Question

Find a 2 digit number such that 2 times the tens digit is three less than thrice the unit digit and four times the number is 99 greater than the number obtained by reversing the digit solve in simultaneous equation

Answer 1

To solve for the two-digit number, we use the provided conditions to form simultaneous equations and solve them step by step to find the values of the tens and unit digits.

Step-by-step explanation:

Let's denote the two-digit number as AB, where A is the tens digit, and B is the unit digit. The conditions given in the problem lead us to two equations:

1. 2A = 3B - 3: 2 times the tens digit (A) is three less than thrice the unit digit (B).
2. If the digits of the number AB are reversed, we get BA. The problem states that 4 × AB = BA + 99. We can represent AB as 10A + B and BA as 10B + A.

Using these equations, we can form a system of simultaneous equations:

1. 2A = 3B - 3
2. 40A + 4B = 10B + A + 99

Now, we need to solve these equations simultaneously:

1. From the first equation, express A in terms of B: A = (3B - 3) / 2
2. Substitute A in the second equation and solve for B:

40((3B - 3) / 2) + 4B = 10B + (3B - 3)/2 + 99

After simplifying, we find the solution for B and subsequently find A using the first equation. The two-digit number can then be determined.