Let \(A = \{1, 2, 3, ..., 20\}\). Let \(R \subseteq A \times A\) such that \(R = \{(x,y): y=2x-7\}\). Then the number of elements in R, is equal to
Step-by-step Solution:
Correct Answer: C (10)
The problem asks for the number of elements in a relation R. An element of R is an ordered pair $(x, y)$ that satisfies three conditions simultaneously.
An ordered pair $(x, y)$ is an element of the relation R if it meets the following criteria:
To find the number of elements in R, we need to find how many integer values of $x$ from 1 to 20 produce an integer value of $y$ that is also between 1 and 20.
We use the constraint on $y$ (condition 2) and substitute the rule for the relation (condition 3) into it.
$$ 1 \le y \le 20 $$
Substitute $y = 2x - 7$:
$$ 1 \le 2x - 7 \le 20 $$
We can solve this compound inequality for $x$ by adding 7 to all parts:
$$ 1+7 \le 2x \le 20+7 $$
$$ 8 \le 2x \le 27 $$
Now, divide all parts by 2:
$$ 4 \le x \le 13.5 $$
We have found the range for $x$. Now we must also apply the first condition: $x$ must be an integer from the set A.
We need to count the number of integers $x$ such that $4 \le x \le 13.5$.
The possible integer values for $x$ are: 4, 5, 6, 7, 8, 9, 10, 11, 12, and 13.
By counting these values, we find that there are 10 possible values for $x$. Each of these values will produce a unique corresponding $y$ value that is within the set A, thus forming a valid element in R.
Therefore, the number of elements in R is 10.