Finding consecutive odd integers

Question

Find the three consecutive odd integers such that the sum of twice the first and three times the second is 43 more than twice the third

Answer 1

Here's what we know:

a = b - 2

c = b + 2 (consecutive odd integers mean that the integers are 2 units away on a number line, and b is the number in the middle)

2a + 3b = 2c + 43 (twice the first plus thrice the second equals twice the third plus another 43)

We have values for variables, so let's plug them in:

2(b - 2) + 3b = 2(b + 2) + 43

Distribute:

2b - 4 + 3b = 2b + 4 + 43

simplify:

5b - 4 = 2b + 47

Now to isolate b. Subtract 2b:

5b - 4 - 2b = 2b + 47 - 2b

3b - 4 = 47

add 4:

3b - 4 + 4 = 47 + 4

3b = 51

Divide 3:

3b / 3 = 51 / 3

b = 17

So this means a = 15 and c = 19

check:

2 * 15 + 3 * 17 = 2 * 19 + 43

30 + 51 = 38 + 43

81 = 81