find 2 consecutive even integers so that 18 times the smaller number is 2
sorry -- there is no even integer x such that 18x = 2
Fix your question and add it to the conversation.
To find two consecutive even integers, let's assume that the smaller number is represented by "x." Since the numbers are consecutive even integers, the next even integer would be represented by "x + 2" since we add 2 to get the next even number.
Now, we have the two numbers: x and x + 2.
According to the problem, 18 times the smaller number (x) is equal to 2. So we can set up the equation:
18x = 2
To find the value of x, we need to isolate it on one side of the equation. To do this, we can divide both sides of the equation by 18:
x = 2/18
x = 1/9
So the smaller even integer (x) is 1/9.
To find the next even integer (x + 2), we can substitute the value of x:
x + 2 = 1/9 + 2
x + 2 = 1/9 + 18/9
x + 2 = 19/9
Therefore, the two consecutive even integers are 1/9 and 19/9.