Hot X: Algebra Exposed

Transcription

1 Hot X: Algera Exposed Solution Guide for Chapter 5 Here are the solutions for the Doing the Math exercises in Hot X: Algera Exposed! (assume that all denominators 0) DTM from p Since they have the same denominator, we can just add these fractions y adding across the top: x 5 + 2x 5 Answer: 5!! x + 2x 5 5. Since it s already reduced, we re done!. Again, we can just add across the top, since the denominators are the same: x y + 2x y!! x + 2x y y. Since it s already reduced, we re done! Answer: y 4. Since they have the same denominator, we can just sutract these fractions y sutracting across the top, ut let s use parentheses so we distriute that negative correctly. Rememer, the numerator or denominators of fractions are understood to e grouped together,, even though they don t usually have parentheses surrounding them.! a +! (a + )!!! a!! a + (!)!a Done! Answer: a

2 (assume that all denominators 0) DTM from p We want to rewrite these so they have the same denominator. What s the LCD of these fractions? In other words, what s the LCM of their denominators, and a? We could use the irthday cake method, ut it s pretty easy to eyeall this one the LCM is just. So what copycat should we multiply times a to rewrite it as an equivalent fraction, ut with the denominator of? We should use 2 2, and we get: a 2 2! a 6. Great, now we can rewrite our original prolem and solve: Answer: 7 + a (samedenominator, yay!) 7. In this case, since the denominators have no common factors, the LCD will just e their product: v. So how can we rewrite each of the fractions to have this denominator? For the first fraction, we ll use the copycat v v, and we get: v v! v. For the v second fraction, we ll rewrite the second fraction y multiplying y the copycat, so we get: 2 v! 2 v 6. Faulous! Now we can rewrite our original prolem with v fractions whose denominators are the same, and do the addition: + 2 v v v + 6 v (samedenominator, yay!) v + 6 v There are no common factors on the top and ottom, so it s reduced. Done! Answer: v + 6 v

3 4. It might seem at first that these denominators oth have a factor of n in them, ut n is a very different value from n! In fact, n and n have no common factors, so the LCD we ll use will just e their product: n( n). That means for the first fraction, we ll use the copycat! n! n in order to get that denominator: n! n! n " n (! n) " (! n) " n n n( n) And for the second fraction in our original prolem, we ll use the copycat n, and we get: n! n n n "! n n " n(! n) Read those slowly and make sure you follow what s going on! n n( n) Great, now we have a common denominator, so we can rewrite our prolem, and add the two fractions together: n +! n! n n(! n) + n! n + n (samedenominator, yay!) n(! n) n(! n) We can also multiply the ottom with distriution to get Both answers are fine! Answer: n( n) or n n 2 n( n), ut it s not necessary. 2 n! n 5. Let s take the hint and rewrite x as a fraction, so our prolem ecomes: x! y. So what s the LCD here? Well, the LCM of and y is y, so that will e our denominator for oth fractions! We ll use the copycat fraction y y on the first fraction, and it ecomes: x y y! x xy y. Great! Now our fractions have a common denominator, and we can finish the prolem: x! y!!!!!! xy y! xy!!!!!!! y y

4 We can t reduce this (don t go thinking we could cancel those y s we can t ecause there is a separate term,, which doesn t share a factor of y), so we re done! Answer: xy y 6. Hm, monster prolem. Okay, we ll take this one step at a time. First jo? To find the LCD, which will e the LCM of 6,, and 4a. As I talked aout on p.25, we can do this with the irthday cake method in two steps: First, we find the LCM of 6 and, take that answer, and find the LCM etween that and the final, unused term, 4a. So for the 6 and, we ll write the common factor of 2 on the outside and we re left with and a on the ottom, resulting in the LCM 6a. As you can see aove, we ll then take 6a and our third, unused term, 4a, and find their LCM in a second cake, and the LCM is the stuff along the ig L of this second cake: a! 2!! 2. Okay, now our challenge is to write each of the original fractions with this new denominator,. Let s figure out which copycats we need to use for each of them! For the first fraction, a, we should use the copycat. Check it out: 6 a 6! a 6 2 That s the denominator we wanted! Okay, for the second fraction, denominator, we should use the copycat 6. And we get: 6 6 6!, to have that same 6.

5 Thirdly, to get the denominator on the fraction, and we get: c 4a! c 4a c. c, we ll need to use the copycat 4a Phew! Now we get to do the easy part. Since they all have the same denominator, we just get to add across. Here s what our prolem has transformed into: a c 4a c c The terms in the numerator have no common factors, so there s no way to factor it, and as a result, there s no way to reduce this fraction, so we re done! Answer: c 7. Okay, let s not get confused y these variales. What s the LCM of 2(y ) and 2y? Let s see, the factors of 2(y ) are 2 & (y ). And the factors of 2y are 2 & y. Make sense? Now, we could use the irthday cake method on this, ut it would look a little strange. Better to just eyeall it in this case. But with these variales, if you don t have much experience with finding LCMs, it can e confusing! The important thing is that we rewrite the fractions of our prolem so that they have the same denominator, and then we ll e allowed to add them together. That s our goal. The truth is, if we end up using a denominator that isn t exactly the LCD, ut if we are ale to write equivalent fractions and attain our goal, then we ll still get the right answer; we ll just end up needing to reduce the answer afterwards. Not the end of the world! Eyealling it, LCM of 2(y ) and 2y is 2y(y ), ut let s say that we don t see that. We can always just take the two original denominators and multiply them together for our new denominator. That s what works when the original denominators don t have any common factors (as on p.6), and it also works any other time! So let s just do that.

6 Our new denominator will e: 2y! 2(y ) 4y(y ). Secretly, you and I know that we could use 2y(y ) as the common denominator, ut let s see how this goes. So, to rewrite our first fraction, 2y 2y, and we get: 7 2(y! ) 7, with the denominator 4y(y ), we ll use the copycat 2(y! ) 2y 2y " 7 2(y! ) 4y 4y(y ). Ok, for the second fraction, 2(y! ), we ll use the copycat, and we ll get: 2y 2(y! ) 2y 2(y! ) 2(y! ) " 2y 2(y ) 4y(y ). And now we add them together! Our original prolem has ecome: 7 2(y! ) + 2y 4y 2(y! ) + 4y(y! ) 4y(y! ) (same denominator, yay!) 4y + 2(y! ) 4y(y! ) Let s distriute that 2, and continue: 4y + 2y! 2 4y(y! ) 6y! 2 4y(y! ) Notice that we can factor the top y pulling out a factor of 2, and then wouldn t you know it, it will cancel with a factor of 2 on the ottom! 6y! 2 4y(y! ) 2(8y! ) 2 " 2y(y! ) 8y 2y(y ) Although it seems like we should e ale to factor another 2 out of the top and ottom, we can t, ecause of that! We re now done, ut if we wanted, we could distriute the 2y in the denominator. Both answers are fine. Answer: 8y 2y(y ) or 8y 2y 2 2y

7 (assume that all denominators 0) DTM from p In order to simplify this complex fraction, we ll need to first make sure that oth the numerator and denominator are written as single fractions. So let s simplify the numerator, and pretend nothing else exists. So, to add x + x, we ll want a common denominator of, and we ll use the copycats and x, respectively, so we get: x x + x! x + x x! x 9 + x2 Now our prolem looks like this: 9 + x 2 (same denominator, yay!) 9 + x2, and it s ready for the means and extremes! (see p. 52 to review this method) Using the means and extremes shortcut, we get: 9 + x 2 (9 + x 2 )! (9 + x2 ). Before we go distriuting that, let s notice that we can cancel a factor of from the top and ottom! Yep, we discovered a hiding kitty cat,. So now our answer has ecome: (9 + x2 ) Answer: 9 + x 2 (9 + x2 ) 9 + x 2. Done!. Let s simply the denominator y adding those little fractions together: c + d. We ll want a common denominator of, so let s use some copycats to make that happen: c + d d d! c + c c! d d + c

8 Great, and also taking the hint and writing the original numerator,, as, our original prolem has ecome:. Using means and extremes, we get: d + c!!(d + c) just ecause it s nicer to write things in alphaetical order (and easier to check our answers against the answer key!), we get Answer: c + d c + d. d + c. And 4. We re not even going to think aout that final 4 until we simplify the ig fraction! Okay, let s just look at just the numerator of the ig fraction first, and let s do the sutraction: 4a!. We ll use a common denominator of 4a, so that means we need to use the copycat fraction 2 2 on the second fraction to make this happen, so the sutraction ecomes: 4a! 4a! 2 2 " 4a! 2! 2 (samedenominator, yay!) 4a 4a 4a Okay, that was the numerator of the original ig fraction. The ig fraction has now ecome: 4a. That s looking a it etter. Now we can use the means and extremes on it, a and we get:! a 4a! a 4a 4. So now our entire prolem has transformed like this: 4a!! 4 a 4! 4 0. How nice! Answer: 0