The sum of the digits of a positive 2-digit number is 12. The units digit is 3 times the tens digit. Find the number

93

Step-by-step explanation:

Let’s call the tens digit x and the unit digit y. From the question we know that x + y = 12 and that x = y^2, giving us the following system of equations:

x + y = 12

x - y^2 = 0

Let’s multiply the second one by -1, to get rid of one of the unknowns in the system.

x + y = 12

-x + y^2 = 0

By adding together the two, we get y^2 + y = 12. As that is a quadratic equation, let’s make it of the form Ay^2 + By + C = 0, so subtracting 12 from each side gives as y^2 + y - 12 = 0. Then let’s solve the quadratic equation to find out the two values of y we get out of it, one of which is legal in the above system of equations.

y = (-B +/- sqrt(B^2–4AC))/2A

This time, both A and B are 1, while C is -12. By entering them in the above formula, we get the inner part of the square root to be 49, and thus the square root to give 7. So solving y = (-1 +/- 7)/2 gives two alternatives, either y = -4 (impossible, as it has to be a digit, ie has to be between 0 and 9 inclusive) or y = 3.

Then let’s return to the system of equations, and as both digits appear without any modifiers in the first one, let’s use it to find x with. So x + 3 = 12, x = 12 - 3, x = 9.

As x is the tens digit and y the unit digit per how the variables were chosen, we can with confidence say that the number is 93.

a. -0.3 > -0.33