  2x +2y = 24
(2) \(\Rightarrow\) \(\cfrac { 2x-y=3 }{ 3y=21 } \)     ...(2)
y = 7
Substitute y = 7 in (1) .
x + 7 = 12
x = 12-7
x = 5
\(\therefore\) The fraction \(\cfrac { x }{ y } =\cfrac { 5 }{ 7 } \)

Let the number of 2 rupee coins be x
Let the number of 5 rupee coins be y
x+y = 80
2x+ 5y = 220
x +y- 80 = 0
2x + 5y - 220 = 0
For cross multiplication method, we write the co-efficients as

\(\cfrac { x }{ -220 } =\cfrac { y }{ -160+220 } =\cfrac { 1 }{ 5-2 } \)
\(\cfrac { x }{ 180 } =\cfrac { y }{ 60 } =\cfrac { 1 }{ 3 } \)
\(\therefore \ \cfrac { x }{ 180 } =\cfrac { 1 }{ 3 } \ \ \cfrac { y }{ 60 } =\cfrac { 1 }{ 3 } \)

x = 60 y = 20
\(\therefore\) No. of 2 rupee coins = 60
No. of 5 rupee coins = 20

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 Raman's age = x
Let the sum of his two sons age = y
now x = 3y \(\Rightarrow\)x - 3y = 0 .........(1)
After 5 years, x + 5 = 2(y + 10)
x + 5 = 2y+ 20
x - 2y = 20 - 5
x - 2y = 15    ...(2)  
Step (1): From equation (1) x = 3y
Step (2): Substitute x = 3y in (2)
3y- 2y = 15
y = 15
Step (3): Substitute y = 15 in (1)
x = 3y = 3 \(\times\) 15
x = 45
\(\therefore\) Raman's age is 45 years.

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

y = 5x + 2, y = 10 +2x