Let’s analyze the question from the uploaded image. The problem is as follows:

"In a two-digit number, the tens digit exceeds the units digit by 1, and the product of the digits is 30. Find the number."

Solution:

Let the tens digit be $x$ and the units digit be $y$.

Step 1: Form the equations

1. The tens digit exceeds the units digit by 1: $x = y + 1$

2. The product of the digits is 30: $x \cdot y = 30$

Step 2: Substitute $x = y + 1$ into $x \cdot y = 30$

Replace $x$ in the second equation: $(y + 1) \cdot y = 30$

Expand the equation: $y^2 + y = 30$

Rearrange into standard quadratic form: $y^2 + y - 30 = 0$

Step 3: Solve the quadratic equation

Use the quadratic formula: $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ Here, $a = 1$, $b = 1$, and $c = -30$. Substitute into the formula: $y = \frac{-1 \pm \sqrt{1^2 - 4(1)(-30)}}{2(1)}$ $y = \frac{-1 \pm \sqrt{1 + 120}}{2}$ $y = \frac{-1 \pm \sqrt{121}}{2}$ $y = \frac{-1 \pm 11}{2}$

Solve for the two possible values of $y$: $y = \frac{-1 + 11}{2} = 5 \quad \text{or} \quad y = \frac{-1 - 11}{2} = -6$

Since $y$ must be a positive digit, $y = 5$.

Step 4: Find $x$

Using $x = y + 1$: $x = 5 + 1 = 6$

Step 5: Form the number

The two-digit number is: $\boxed{65}$

Would you like a detailed verification or further explanation?

Related Questions:

1. What if the product of the digits was 56 instead of 30? What would the number be?
2. What are the possible two-digit numbers where the digits differ by 1?
3. How would the solution change if the product of the digits was negative?
4. Can you generalize this problem for $x - y = k$ and $x \cdot y = p$?
5. What happens if both digits are the same? Can such a case satisfy similar conditions?

Tip:

For problems involving digits, define them as variables and build equations based on their relationships (e.g., sum, difference, or product).