Chapter 1.7, Problem 19HP

To determine

The range of the relation.

Answer to Problem 19HP

The range of the relation is $\left\{8,10,12,14\right\}$

Option B is correct.

Explanation of Solution

Given:

The relation that is represented by the equation $y=20-2x$ has domain $\left\{3,4,5,6\right\}$ .

Concept Used:

• To get rid of a number in addition from one side, subtract the same number from both sides of equal sign.
• To get rid of a number in subtraction from one side, add the same number both sides of equal sign.
• To get rid of a number in multiplication from one side, divide the same number from both sides of equal sign.
• To get rid of a number in division from one side, multiply the same number both sides of equal sign.

Rules of Addition/ Subtraction:

• Two numbers with similar sign always get added and the resulting number will carry the similar sign.
• Two numbers with opposite signs always get subtracted and the resulting number will carry the sign of larger number.

Rules of Multiplication/ Division:

• The product/quotient of two similar sign numbers is always positive.
• The product/quotient of two numbers with opposite signs is always negative.

Calculation:

In order to find the range of the relation . Here to solve the table and simplify as shown below,

  $\begin{array}{c}y=20-2x\\ x=3\\ y=20-2\cdot 3\\ y=20-6\\ y=14\\ x=4\\ y=20-2x\\ y=20-2\cdot 4\\ y=20-8\\ =12\\ x=5\\ y=20-2x\\ y=20-2\cdot 5\\ y=20-10\\ y=10\\ x=6\\ y=20-2x\\ y=20-2\cdot 6\\ y=20-12\\ y=8\end{array}$

Thus, the range of the relation is $\left\{8,10,12,14\right\}$

Option B is correct.