al g e b r a 2...al g e b r a 2 re mo te l e a r n i n g fo r ap r i l ap r i l 1 5 th - s u b s ti...

Algebra 2 Remote Learning for April

April 13th - Rational operation review We will both be review this month and learning some new concepts. The goal for today is to review some rational rules!! Remember, when you multiply and divide, you do NOT need a common denominator. When you add or subtract, you do!! When multiplying or dividing, you may want to factor (use the box) to help you simplify some problems!! For #1-4, just simplify. (Hint: you must factor #3 and #4)

Add/Sub. The first two already have common denominators!

7) 6x−1 − 4

5x 8) 3x+7 + 4

x−8 Multiply and Divide! (Remember, you may have to factor first!)

9) 12x+3 · x+3

3(x−6)10) 1

n+9 ÷ 6−n2n+18

11) x−9x −162

· x +x−902

x +14x+402 12) x +x−562

x −2x−802 ÷ 1v−10

And, just for fun, graph this on your graphing calculator (or on desmos.com) and see if you remember where the asymptotes are!

13) y = x+5x −3x−402


April 14th - Introduction to substitution There will be an online group teaching time for this page if you have internet access! Just check Google classroom or your email for details. If not, here are the filled out notes on Substitution. Read them to understand part of our concepts this week! You can also youtube search “Solving systems of linear equations by substitution” on your phone and watch a video or two! You actually learned this in Algebra 1! So, even though it is new this year, it should be familiar! We just do a few more fun things with it in Algebra 2!!


April 14th (cont.)

Try these next two on your own, and then check your answers on the next page!! 2x + 2y = 3 x - 6y = 6 x - 4y = -1 -3x + 2y = -2


April 14th (cont.)


April 15th - Substitution Practice Here are some more substitution problems, giving you practice on yesterday’s topic!! And a word problem tying into the school theme of the 1940s - try it out! Solve by substitution. 1. 2x - 5y = 9 2. -3x + 4y = -1 -3x + y = -7 x - 2y = 1 3. 6x + 2y = 11 4. x - 2y = -1 4x + y = 6 5x - 7y = 4 5. 4x + 3y = 3 6. 10x - 16y = 17 2x + y = -3 x + y = 3 7. In World War 2, as the Allies finally made progress across the war-torn lands of France, they aided those that they encountered as they fought toward Germany. They had two types of utility vehicles, and in one camp, there were 124 total utility vehicles. The first type of utility vehicle could hold 3,850 pounds, the second, 2,250 pounds. All these vehicles combined were carrying 343,000 pounds. How many of each vehicle were there? Here is the set of equations. Can you see how they come from the information in the problem? Solve for x and y, which will show how many of each type of vehicle there was. x + y = 124 3,850x + 2,250y = 343,000


April 16th...More substitution!(we’ve been quarantined a whole month...but we can make it!) Here are a few substitution problems that are a bit trickier! It is harder to get a letter by itself because they all have coefficients! But, I have faith that you can do it! Don’t forget to divide everything in the equation by the coefficient when you are getting a variable by itself!! Solve by substitution. 1. 5x + 2y = 10 (Hint: on this problem, I would get y by itself in

-5x + 3y = 5 the first equation. Then, it would be:

y = -2.5x + 5 )

2. 4x + 7y = -10 (Hint: on this problem, if you get x by itself on

3x - 7y = -4 the second equation, you will want to keep it as

fractions instead of decimals, or you will have a

slightly wrong answer due to rounding. So, your

second equation should look like this:

x = y - )37

34

Now you can try a few on your own! 3. -3x - 4y = 2 4. -2x + 6y = 6 3x + 3y = -3 -7x + 8y = -5 5. -5x - 8y = 17 6. -2x - y = -9 2x - 7y = -17 5x - 2y = 18 Here’s something fun! See if you can solve these two 3-variable systems! You already have the answer for z, just plug it in to find y. Then, plug both in to find x! 7. z = -1 8. z = -4

y – z = 3 y – 3z = 7

x – 2y + 3z = -4 x + 5y – 7z = 6


April 17th - Graphing Again today, there will be an online group teaching time for this page if you have internet access! Just check Google classroom or your email for details. If not, here are the filled out notes! Let’s talk about what we’ve actually been doing when we perform substitution to solve a problem! Why is our answer always a point ( , )? Here’s why:

The two equations we are given for substitution are two lines! They meet somewhere, and that meeting point is the answer to our substitution problem! We just did them without graphs! Here’s another one!


April 17th (cont.) By the way, if you want to use your graphing calculator to check yourself, you can! But, you still have to get y by itself! Here is how you would type them in, and the graph that would show up (this is the same as the previous example):

They meet at (-3, 7) on the calculator too!! Now, let’s look at another example.

If lines are parallel, they don’t meet! So, your answer would be no solution! This is because there is no point where they cross, and that is what we are looking for whether we graph the problem or use substitution. (We haven’t done any parallel lines by substitution yet, but we will!) There is one more scenario, given on the next page!


April 17th (cont.)

Sometimes the equations are actually the exact same line, which means that the point where they meet isn’t just one point, it is EVERYWHERE on the line - and that’s infinite points! So, your answer would be infinite solutions! Again, we haven’t done any of these by substitution, but we will! Here is a quick recap:


April 17th (cont.) To end the week out, try these three problems by graphing! (Hint: one of the answers is a point, another answer is no solution, and another answer is infinite solutions. You just have to figure out which one is which!!) 1. x - 2y = 6 -2x + 4y = -12 2. y = -2x + 9 -3x + y = -1 3. 3x + 2y = 4 6x + 4y = 10


April 20th - Substitution and Graphing review Let’s review what we did last week, and we will have an online group teaching time tomorrow! Today, you are going to solve each of the systems given by graphing them, and then by solving them by substitution, and you should get the same answer!! (I’ll give you a hint when you get to a parallel or same line problem!) Solve by substitution AND graphing! (Hint: #1-3 are easy to graph because they are already in y = mx + b form. To solve by substitution, you can use set the two equations equal because they both already have y by itself! For instance, to start solving #2, your equation would be: 4x + 3 = -x - 2 )


April 20th (cont.) Watch out #3 should be no solution. Make sure your graph depicts that fact and you see why the equation by substitution has no solution as well! 3. 4. -6x + 6y = 6 -6x + 3y = -12 5. 3x - 2y = 2 5x - 5y = 10


April 21st - Introduction to elimination Again today, there will be an online group teaching time for this page if you have internet access! Just check Google classroom or your email for details. If not, here are the filled out notes! This is another time that if you have no internet, but a cell phone with data, you can search on youtube, “solving systems of linear equations by elimination” and something helpful will come up! So, why do we have more than one method...are you sick of solving for y or x and getting fractions? This method helps with this! No more getting x or y by itself first!!

Why don’t you try this example on your own, and then check yourself on the next page! x + 3y = 2 -x + 2y = 3


April 21st (cont.)

Now, both of those examples were nicely set up. A variable conveniently cancelled when we added the equations together! But, not all problems are set up like that. So, what do we do when they aren’t set up like that?


April 21st (cont.) To end today, here are two more examples that are not conveniently set up. I’m hoping that by now you remember some of this from Algebra 1! So, try your hand at these two examples, and then check your answers and be ready for more elimination tomorrow! y = x - 9 -3x + 6y = 4 x + 8y = 0 4x - 12y = -8 Notice that these could also be solved by substitution or elimination! So, if you want to check your work that way, feel free! But, here are the answers!


April 22nd - Elimination Practice Here’s the day that you can practice your elimination skills! Although, the first two, I want to show you something! Try #1 and #2 and see where you get! 1. 3x - 2y = 7 2. 4x + 5y = -3 -3x + 2y = 12 -12x - 15y = 9 Do you see what happens in #1? All the variables cancel, and you are left with “0 =19” (which is both a nonsensical and untrue equation). What does this mean? There would be no solution! The lines would be parallel. In #2, everything cancels, and you get “0 = 0” (which is weird, but true!). This means they are the same line, and there are infinite solutions. Try the rest of these elimination problems: 3. 6x - 7y = 4 4. x + 2y = 7 x + 7y = 17 3x + 5y = 17 5. 3x - 4y = 6 6. 6x + y = 0 4x + 7y = 1 2y + 15x = 1 7. -2x + 3y = 8 8. 4x + 7y = -10 x - 5y = -4 3x - 7y = -4


April 22nd (cont.) Here is a word problem, relating to our theme again! 8. By the 1940s, there will all types of speed boats for ocean and river travel. Let’s say a boat on the Mississippi traveled downriver 210 miles in 10 hours. The return trip of the same distance took 70 hours. How fast was the boat’s steady speed? How fast was the current? Using Distance = (Rate)(Time), we know that the equations would be: 210 = 10(rate) for the first trip, and, 210 = 70(rate) for the return trip. However, the rate for the first trip would be x + y, which is the boat’s rate added to the current’s rate. Whereas, the return trip’s rate would be x - y,, as it is going against the current. So, our system is 10(x + y) = 210 or 10x + 10y = 210 70(x - y) = 210 70x - 70y = 210 Solve by elimination! Solve a 3-way system? 9. 2x + y – z = -7 (Hint: add first two equations together to find z because -2x – y + 3z = 17 the x and y terms all eliminate. Then, plug z into both 2x + 3y – 2z = -12 of the bottom equations and then eliminate the x terms!

You’ll be able to find y. Once you have both y and z, you Can plug in both y and z to any equation to get x!)


April 23rd - Choose your method! You may have been wondering this whole time - why are we doing Algebra 1 type stuff? Well, it is half to get ready for the SAT (which I realize is a moot point right now). But, it is half to get us ready for Algebra 2 type substitution and elimination! That is coming your way tomorrow!! For today’s problems, it is review, but you choose! For each problem, solve the system, but use whatever method you want: graphing, substitution, or elimination! Solve the system! 1. y = 6x - 11 2. -7x - 2y = -13 -2x - 3y = -7 x - 2y = 11 3. -4x + y = 6 4. -5x + y = -2 -5x - y = 21 -3x + 6y = -12 5. x + 3y = 1 6. -3x - 8y = 20 -3x - 3y = -15 -5x + y = 19 7. -2x + 6y = 6 8. y = -2 -7x + 8y = -5 4x - 3y = 18 9. 2x - y - 3z = 20 (You don’t have to do this one, but we’ll go over 3x + y + 6z = 4 it tomorrow...for you brave ones!) x + 2y + 9z = -16


April 24th - Other types of systems! Again today, there will be an online group teaching time for this page if you have internet access! Just check Google classroom or your email for details. If not, here are the filled out notes! So, let’s get into Algebra 2. So far, we’ve seen how to solve by substitution, elimination, and graphing. We’ve seen that there may be one solution, no solutions, or infinite solutions. However, it has only related to lines. Think of all the other functions we’ve covered this year! Today, we will experiment with some parabolas and lines! Check out this first example. It is to figure out how many solutions the system has!

So, before, there was no way for two (straight) lines to touch more than once unless they were they same line. But, parabolas and lines can cross twice! There are other possibilities! Like this:


April 24th (cont.) Now, remember, if you are having trouble remembering how to graph parabolas, you can do one of two things: #1) Look in your notes from last semester, or #2) Just use your graphing calculator. (or desmos.com!) Here’s a brief review of what it will look like to graph both on the same screen! Use you y = button to get to the first screen, and the graph button to get to the second!

So, these have no solution! Are there other options?

You try this example on your own, and then check your work on the next page. I make it easy in case you are graphing by hand! y = x² y = -1


April 24th (cont.)

Lastly, it can get crazy if we do two parabolas!

Two parabolas could have two solutions, like you see here. They could also have one solution, if they meet at a point, but one opens down and the other opens up. They could have no solution if they never touch, or infinite solutions if they are the same parabola!! Think of other functions we’ve done this year - like logs! (If you can remember what it looks like, graph y = log(x) in your calculator!) Brainstorm to end the week: How many solutions are possible if a line and a log are on a graph? How many times would they potentially meet?


April 27th - Review of Solving Rationals Before we head into substitution and elimination with parabolas, I thought we’d solve some equations that involve factoring (or using the box), in order to keep our brains fresh! Here’s some equations you would have seen at the beginning of the semester. See if you can solve them! Solve! (Hint: On #1 and #2, cross multiply and then set them equal Like #1 would be:

x² + 3x = 28 . Subtract the 28 and then factor/box!)

1) 7x + 3 = x

4 2) 125x + 4 = 10

3x − 2 Solve! (Hint: on these, use the common denominator rule. Remember that when you get a common denominator, then you can cross it off? So, once you get a common denominator on #3, it would look like this: 1 + b = 6, which is really easy to solve!)

7) 3

x + 1 = 3x + x2 8) x

3 − 1x − 5 = 2

x−5

9) 6 3x − 1 − = 5x

x − 1 10) x

2x + 1 = 54 − x


April 28th - Substitution with Parabolas Again today, there will be an online group teaching time for this page if you have internet access! Just check Google classroom or your email for details. If not, here are the filled out notes! Since graphing parabolas is kind of a pain, it’s time to talk about how we can use substitution or elimination to help solve systems with parabolas and lines! So, that is what today’s notes are about! Substitution is definitely the easiest. Most of the time the equations are given with y by itself, so you can just set them equal to each other and solve! Here’s an example:

Notice that both times that you get an answer for x, you have to plug it in to get an answer for y. This is because all our answers to these types of questions are points where a parabola and a line meet!


April 28th (cont.) Here’s another example:

Elimination isn’t quite as handy. I usually only use it on these types of problems when there are two parabolas and the squared terms will cancel. Like this:

It’s crazy that in this case, there is only one place that these two parabolas meet! It is kind of an interesting graph if you wanted to check it out!!

Think you’ve got it? Solve this one and check yourself on the next page: y = 3x -20 y = -x² + 34


April 28th (cont.)

So, there’s today’s lesson. Tomorrow you’ll have a chance to try your hand at more of these!


April 29th - Solving Linear/Quadratic systems Solve by substitution or elimination (I would probably use substitution though!) Hint: Everything on this worksheet is factorable, so if you can’t box something, you may have made a mistake! 1. y = 2x² + 13x 2. y = -8x y = -6x - 9 y = 1 + 16x²

3. y = 3x 4. y = x² -4x - 5 y = x² + 9x - 91 y = -x - 7 5. y = 2x² + 4x - 1 6. y = x² + 11 -5x + y = 5 y = -12x 7. y = 5x - 20 8. y = x² - x - 90 y = x² - 5x + 5 y = x + 30


April 30th - Solving all kind of systems! So, the last two days of this unit are pretty crazy! They are not for the faint of heart. If you have been following up until now, you get full credit, but if you want a challenge, today and tomorrow are for you! (All that tomorrow will be is an online to go over the craziness of today!) First, here are some systems that may take some foiling or the quadratic formula to solve! Try your best and leave your answers as decimals if you use the quadratic formula! 1. y + x² = 4 2. 5x² + 14x + 1 = y 2y + 4 = x y + 40x = 12 3. y = x² - 12x - 20 4. y + 20x = 39 y = 100 - 25x 15 + 4x² + 9x = y Now, what is left from here is what I would have given as the extra credit after the test, if we had been in school. It is pretty crazy!! 5. Solve where a cubic and a parabola meet! y = x³ - 20x + 16 y = -x² + 6x - 8


6. Solve where a circle and a parabola meet! (x - 2)² + y² = 4 2y = x² 7. Solve where a cubic and a line meet! y = 3x - 5 y = x³ - 5x² - 5x + 7 April 30th/May 1st There will be an online group teaching time for this page if you have internet access on one of these days to go over this last assignment! Just check Google classroom or your email for details!

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