Find The Domain Of A Simplified Algebraic Expression

by Andrew McMorgan 53 views

Hey guys, welcome back to Plastik Magazine! Today, we're diving deep into the fascinating world of mathematics, specifically focusing on algebraic expressions. You know, those things that look like a jumble of letters and numbers but hold so much power when you crack them open. We're going to tackle a doozy today: finding the domain of a simplified expression derived from a more complex one. This isn't just about crunching numbers; it's about understanding the very foundation of where a mathematical statement makes sense. Think of the domain as the VIP list for your variables – only certain values get in. If a value makes any part of the original expression undefined (like dividing by zero), it's kicked off the list, even if it magically reappears after simplification. So, buckle up, grab your calculators, and let's get this mathematical party started!

Our mission, should we choose to accept it, is to find the domain of the simplified form of the expression (aβˆ’1βˆ’2a21βˆ’a+1)Γ·(1βˆ’11βˆ’a)\left(a-\frac{1-2 a^2}{1-a}+1\right) \div\left(1-\frac{1}{1-a}\right). Now, before we even think about simplifying, the golden rule of domains is that we must consider the original expression. Any value of 'aa' that makes any denominator in the original expression equal to zero is immediately excluded from our domain. This is super important, guys, because sometimes, after a lot of algebraic wizardry, a problematic denominator might disappear. But that doesn't mean the original restriction is gone! It's like inviting someone to a party and then realizing they can't actually get through the door – just because you smoothed over the doorway later doesn't mean they could ever enter. So, let's identify those troublemakers. We have denominators '1βˆ’a1-a' appearing twice. For these to be non-zero, we need 1βˆ’aβ‰ 01-a \neq 0, which means aβ‰ 1a \neq 1. This is our first, and in this case, only restriction coming directly from the denominators. Keep this restriction in your back pocket; we'll need it later!

Now, let's get down to the nitty-gritty of simplifying this beast. We'll tackle the expression in two parts: the numerator part (aβˆ’1βˆ’2a21βˆ’a+1)\left(a-\frac{1-2 a^2}{1-a}+1\right) and the denominator part (1βˆ’11βˆ’a)\left(1-\frac{1}{1-a}\right). Let's start with the first part. To combine these terms, we need a common denominator, which is 1βˆ’a1-a. So, we rewrite 'aa' as a(1βˆ’a)1βˆ’a\frac{a(1-a)}{1-a} and '11' as 1(1βˆ’a)1βˆ’a\frac{1(1-a)}{1-a}. This gives us:

a(1βˆ’a)1βˆ’aβˆ’1βˆ’2a21βˆ’a+1(1βˆ’a)1βˆ’a\frac{a(1-a)}{1-a} - \frac{1-2 a^2}{1-a} + \frac{1(1-a)}{1-a}

Now, combine the numerators over the common denominator:

aβˆ’a2βˆ’(1βˆ’2a2)+(1βˆ’a)1βˆ’a\frac{a - a^2 - (1 - 2 a^2) + (1 - a)}{1-a}

Let's distribute the negative sign and combine like terms in the numerator:

aβˆ’a2βˆ’1+2a2+1βˆ’a1βˆ’a\frac{a - a^2 - 1 + 2 a^2 + 1 - a}{1-a}

Simplify further:

(aβˆ’a)+(βˆ’a2+2a2)+(βˆ’1+1)1βˆ’a=0+a2+01βˆ’a=a21βˆ’a\frac{(a - a) + (-a^2 + 2 a^2) + (-1 + 1)}{1-a} = \frac{0 + a^2 + 0}{1-a} = \frac{a^2}{1-a}

Awesome! We've simplified the first part. Now, let's move on to the second part, the denominator of the main division: (1βˆ’11βˆ’a)\left(1-\frac{1}{1-a}\right). Again, we need a common denominator, which is 1βˆ’a1-a. So, we rewrite '11' as 1βˆ’a1βˆ’a\frac{1-a}{1-a}.

1βˆ’a1βˆ’aβˆ’11βˆ’a\frac{1-a}{1-a} - \frac{1}{1-a}

Combine the numerators:

(1βˆ’a)βˆ’11βˆ’a\frac{(1-a) - 1}{1-a}

Simplify the numerator:

1βˆ’aβˆ’11βˆ’a=βˆ’a1βˆ’a\frac{1 - a - 1}{1-a} = \frac{-a}{1-a}

Fantastic! We've successfully simplified both parts of the original expression. Now, the original expression is (a21βˆ’a)Γ·(βˆ’a1βˆ’a)\left(\frac{a^2}{1-a}\right) \div\left(\frac{-a}{1-a}\right).

Division by a fraction is the same as multiplication by its reciprocal. So, we flip the second fraction and multiply:

a21βˆ’aΓ—1βˆ’aβˆ’a\frac{a^2}{1-a} \times \frac{1-a}{-a}

Now, we can perform the multiplication. Notice that we have (1βˆ’a)(1-a) in the numerator and the denominator, so they cancel out, provided 1βˆ’aβ‰ 01-a \neq 0 (which we already established as aβ‰ 1a \neq 1).

a2βˆ’a\frac{a^2}{-a}

We can simplify this further by canceling out an 'aa' from the numerator and the denominator. This gives us:

aβˆ’1=βˆ’a\frac{a}{-1} = -a

So, the simplified expression is simply βˆ’a-a. Pretty neat, right? But remember our mission: find the domain of the simplified expression, taking into account the original expression's restrictions. We found that aβ‰ 1a \neq 1 from the original denominators. Now, let's look at the simplified expression, which is βˆ’a-a. Are there any restrictions on 'aa' in βˆ’a-a? Not inherently. However, the crucial part is that the simplification process itself introduced a step where we divided by βˆ’a1βˆ’a\frac{-a}{1-a}. For this division to be valid, the divisor cannot be zero. So, βˆ’a1βˆ’aβ‰ 0\frac{-a}{1-a} \neq 0. This means βˆ’aβ‰ 0-a \neq 0, which implies aβ‰ 0a \neq 0. This is another restriction we need to consider! It arose because we performed a division, and the divisor cannot be zero.

Therefore, the domain of the simplified expression must exclude any value of 'aa' that made the original expression undefined or made any denominator in the simplification steps equal to zero. We found two such values: aβ‰ 1a \neq 1 (from the original expression's denominators) and aβ‰ 0a \neq 0 (from the division step in simplification). Thus, the domain of the simplified expression is all real numbers except for 11 and 00. We can write this in set notation as { a∈R∣aβ‰ 0Β andΒ aβ‰ 1a \in \mathbb{R} \mid a \neq 0 \text{ and } a \neq 1 }. In interval notation, this would be (βˆ’βˆž,0)βˆͺ(0,1)βˆͺ(1,∞)(-\infty, 0) \cup (0, 1) \cup (1, \infty). So, there you have it, guys! We've not only simplified a complex expression but also rigorously determined its domain by being super careful about where the original expression and the simplification steps are valid. Keep practicing, and you'll be a domain-finding pro in no time. Catch you in the next one!