Find two consecutive odd integers such that 44 more than the lesser is three times the greater.

Answer 1

Find two consecutive odd integers such that 44 more than the lesser is three times the greater.

Want two consecutive odd integers.

Call the first one "x", the next ODD integer is "x+2".

E.g., if you have the ODD integer 3, then next ODD integer is 5 (or 3 + 2).

1) So we have our integers x and x + 2

2) We want "44 more than the lesser is three times the greater."

so x + 44 = 3*(x+2)

=> x + 44 = 3x + 6

-x -x

______________

44 = 2x + 6

-6 -6

_________________

38 = 2x

/2 /2

_______________

19 = x (smaller integer), remember the bigger integer was equal to the smaller one + 2

so the bigger integer = 19+ 2 = 21

=>

Answer 2

Problem: Find two consecutive odd integers such that 44 more than the lesser is three times the greater.

Solution:

When beginning any problem, let's address the 'setup.'

Here, we need to express two consecutive odd integers. The best way to achieve this is to select n ∈ Z {the set of integers}, representing the first odd integer as 2n + 1, because we know that any integer times 2 is even. Since we are going with consecutive 'odd' integers, select 2n + 3 as the next odd integer. Try substituting any number into the two equations to verify that your result is odd!

Now we need to set up the equation.

From the instructions, 2n + 1 is the lesser, since (2n + 1) < (2n +3),

so (2n + 1) + 44 = 3(2n + 3).

Now solve for n, and once found, insert into the original two integers [which we will accomplish shortly].

(2n + 1) + 44 = 3(2n + 3) ≅ 2n + (1 + 44) = [(3*2n) + (3*3)] ≅ (2n + 45) = (6n + 9)

2n + 45 - 9 = 6n + 9 -9 ≅ 2n + 36 = 6n ≅ 2n -2n + 36 = 6n -2n ≅ 36 = 4n ≅ (1/4)*36 = 4n(1/4)

9 = n

Inserting n = 9 into both 2n + 1 and 2n + 3 yields 19 and 21 respectively.

Let's verify our work

19 + 44 = 63.

Now 63/21 = 3, which satisfies one of our original conditions!

Answer 3

Let's represent the two consecutive odd integers as "x" and "x + 2," where "x" is the lesser integer.

According to the given information, 44 more than the lesser integer is three times the greater integer. Mathematically, this can be written as:

x + 44 = 3(x + 2)

Now, let's solve this equation to find the values of "x" and "x + 2":

x + 44 = 3x + 6

Rearranging the equation:

3x - x = 44 - 6

2x = 38

Dividing both sides by 2:

x = 38 / 2

x = 19

Therefore, the lesser integer is 19, and the greater integer is 19 + 2 = 21.

Hence, the two consecutive odd integers that satisfy the given condition are 19 and 21.

Answer 4

Create values for the two consecutive odd integers

First odd integer = x

Second odd integer = x+2

Make equation and solve

x+44 = 3(x+2)

x+44 = 3x+6

44 = 2x+6

38 = 2x

19 = x

Therefore, the smaller number is 19 and the greater number is 21