There are two numbers. Four times the first exceeds seven times the second by 2. Sum of two times the first and three times the second in 92. Find the numbers.

Answer

To solve the problem, we need to find two numbers based on the given conditions. Let's denote the first number as \( x \) and the second number as \( y \).

### Step 1: Set up the equations based on the problem statement.
From the first part of the problem, we know:
- Four times the first number exceeds seven times the second number by 2. This can be expressed as:
\[ 4x - 7y = 2 \quad \text{(Equation 1)} \]

From the second part of the problem, we know:
- The sum of two times the first number and three times the second number is 92. This can be expressed as:
\[ 2x + 3y = 92 \quad \text{(Equation 2)} \]

### Step 2: Solve Equation 2 for one variable.
Let's solve Equation 2 for \( x \):
\[ 2x + 3y = 92 \]
Subtract \( 3y \) from both sides:
\[ 2x = 92 - 3y \]
Now divide both sides by 2:
\[ x = \frac{92 - 3y}{2} \quad \text{(Equation 3)} \]

### Step 3: Substitute Equation 3 into Equation 1.
Now we will substitute \( x \) from Equation 3 into Equation 1:
\[ 4\left(\frac{92 - 3y}{2}\right) - 7y = 2 \]
Multiply through by 2 to eliminate the fraction:
\[ 4(92 - 3y) - 14y = 4 \]
Distributing the 4:
\[ 368 - 12y - 14y = 4 \]
Combine like terms:
\[ 368 - 26y = 4 \]
Subtract 368 from both sides:
\[ -26y = 4 - 368 \]
\[ -26y = -364 \]
Now divide by -26:
\[ y = \frac{-364}{-26} = 14 \]

### Step 4: Substitute \( y \) back to find \( x \).
Now that we have \( y = 14 \), substitute it back into Equation 3 to find \( x \):
\[ x = \frac{92 - 3(14)}{2} \]
Calculate:
\[ x = \frac{92 - 42}{2} = \frac{50}{2} = 25 \]

### Conclusion:
The two numbers are:
\[ x = 25 \quad \text{and} \quad y = 14 \]

### Final Answer:
The two numbers are 25 and 14.

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