I have been working this one for awhile and have become more confused. Please assist.
Problem:
Find two consecutive odd integers such that 5 times the first interger is 12 more than 3 times the second.
The key thing is to realize that odd numbers, just like the even numbers, are 2 apart
So if you let the first odd number be x, then the second odd number is x+2
Now all you do is translate the English into math.
"...such that 5 times the first interger is 12 more than 3 times the second."
5x > 3(x+2) by 12
so we add 12 to the smaller side to make it into an equation.
5x = 3(x+2) + 12
easy to solve
To solve this problem, you can follow these steps:
1. Let the first odd integer be represented by x.
2. Since the problem states that the first odd integer is followed by the second odd integer, the second odd integer can be represented as x + 2 (since odd numbers are 2 units apart).
3. The problem also states that 5 times the first integer is 12 more than 3 times the second integer. Translating this into an equation, you get: 5x = 3(x + 2) + 12.
4. Simplify the equation. Distribute the 3 inside the parentheses: 5x = 3x + 6 + 12.
5. Combine like terms on the right side: 5x = 3x + 18.
6. Subtract 3x from both sides to isolate the variable: 2x = 18.
7. Divide both sides by 2 to solve for x: x = 9.
8. Substitute the value of x back into the equation to find the second odd integer: x + 2 = 9 + 2 = 11.
9. Therefore, the first odd integer is 9, and the second odd integer is 11.