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?
16
How did you get this answer (what are the steps involved)?
Two more than four times more than the tens digit
The tens digit cannot be 2, so it must be 1.
What are the algebraic equations you used to find this answer?
I didn't use any algebraic equations.
Oh, well...do you think you'd be able to come up with any?
Please repost your question and ask for an algebraic equation that will solve it. Perhaps a math tutor can help you.
4x + 2 < 9
To solve this problem, we can represent the original two-digit number as (10a + b), where 'a' represents the tens digit and 'b' represents the units digit. The reversed number can be represented as (10b + a).
According to the problem, the units digit (b) is two more than four times the tens digit (a). This can be expressed as:
b = 4a + 2
Additionally, the reversed number is thirteen more than three times the original number. This can be expressed as:
10b + a = 3(10a + b) + 13
Now, let's solve this system of equations to find the values of a and b.
Substituting the value of b from the first equation into the second equation, we have:
10(4a + 2) + a = 3(10a + 4a + 2) + 13
Expanding and simplifying both sides of the equation:
40a + 20 + a = 30a + 12 + 6 + 13
41a + 20 = 30a + 31
Moving the terms involving 'a' to one side and the constant terms to the other side:
41a - 30a = 31 - 20
11a = 11
Dividing both sides of the equation by 11:
a = 1
Substituting the value of a = 1 into the first equation to find b:
b = 4(1) + 2
b = 6
Therefore, the original number is 10a + b = 10(1) + 6 = 16.