Simplifying and Multiplying Rational Expressions

On tuesday we started out the class everyone's favorite way - a surprise homework collection! But once everyone was done stressing about how they did, we moved onto a new topic, simplifying rational expressions. This is similar to simplifying fractions, but with variables involved. You can cancel out any terms that the numerator and denominator have in common, just like one would if you were simplifying even the simplest of fractions, such as 5/20. 5 and 20 are both divisible by five, so if you divide both by five the fraction simplifies to 1/4.

Once we added variables to the mix, it became more complex, because we almost always had to factor the numerator and the denominator in order for the terms to cancel. The first example we used is in the picture below.

As you can see, once you factor both expressions, any factors that they have in common will cancel. It is important to remember, however, that you can't cancel any terms until each term is connected by multiplication. If one of the terms has addition or subtraction in it, such as (x-2), you can only cancel the full (x-2) if both expressions have it in common. You can't cancel just the x or just the -2.

After everyone understood the idea of simplifying rational expressions (with variables), we moved on to also finding the domain and zeros of the simplified expression. We had worked with domain before, when we were talking about graphing earlier in the year, but just as a review we defined it again.

Domain: The set of all x values that work for the expression (in order to have a valid output).
An example of an invalid output would be any x values that would make the denominator 0, because then the expression would be "undefined".

We have not worked with zeros before, so this was a new definition for us.

Zeros: All x values that make a fraction equal 0.
This means any x value that will make the numerator 0, because any fraction with a numerator of 0 is equal to 0.

In this example, the two expressions simplified to (x+1)/(x-1). This means that the number that would make the denominator 0 is 1, which is why it is excluded from the domain. All other numbers work with x-1. The zero, or the number that makes the numerator 0, was -1, because the numerator was x+1, and if you add 1 to -1, the difference is 0.

We spent the rest of the class working on examples and finding the domain and zeros of different rational expressions. This was the gateway into Wednesday's class.


We started the class on Wednesday by going over the homework, which was all either simplifying fractions with variables or finding the domain and zero. One that people had a particular amount of trouble with was #31. Mainly the problem was factoring each of the expressions, because once they were factored the problem was actually fairly simple.

The denominator of the fraction could be factored using the difference of squares pattern that we learned earlier: a^2-b^2=(a-b)(a+b). Eventually, part of the numerator could be factored using the same pattern. Once everything was completely factored, everything in the denominator factored out, and only x+y remained in the numerator. That meant that the simplified answer was (x+y)/1, which is equal to just x+y.

On Wednesday, we expanded on the idea of simplifying these expressions by adding multiplication and division to the mix. In order to refresh our memory, Lisa gave us an example of a multiplication problem with two fractions without any variables. She showed us that you could simplify the fraction after you multiply, or you could simplify the fractions before multiplying in order to save yourself time and work.

Dividing fractions is very similar to multiplying fractions, because you can turn those problems into multiplication problems. The process is KCF, or Keep Change Flip. You keep the first fraction in your equation, change your sign from division to multiplication, and then flip your second fraction.

Since we all understood the basic principles of multiplying and dividing fractions, we quickly moved on to adding variables into the fractions. The process is similar, but you just need to factor the numerators and denominators of each fraction. That way you can cancel the terms before you do the actual multiplication, like in the second example in the picture above. A term that is in the numerator and the denominator of the same fraction will cancel, or a term that is in the numerator of one fraction and the denominator of another. A term that is on the same side of the fractions will never cancel. Once everything is cancelled, you can multiply to find your answer. I found a video on Youtube that explains multiplication of "algebraic fractions."

 Hope this clarified any questions you had!

- Molly Brock

Polynomial Equations for Problem Solving and Review

On thursday, our class schedule was organized to help us review the previous concept we had learned, quadratics, and learn a final concept before it would be time to review for our upcoming test.

Some of the problems in the homework that many students had difficulty with were the more complex problems, such as problem number 41, in which we had to figure out what possible numbers could be represented by the variable, t.

Lisa explained that to solve this problem, we had to first expand (t+1)^3. An easy way to expand exponents is to apply the problem to pascals triangle. For an exponent of 3, the numbers that the  variables are multiplied by are 1, 3, 3, and 1. The number of times that 1, 3, 3, and 1 are multiplied by the variable decreases by 1 with each set in the series, starting by the largest original exponent, 3.
After expanding (t+1)^3 to t^3+3t^2+3t+1, we found that t^3 could be eliminated from both sides of the equation, as the other side of the equation was t^3 + 7. We also subtracted 7 from the right side of the equation so it could be left with 0. The equation now read 3t^2+3t-6=0, which was divisible by 3. Dividing the problem by 3 left us with a problem that we could factor to (t+2)(t-1)=0. Therefor, t was either equal to -2 or 1, or {-2,1}


For our previous homework assignment we had also been introduced a word problem. We were unfamiliar with the concept of forming an equation based on measurements in this way, so Lisa further introduced the concept in class, starting by explaining the word problem from our homework:
#5, page 189: Find the dimensions of a rectangular lot if its length is 7 m greater than its width and each of its diagonals is 13 m long.

Lisa showed us that an easy way to approach a problem like this was to draw a picture. We drew the rectangle and labelled the dimensions. The width was labelled x to represent its length and the length was labelled x+7 because we knew that the length was 7 m greater than the width. Next we labelled the diagonals as 13 because we knew that the diagonals were 13 m long. The length, width, and diagonal formed a triangle that we could use to solve the problem. We plugged in the information we had to the equation to find the area of a triangle:

a^2+b^2=c^2   --->   x^2+(x+7)^2=13^2 

We then expanded the problem and subtracted 169 from each side of the equation to bring us to 2x^2+14x-120=0, which was divisible by 2. We were able to factor x^2+7x-60=0 to (x+12)(x-5)=0. Therefor, x was equal to -12 or 5, and because the problem dealt with measurement, x had to be positive. 

So the width being x, was 5, and the length, being x+7, was 12.
Answer: The lot is 5x12 m.

We continued to practice similar word problems to number 5, such as number 7:
Two ships leave port, one sailing due west and the other due south. Some time later they are 17 miles apart and one is 7 miles farther from the port than the other. How far is each from port?

We started by drawing the port as a point and drawing the distance that each ship had travelled. We then plugged in the distance each boat had travelled from the port and the boats' distances from each other to the formula for the area of a triangle. The equation was:

x^2+(x+7)^2=17^2

We then simplified the equation as we had done with the previous one until we figured out that x was equal to 8, so the answer was: one boat is 15 miles and the other is 8 miles from the port. 

On Friday, our class schedule was similar to Thursday's. We first reviewed our homework, which had been review for the test. Lisa presented us with the topics that the test would cover:

We had learned all the topics in previous weeks and had solidified them through practice. We had not, however, fully explored word problems, so we next focused on some more difficult problems than what we had worked on on Thursday, such as problem number 15 on page 190:

When a 0.5 cm was planed off each of the six faces of a wooden cube, its volume was decreased by 169 cm^3. Find its new volume. 

Students had difficulty approaching this problem, so Lisa gave us pieces of paper and told us to physically cut off a square from each of their corners so they could be folded into a box. 

We then drew the box and established that each side had originally been equal to x but was now equal to x-1, because 0.5 cm had been planed off of each side. The area of a cube is length times hight times width, so we plugged in the information we had to establish the volume:

(x-1)^3

We used the exponent three because we knew that each side of a cube is the same.

We also knew that the new volume was equal to the old volume (x^3) minus 169cm^3. We combined the two equations to form a final equation: 

x^3-169=(x-1)^3

We were able to establish our answer having been given this strategy to approach the word problem.

Lisa explained that the test we were about to take would not cover word problems that were this complicated, but it was helpful and interesting to explore them in such a way. We left class prepared to review for our test on Tuesday.

When struggling with a concept we learn in class during homework, I find that the Algebra and Trigonometry book have clear explanations to the subjects we are covering. 

I found that the examples in the book on page 188 laid out all the steps to solving the word problems we were working on in class in a clear way, and I was able to apply it to my own thinking.

Thanks for reading! I hope you found this helpful.

Sadie Grant