**Problem Statement:**

Let's assume the first number as 'x' and the second number as 'y'.

According to the given conditions:

1. The sum of twice the first number and thrice the second number is 92:
2x + 3y = 92

2. Four times the first number exceeds seven times the second number by 2:
4x = 7y + 2

We need to find the values of 'x' and 'y' that satisfy both of these equations simultaneously.

**Solution:**

To solve this system of equations, we can use the method of substitution or elimination. Let's use substitution.

**Step 1:** Solve one equation for one variable in terms of the other variable.

From equation (2), we can isolate 'x' as follows:

4x = 7y + 2
x = (7y + 2) / 4

**Step 2:** Substitute the expression obtained in step 1 into the other equation.

Substituting the expression for 'x' in equation (2) into equation (1), we have:

2((7y + 2) / 4) + 3y = 92

Simplifying the equation:

7y + 2 + 6y = 92
13y + 2 = 92
13y = 90
y = 90 / 13

Therefore, the value of 'y' is 6.92 (rounded to two decimal places).

**Step 3:** Substitute the obtained value of 'y' into either of the original equations to find the value of 'x'.

Using equation (2), we substitute y = 6.92:

4x = 7(6.92) + 2
4x = 48.44 + 2
4x = 50.44
x = 50.44 / 4
x = 12.61 (rounded to two decimal places)

Therefore, the value of 'x' is 12.61 (rounded to two decimal places).

**Conclusion:**

The two numbers that satisfy the given conditions are approximately 12.61 and 6.92.

25 and 14 are required numbers
the equation is 2a+3b=92
and 4a -7b =2
solve both equation
2a =92-3b
a=92-3b÷2
put a in 4a-7b =2