c Avoid a values that would make the denominator zero.
A
a A = 6 (a+1)/a-3
B
b See solution.
C
c a≠- 3, 3
Practice makes perfect
a We are given rational expressiona for the length l and width w of a rectangle.
l = 3a + 9/2a -6 w = 4a+4/a+3
We need to find the area A of this rectangle in its simplest form. To do this we can substitute the expressions for the length and width in the formula for the area of a rectangle A= w l. Then, we will be able to find A and get it to its simplest form.
b Recall that any rational expression and its simplified form must have the same domain. Note that a rational expression is undefined whenever the denominator has a real zero, as division by zero is an operation that is undefined. Thus, this is the part of the expression we should check.
c Let's analyze the expression for the area we found at Part A and its simplest form, following the ideas discussed in Part B. Note that the original expression (before simplifying) for the area is undefined not just for a = 3 but also for a = -3.
A = 3(a+3) 3( a-3) * 2(a+1) a+3
The restriction in the domain, a ≠-3, is no longer visible in the simplest form. Therefore, we must explicitly state this.
A = 3(a + 3)/2(a -3)* 4(a+1)/a+3, a≠- 3, 3 [1.2em]
⇕ [0.8em]
A = 6 (a+1)/a-3 , a≠- 3, 3
