Exercise 9 Page 260

To determine if an ordered pair satisfies an equation, we substitute the point into the equation and simplify. If the resulting statement is true, then the point is contained in the solution set. For systems of inequalities we can use the same method. However, substituting the point must create true statements in every inequality in the system. Let's test the given point.

y≤ x+7 & (I) y≥ 2x+3 & (II)

SubstituteII

x= 0, y= 0

( 0) ? ≤ ( 0) + 7 ( 0)? ≥2( 0)+3

ZeroPropMult

(II): Zero Property of Multiplication

0 ? ≤ 0 + 7 0? ≥0+3

(I), (II): Add terms

0 ≤ 7 ✓ 0 ≱ 3 *

One of our resulting statements, 0≱3, is not true. The point (0,0) satisfies (I) but does not satisfy (II). Because both statements are not true, the point (0,0) is not contained in the solution set of the system.