Rationals problem
7 5
4x+3 - x+4
----------------------
4x+3 + x+4
2 4x+3
1) The first thing you want to do is get a common denominator for the top set of fractions.
2) The common denominators for this first set are (4x+3) and (x+4).
3) Now what you’ll have to do is multiply the numerators by the same factor. So, it would be
7 * x+4, which equals, 7x + 28 because you have to distribute thoroughly. Now you’ll do the same on the other side. So you’ll take the -5 * 4x+3 = -20x -15.
4) You now can combine the numerators all over the same denominator. As shown below:
7x + 28 - 20x -15
(4x+3)(x+4) Remember to simplify as much as possible!
5) -13x + 13
(4x+3)(x+4) Is now what you are left with on the top! We are halfway done.
6) We can start by repeating step one. Try and get a common denominator for the fractions.
7) Let’s take 4x+3 and multiply by 4x+3
- 4x*4x = 16x^2
- 4x*3 = 12x
- 4x*3 = 12x
- 3*3 = 9 So, we end up with, 16x^2 + 24x + 9
8) The next thing you’ll want to do is multiply x+4 by 2.
x+4 * 2 = 2x + 8.
9) Combine the 16x^2+24x+9 and 2x+8 over the common denominator.
16x^2+26x+17
2(4x+3)
10) You now can combine it into one whole fraction! -->
-13x + 13
(4x+3)(x+4)
--------------------------
16x^2+26x+17
2(4x+3)
11) To undo division, you can now multiply by the reciprocal.
12) -13x+13 2(4x+3)
(4x+3)(x+4) * (16x^2+26x+17)
13) Notice that we have a common factor of 4x+3 in both the numerator and denominator allowing us to get rid of it because they cancel each other out.
14) -13x+13 2
(x+4) * (16x^2+26x+17)
15) We have to make sure we factor everything out completely.
we can factor out a -13 from -13x+13 leaving us with -13(x-1)
16) -13(x-1) 2
(x+4) * (16x^2+26x+17)
17) All that is left to do now is multiply straight across leaving you with the final answer of:
-26(x-1)
(x+4)(16x^2+26x+17).
An Extreme solving problem
12 - 27sin(3Θ+5π/4) = 21 - 9sin(3Θ+5π/4)
1) We can start solving by first subtracting 21 from both sides.
-9 - 27sin(3Θ+5π/4) = -9sin(3Θ+5π/4)
2) The next step is to add 27sin(3Θ+5π/4 ) to -9sin(3Θ+5π/4).
-9 = 18sin(3Θ+5π/4)
3) Now that we almost have sin alone, our next step is to divide the -9 by 18, which will leave us with → -1/2 = sin(3Θ+5π/4)
4) Now you find all values that make this statement true. Where on the unit circle is sin -1/2? Well it happens to be at 7π/6 and 11π/6.
5) What you want to do next is take the radian measures you found and add or subtract 2πn and then set it equal to our theta expression in our parenthesis.
- 11π/6 +/- 2πn= 3Θ+5π/4
6) We are trying to get theta by itself so the first thing you do is subtract 5π/4 from 11π/6 by getting common denominators.
- 2* 11π/6 - 5π/4 *3
- 22π/12 - 15π/12 = 7π/12
7) So now you can take 7π/12 +/- 2πn= 3Θ and divide it by 3 to get theta alone.
8) After dividing by 3 you should end up with, 7π/36 +/- 2/3πn.
9) One possible solution is 7π36. To find more solutions (this can be done in a calculator or just on paper) you can take (7/36) and add (1/2) which equals (25/36). 2π is equivalent to 72π/36 so you find as many solutions as possible without going over 2π.
- (25/36) + (1/2) = (43/36)
- (43/36) + (1/2) = (61/36)
10) Our last solution for this particular radian would be 61/36 because if you add another 1/2 you exceed 72π/36.
11) Now we can start the other radian measure.
- 7π/6 +/- 2πn = 3Θ+5π/4
12) We are aiming at getting the theta value by itself so we’re going to subtract 5π/4 from 7π/6 by getting common denominators.
- 2* 7π/6 - 5π/4 *3
- 14π/12 - 15π/12 = -π/12
13) Now you can take your radian measure of -π/12 +/- 2πn = 3Θ and divide it by 3 to get theta alone.
14) After you have divided by 3 you should end up with, -π/36 +/- 2/3πn.
15) You already have one of your solutions, that being -π/36, to find more solutions all you have to do is add 1/2 to your previous solution but keep in mind that 2π is equivalent to 72π/36.
- (1/36) + (1/2) = (19/36)
- (19/36) + (1/2) = (37/36)
- (37/36) + (1/2) = (55/36)
16) Our last solution is 55π/36 because if we were to add another 1/2 we would exceed our limit of 2π.
17) So all of the possible solutions for this particular problem are:
7π/36, 25π/36, 43π/36, 61π/36, -π/36,19π/36, 37π/36, and 55π/36.
I chose to do these problems because I had a hard time understanding them when we first went over them. There are so many steps in figuring out a rationals problem. You have to be careful and make sure you completely simplify before you take further steps. I would often forget that to undo multiplication you would have to multiply by the reciprocal of the fraction and it would get me every time. I finally got it though! I also chose to do an extreme solving problem because I remembered how annoying and long and hard they were. It's extremely rewarding going through and being able to create and solve a problem you once thought was nearly impossible. Anyways, I'd also like to thank you for always being there for me when I needed it! See ya! :)
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