Find two consecutive odd integers such that 5 times the first integer is 12 more than 3 times the second.

let X = 1st odd number.
and Y = 2nd odd number = X + 2 since it is consecutive.
Now set up two equations (the first is Y = X + 2). The second is
5X = 3Y + 12.
Substitute in this last equation for Y = X + 2 and solve for X.
I hope this helps. If you get stuck please post your work to that point and tell us why you are confused as to the next step to take.

Let A and (A + 2) be the two consecutive integers.

Then,5A = 3(A + 2 + 12.

Can you take it from here?

-6 -(-5 * 3 + 2}

To solve the equation, let's start by distributing the 3 to the terms inside the parentheses:

5A = 3(A + 2) + 12

Next, simplify the equation by multiplying the 3 inside the parentheses:

5A = 3A + 6 + 12

Combine like terms:

5A = 3A + 18

Now, let's isolate the variable A by subtracting 3A from both sides of the equation:

5A - 3A = 18

2A = 18

Divide both sides of the equation by 2 to solve for A:

A = 18 / 2

A = 9

So, the first odd integer is 9.

To find the second odd integer, we can substitute the value of A (9) into the expression (A + 2) to get:

9 + 2 = 11

Therefore, the second odd integer is 11.