The sum of the two numbers is 11 and their product is 30, then the numbers are

Options

A. 8,3
B. 9,2
C. 7,4
D. 6,5

Answer

To solve the problem, we need to find two numbers whose sum is 11 and whose product is 30. Let's denote these two numbers as \( x \) and \( y \).

### Step 1: Set up the equations
From the problem, we have two equations:
1. \( x + y = 11 \) (Equation 1)
2. \( xy = 30 \) (Equation 2)

### Step 2: Express one variable in terms of the other
From Equation 1, we can express \( y \) in terms of \( x \):
\[ y = 11 - x \]

### Step 3: Substitute into the second equation
Now, we substitute \( y \) in Equation 2:
\[ x(11 - x) = 30 \]

### Step 4: Expand and rearrange the equation
Expanding the equation gives us:
\[ 11x - x^2 = 30 \]
Rearranging it leads to:
\[ -x^2 + 11x - 30 = 0 \]
Multiplying through by -1 to make the leading coefficient positive:
\[ x^2 - 11x + 30 = 0 \]

### Step 5: Factor the quadratic equation
Now we need to factor the quadratic equation \( x^2 - 11x + 30 \). We look for two numbers that multiply to 30 and add up to -11. The factors are:
\[ (x - 6)(x - 5) = 0 \]

### Step 6: Solve for \( x \)
Setting each factor to zero gives us:
\[ x - 6 = 0 \quad \Rightarrow \quad x = 6 \]
\[ x - 5 = 0 \quad \Rightarrow \quad x = 5 \]

### Step 7: Find the corresponding \( y \) values
Using \( y = 11 - x \):
- If \( x = 6 \), then \( y = 11 - 6 = 5 \).
- If \( x = 5 \), then \( y = 11 - 5 = 6 \).

### Conclusion
Thus, the two numbers are \( 5 \) and \( 6 \).