The second of two numbers is five more than the twice first. The sum of the numbers is 44. Find the numbers. Show all work.

1st number --- x

2nd number --- y

"The second of two numbers is five more than the twice first." ---> y = 2x+5

"the sum of the numbers is 44" ---> x+y = 44

x+y = 44
x + (2x+5) = 44
x + 2x + 5 = 44
3x = 39
x = 13

1st is 13, 2nd is 31

check:
is 31 five more than twice 13 ? YES
is 13 + 31 = 44 ? , YES

Let's begin by assigning variables to the numbers.

Let's call the first number "x" and the second number "y".

From the problem, we know that the second number is five more than twice the first number. This can be expressed as:

y = 2x + 5

We are also given that the sum of the numbers is 44. So we can write an equation for the sum of the numbers:

x + y = 44

Now we have a system of two equations with two variables.

To solve this system, we can use the method of substitution. We can rearrange the first equation to solve for x:

y = 2x + 5
=> 2x = y - 5
=> x = (y - 5)/2

We can substitute this value of x into the second equation:

x + y = 44
=> ((y - 5)/2) + y = 44
=> (y - 5) + 2y = 88
=> 3y - 5 = 88
=> 3y = 93
=> y = 31

Now, we can substitute this value of y back into the first equation to find x:

x = (y - 5)/2
=> x = (31 - 5)/2
=> x = 26/2
=> x = 13

Therefore, the first number is 13 and the second number is 31.