Algebra I and Algebra II are basically the same course with some additional material being presented in Algebra II. You basically go through Algebra I all over again.
pancho, what's the time marker and what value are you speaking of? If you include these in a reply I'm happy to show you what's going in the explanation of the problem. Remember with word problems there are multiple ways to achieve the same answer.
For the downstream question, I had a similar question in my class but they did it so they split the total distance number in half. I tried doing the same problem without this method, like the way u did it in the video, but i got a different answer than my classmates did before. I am doing it wrong?
@@Greenemath Sorry if i'm bothering you! The question was: A boat traveled 336 miles downstream and back. The trip downstream took 12 hours. The trip back took 14 hours. What is the speed of the boat in still water? What is the speed of the current? My classmates got: 1 mph as the speed of the current and 13 mph as the boat speed. I got: 2 mph for current and 26 mph for the boat. My class and I did the same method, but I used 336 as the whole distance, and the class divided it by 2 to and did the equation like this.
@@acec3080 No one is ever bothering me, I'm happy to try and help anyone. So here is how I would solve the problem. let x = speed of the boat in still water then y = speed of the current We know that distance = rate x time And in this case, the distance for round trip travel is given as 336 for round trip or 168 for a single trip (in miles of course) So we can set up two equations. Downstream: 12(x + y) = 168 Upstream: 14(x - y) = 168 This is just time multiplied by the rate of speed. The + y is for downstream and the -y is for upstream. So now just solve the system and you get: x = 13, y = 1 So basically the speed of the boat in still water is 13mph and the speed of the current is 1mph. I have some written tutorials and practice on this topic at the website which is free: www.greenemath.com/Algebra1/24/SystemsofEquationsWordProblemsLesson.html There are links at the bottom for the video and the practice test if you want to do more practice. Let me know if you have other questions.
I have a question about the variables used in the first problem (the one about the senior tickets and child tickets being sold). I noticed that you gave the child tickets the x variable and the senior tickets the y variable. I tried to make an equation on my own first to see if I would nail it before watching the tutorial and I gave the x variable to the senior tickets and the y variable to the child tickets. This is because I saw that the senior tickets came first in the word problem. I was wondering why you gave the x variable to the child tickets instead of the senior tickets. Am I making a big deal about this? Probably, but because I switched the variables I got a different answer in the end. Here's the question just in case: Beth's school is selling tickets to a fall musical. On the first day of ticket sales, the school sold 12 senior tickets and 16 child tickets for a total of $336. The school took in $312 on the second day by selling 15 senior tickets and 11 child tickets. Find the price of a senior ticket and the price of a child ticket.
When you solve word problems such as this one, you can change how you assign things. Your approach is just as valid as what I did in the video. At the end, you will get the same answer for each senior ticket and each child ticket.
Everything was going well until the upstream (X-Y). Completely lost me. I have zero problem working most problems in algebra. It's usually the set up that kills me. Thanks for the video. The ticket sales was easy.
Please leave a time marker for your question. Also, please explain why you think there should be subtraction. If you do both of those things, I'll watch that part of the video and give you an explanation as to what's going on.
