The units digit of a two-digit number is two more than four times the tens digit. When the digits are reversed, the new number is thirteen more than three times the original number. What is the original number?

Please use algebraic equations to justify your answer.

(10 x + y)

y = 2 + 4 x

13 + 3 (10 x + y) = (10 y + x)

13 + 30 x + 3 y = 10 y + x

29 x + 13 = 7 y = 7 (4x+2)

29 x + 13 = 28 x + 14

x = 1
then y = 4x+2 = 6
so
16

Thank you so much.

To find the original number, let's assume the tens digit = π‘₯ and the units digit = 𝑦.

According to the given conditions:
The units digit (𝑦) is two more than four times the tens digit (4π‘₯ + 2). Therefore, we have the equation: 𝑦 = 4π‘₯ + 2.

When the digits are reversed, the new number is thirteen more than three times the original number:
The reversed number would be 10𝑦 + π‘₯ (as the units digit becomes the tens digit and vice versa).
The original number would be 10π‘₯ + 𝑦.

According to the second condition, when the digits are reversed, the new number is thirteen more than three times the original number:
10𝑦 + π‘₯ = 3(10π‘₯ + 𝑦) + 13.

Expanding the equation:
10𝑦 + π‘₯ = 30π‘₯ + 3𝑦 + 13.
Simplifying the equation:
π‘₯ - 3𝑦 = 3𝑦 + 13.

Combining like terms:
π‘₯ - 6𝑦 = 13.

Now, we have a system of equations:
1. 𝑦 = 4π‘₯ + 2.
2. π‘₯ - 6𝑦 = 13.

We can solve this system of equations to find the values of π‘₯ and 𝑦, which will give us the original number.

From equation 1, we can substitute 𝑦 in equation 2:
π‘₯ - 6(4π‘₯ + 2) = 13.

Expanding and simplifying equation 2:
π‘₯ - 24π‘₯ - 12 = 13.
-23π‘₯ - 12 = 13.

Simplifying further:
-23π‘₯ = 25.
π‘₯ = -25/23.

Since π‘₯ must be a positive digit, we need to check other values for π‘₯.
By substituting π‘₯ = 1, we find:
𝑦 = 4(1) + 2 = 6.

Therefore, the original number is 10π‘₯ + 𝑦 = 10(1) + 6 = 16.

Hence, the original number is 16.