Consider three consecutive odd numbers. Three times the largest is equal to 19 less than 4 times the second. Find the three numbers.

Answer 1

Okay, Pricila, let's take a long, careful look at this problem.

Let x be the smallest of the three consecutive odd numbers. Then the others are x+2 and x+4.

Now the problem says that three times the largest, which here is x+4, so that would be 3(x+4)...

is equal to nineteen less than four times the second. Here, the second is x+2, so we have the equation

3(x+4)=4(x+2)-19

Now using the distributive law, we get 3x+12=4x+8-19, or 3x+12=4x-11.

Subtract 3x from both sides of the equation, and then add 11 to both sides and we get 23=x, or x=23.

Since x = the smallest of the three consecutive odd numbers, the three numbers are 23, 25 and 27.

You can verify these numbers by substituting these three values back into the original question.

I hope this helps.

Answer 2

Hi Pricila!

In the problem, we have to find three consecutive odd numbers where the first and last follow this relationship:

3x = 4y-19

Now we can also simplify this since we know that the smallest number is equivalent to the largest number plus 3, since we know that for any set of three consecutive odd numbers (say 11, 13, 15), the difference is 4. So we can rewrite y (the smaller number) as 4 less than the bigger number, or x-4:

3x = 4(x-4) - 19

We can then simplify this as done below:

3x = 4x - 16 - 19

3x = 4x - 35

35 = x

This means that the larger number x is 35, so the number set would be 31, 33, 35