Let the number of matches lost = x.
Then number of matches won = x + 2.
Given that, 5(x + 2) + (−3)x = 50
5x + 10 − 3x = 50
2x + 10 = 50
2x + 10 − 10 = 50 − 10 (Subtract 10 on both sides)
2x = 40
Divide by 2 on both the sides, x = 20.
Therefore, the number of
matches played is x + x + 2 = 2x + 2 = (2 x 20) + 2
= 40 + 2
= 42

Since we do not know the number of students who get down at the first stop, let us take the number as x. The number of students get down at the second bus stop is 2x.
x + 2x + 6 + 3 = 60
(1 + 2) x + 9 = 60
3x + 9 = 60
3x + 9 – 9 = 60 − 9 [Subtract 9 on both sides]
3x = 51
\(\frac { 3x }{ 3 } =\frac { 51 }{ 3 } \) [Divide by 3 on both sides]
Therefore, x = 17.
Thus, the number of students got down in the first bus stop is 17.

Let the number of correct answers be x
Thus the number of wrong answers = 60 − x
Then, 4x − 1 (60 − x) = 130
4x − 60 + x = 130
4x + x − 60 + 60 = 130 + 60 [Add 60 on both sides]
5x + 0 = 190
5x = 190
Divide by 5 on both the sides,
\(\frac { 5x }{ 5 } =\frac { 190 }{ 5 } \)=38
x = 38
Hence, the number of correct answer is 38.

Let the number of notes in each denomination be x.
Then x + 5x + 10x = 960
(1 + 5 + 10) x = 960
16x = 960
Divide by 16 on both the sides,
\(\frac { 16x }{ 16 } =\frac { 960 }{ 16 } \) = 60
Therefore, x = 60
Thus, the number of notes in each denomination is 60.

The numbers are natural and consecutive. Let the numbers be x and x + 1.
Given that,
x + (x + 1) = 75
2x + 1 = 75
2x + 1 − 1 = 75 − 1 [Subtract 1 on both sides]
2x + 0 = 74
\(\frac {2x }{2 } =\frac { 74 }{ 2 } \) [Divide 2 at both sides]
x = 37 and x + 1 = 38
Therefore, the required numbers are 37 and 38.