When the smaller of 2 consecutive integers is added to three times the larger, the result is 43.?

Question

What is the numbers?

Answer 1

Consecutive integer problems can always be set up by the following:

Let n = the smaller integer. Then
n + 1 = the larger integer.

According to the information in the problem we get

n + 3(n + 1) = 43
n + 3n + 3 = 43
4n + 3 = 43
4n = 40
n = 10 and n + 1 = 11

Check: 10 + 3*11 = 10 + 33 = 43. Bingo, we have the correct integers.

An extension of this type of problem would involve 3 or 4 consecutive integers. So, extend the setup so that n + 2 = the third integer, n + 3 = 4th integer, etc.

Consecutive odd (or even) integers would have the set as follows:

Let n = 1st odd (even) integer
n + 2 = 2nd odd (even) integer
n + 4 = 3rd odd (even) integer, etc.

Answer 2

You have 2 integers (x and y).
Since they're consecutive, y = x+1.

Your statement can be rewritten as x + 3y = 43.
Since y = x + 1:
x + 3(x+1) = 43
x + 3 x + 3 = 43
4x + 3 - 3 = 43 -3
4x/4 = 40/4
x = 10

y = x + 1
y = 10 + 1
y = 11

Answer 3

x = first integer
x+1 = second integer (larger)
x + 3(x+1) = 43
x + 3x + 3 = 43
4x + 3 = 43
4x + 3 - 3 = 43 - 3
4x = 40
4x/4 = 40/4
x = 10
x +1 = 11
10 + 3(11) = 43
10 + 33 = 43
43 = 43

Answer 4

10 and 11

Answer 5

We know:
x + 3y = 43
and
y = x +1

For the y, plug in x+1, and get:

x + 3(x+1) = 43
x + 3x +3 = 43
4x = 40
x = 10, and therefore y = 11

Answer 6

So it looks like:

(x-1)+3x=43
4x=44
x=11

and (x-1)=10