the general expression for consecutive multiples of 6 is 6N, 6(N + 1), 6(N +2), etc. find three consecutive multiples of 6 such that 4 times the first exceeds twice the third by 12.
Let N be the first, so n+1 is next, etc.
4(6(n+1))-2(6(n+3))=12
so, find n, then 6(n+1) for the first, 6(n+2) for the second, and 6(n+3) for the third.
i don't understand what you mean by that
What's some things I'll need for this class?
To solve the problem of finding three consecutive multiples of 6 such that 4 times the first exceeds twice the third by 12, we need to follow these steps:
Step 1: Express the multiples in terms of N
The general expression for consecutive multiples of 6 is given as 6N, 6(N + 1), 6(N + 2), etc. Let's assume N represents the first multiple.
Step 2: Formulate the equation
We are given that 4 times the first multiple (4 times 6N) exceeds twice the third multiple (2 times 6(N + 2)) by 12. This can be written as:
4(6N) = 2(6(N + 2)) + 12
Step 3: Solve for N
Simplify the equation and solve for N:
24N = 12N + 24 + 12
12N = 24 + 12
12N = 36
N = 3
Step 4: Find the three multiples
Now that we have found N as 3, we can substitute it back into the expression for multiples to find the three consecutive multiples of 6:
First multiple = 6N = 6(3) = 18
Second multiple = 6(N + 1) = 6(3 + 1) = 24
Third multiple = 6(N + 2) = 6(3 + 2) = 30
Therefore, the three consecutive multiples of 6 that satisfy the given condition are 18, 24, and 30.