Find to consecutive integers that the sum of 3 times the first integer and 6 times the second integer is 24. Define the variabls you use. If somebody can show me how to work this out, I will be very greatful for the help?
consecutive integers, n, n+1
3n + 6(n+1)=24 You need to put some thinking into these, it is not of much value to you if I do the work
I am not asking you to do the work .Just to show me the walk-through of how to get the answer. That's is all. I still appreciate the help that you have given me.
Of course! I apologize for any confusion. Let's work through the problem step by step.
We are asked to find two consecutive integers. Let's represent the first integer as "n" and the second integer as "n+1".
According to the given information, the sum of 3 times the first integer and 6 times the second integer is 24. We can write this as an equation:
3n + 6(n+1) = 24
Now, let's solve this equation for n.
Distribute the 6 to the terms inside the parentheses:
3n + 6n + 6 = 24
Combine like terms:
9n + 6 = 24
To isolate the term with n, subtract 6 from both sides of the equation:
9n = 24 - 6
9n = 18
Now, divide both sides of the equation by 9 to solve for n:
n = 18/9
n = 2
Therefore, the first consecutive integer is 2. To find the second consecutive integer, we can substitute this value back into the expression n+1:
n+1 = 2+1
n+1 = 3
So, the second consecutive integer is 3.
Hence, the two consecutive integers that satisfy the given conditions are 2 and 3.
I hope this walkthrough helps! Let me know if you have any further questions.