# More worked out problems

Example-1

Solve the equation 3

*x*âˆ’ 2*y*= 7Solution

Let us first find one particular solution and then we will extend that. Its not too difficult to find one solution using hit and trial. One such solution is (3, 1). Alternatively, we can find it using [
To find the next solution, let us re
Using the concepts of remainder, above written equation can be explained as:

LHS = A multiple of 3
RHS = A number that gives remainder 7 when divided by 2 which means a number that gives remainder 1 when divided by 2.

Obviously the next number will be obtained after the LCM (2, 3) = 6. To move ahead by 6, we will cover 2 multiples of 3 (2 Ã— 3) and 3 multiples of 2 (3 Ã— 2). Hence, the next solutions are (5, 4), (7, 7), (9, 10), (11, 13), etc.

*x*= (7 + 2*y*)/3]*-*write the equation as 3*x*= 2*y*+ 7.LHS = A multiple of 3

Obviously the next number will be obtained after the LCM (2, 3) = 6. To move ahead by 6, we will cover 2 multiples of 3 (2 Ã— 3) and 3 multiples of 2 (3 Ã— 2). Hence, the next solutions are (5, 4), (7, 7), (9, 10), (11, 13), etc.

**Note***â€”*Since it is a linear equation, there will be obviously infinite solution.

Example-2

Find all integral roots of the equation 21

*x*+ 48*y*= 5Solution

LHS is divisible by 3, but RHS is not divisible by 3. Hence, no integral solution is possible.

**Statement I**

- 1990
*x*âˆ’ 173*y*= 11 has no solution in integers for*x*and*y*. - 3
*x*âˆ’ 12*y*= 7 has no solution in integers for*x*and*y*.

Which of the following is true?

- Statement I is wrong and statement II is right.
- Statement I is right and statement II is wrong.
- Statement I and II both are right.
- Statement I and II both are wrong.

**Statement II**

The co-efficients in the equation are large enough to find a particular solution. However, we can see that the numbers 1990 and 173 are relatively prime and this helps us reach the conclusion.

Hence, we can find the solution of 1990

*x*âˆ’ 173*y*= 11 in integers for*x*and*y*.Hence, statement I is wrong.

**Statement II**

In the linear equation 3

*x*âˆ’ 12*y*= 7; co-efficients of*x*and*y*, i.e., 3 and 12, are not relatively prime. HCF of (3, 12) = 3. Also, constant 7 is not exactly divisible by 3. In other words, LHS is divisible by 3, but RHS is not divisible by 3.Hence, 3

*x*âˆ’ 12*y*= 7 has no solution for*x*and*y*in integers.Hence, statement II is wrong.

Therefore option (d) is true.