How do you find 3 consecutive odd integers where the sum is -141?

2 Answers
May 24, 2018

Answer:

See a solution process below:

Explanation:

Let's call the first odd integer: #n#

Then the next two consecutive odd integers will be:

#n + 2# and #n + 4#

We can then write this equation and solve for #n#;

#n + (n + 2) + (n + 4) = -141#

#n + n + 2 + n + 4 = -141#

#n + n + n + 2 + 4 = -141#

#1n + 1n + 1n + 2 + 4 = -141#

#(1 + 1 + 1)n + (2 + 4) = -141#

#3n + 6 = -141#

#3n + 6 - color(red)(6) = -141 - color(red)(6)#

#3n + 0 = -147#

#3n = -147#

#(3n)/color(red)(3) = -147/color(red)(3)#

#(color(red)(cancel(color(black)(3)))n)/cancel(color(red)(3)) = -49#

#n = -49#

Therefore:

#n + 2 = -49 + 2 = -47#

#n + 4 = -49 + 4 = -45#

The 3 consecutive odd integers summing to -141 are: -49, -47, -45

Solution Check:

#-49 - 47 - 45 = -96 - 45 = -141#

May 24, 2018

Answer:

the three numbers are -49,-47 and -45

Explanation:

Let the first odd number be #x#
Then, the consecutive odd number is #x+2#
The last consecutive odd number is #x+4#

Why do we add 2 to each odd number?
Well, think of any three consecutive odd numbers.

135,137,139
186395,186397,186399

From the above, you can see that when you minus 2 consecutive numbers, you will always get a difference of 2

Since their sum equals to #-141#,

#x+(x+2)+(x+4)=-141#
#3x+6=-141#
#3x=-147#
#x=-49#

Therefore, your first odd number is -49
Your second number is #-49+2=-47#
Your third number is #-47+2=-45#

Now, just to make sure you are right, add the numbers together

#-49-47-45=-141#