Determine whether there are 2 consecutive odd integers such that 5 times the first exceeds three times the second by 54.
Can anyone help???
We're asked:
Determine whether there are 2 consecutive odd integers such that 5 times the first exceeds three times the second by 54.
Let the first odd number be 2k+1, then the next odd number is is 2(k+1)+1=2k+3.
Then 5*(2k+1)=3*(2k+3)+54 so
10k+5=6k+9+54 thus
4k=58
Since 4 does not divide 58 there is no k to satisfy the integers, so the answer is no, the integers don't exist.
The closest k's we could use are 14 and 15.
Consider k=14, then 2k+1=29 and 2k+3=31
We find 5*29=145, 3*31=93 and 145-93=52
Consider k=15, then 2k+1=31 and 2k+3=33
We find 5*31=155, 3*33=99 and 155-99=56
Check that the problem was entered correctly and that I read it correctly.
Since 5x = 3(x + 2) + =54, 2x = 60 making x = 30 and (x + 2) = 32.
I guess not.
To determine whether there are 2 consecutive odd integers that satisfy the given condition, we can follow these steps:
1. Assume the first odd integer as 2k+1, where k is an integer. This is because odd integers can be represented as 2k+1, where k takes different integer values.
2. The second odd integer would be the next consecutive odd number, which can be represented as 2(k+1)+1 = 2k+3.
3. Set up the equation based on the given condition: 5 times the first integer exceeds three times the second integer by 54. So, 5(2k+1) = 3(2k+3) + 54.
4. Simplify the equation using algebraic operations. Expand and group like terms: 10k+5 = 6k+9+54.
5. Further simplify the equation: 10k+5 = 6k+63.
6. Solve for k: Subtract 6k from both sides and subtract 5 from both sides: 4k = 58.
7. Determine if the equation has a solution by checking if 4 divides evenly into 58. In this case, it does not, so there is no integer solution for k.
Therefore, there are no 2 consecutive odd integers such that 5 times the first integer exceeds three times the second integer by 54.