Learn about California Standards Test: Algebra II Video

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Learn about California Standards Test: Algebra II

So we’re on problem number one on the California’s Standard Test Problems for Algebra II and so the theory being is if you were able to do all or at least understand all 80 of these question that I think you have a pretty good understanding of Algebra II at least from the perspective of California. So in problem number one, what is a complete solution to the equation of the absolute value of 3-6x=15? So let’s think about it, 3-6x could clearly be 15 because the absolute value of 15 is 15. So we could say that 3-6x=15 but 3-6x can also be equal to minus 15 because the absolute value of -15 is also 15. So let us write that down so or 3-6x=-15, right? If this is equal to -15, you take the absolute value you get positive 15 so either of this could be true. So let solve for x. Subtract three from both sides, you get -6x=12, divide both sides by -6, you get x=-2 and so here if we subtract three from both sides you get -6x=-15-3 that’s -18, divide both sides by -6, you get x=3, right? Negative divide by negative so x=3. So that’s actually could be -2 or 3 and that’s choice B. Problem two, I’ll switch colors to make things interesting and there’s a similar type problem where the possible values of x in the absolute value of 12-4x=2. Same logic, 12-4x could be two, right? The absolute value of two is two so you get 12-4x=2 or 12-4x could be -2 because the value absolute value of -2 is 2. So then you get 12-4x could be equal to -2. So this is an oral proposition. Subtract 12 from both sides, you get -4x, 2-12 is -10 but when you get x divide both sides by -4 is -10/-4. Here the negatives canceled out and that’s equal to 5 halves or two and a half or 2.5 and just look at the choices they have everything is decimals. So let’s look at the other possibilities. Subtract 12 from both sides, you get -4x is equal to -2-12 is -14, x is equal to -14/-4. Negatives was cancel out which is equal to 7 halves which is equal to 3-1/2, which is equal to 3.5. So x could be 2.5 or 3.5 and that is choice D. Problem 3, for a wedding, Sharita bought several dozen of roses and several dozen carnations. The roses cost $15 per dozen. The carnations cost $8 per dozen. If Sharita bought a total of 17 dozen flowers and paid a total of $192.00. How many roses did she buy? Alright, let’s set some variables that are equal dozens of roses and let’s see, equal to dozens of carnations. Alright, so they said that she bought a total of 17 dozen flowers. So the dozens of roses she bought plus the dozens of carnations she bought must be equal to 17 because she bought a total of 17 dozens. It also tells that she spend a $192.00. So how do we figure out how much money she spends? So she bought r dozen of roses, how much she spend on roses? Well, each dozen of rose is $15.00 so if she spends 15 times the number of dozens, that how much she spend on roses. How much she spend on carnations? Well, she bought c dozen of carnation, each dozen was $8.00. She spends 8 c dollars on carnations. So her total amount of money that she spends was the amount she spends on roses plus the amount she spent on carnations, and that is equal to 192. Now we have two linear equations and two unknowns, so we can just solve them. Let’s say we want to solve for a number of roses. That was the question. So let’s try to cancel out the number of carnations. So if we multiply this top equation by -8, we got a 8-c that’ will cancel out with that. So let us do that. So let’s multiply that top equation by -8, you get -8r-8c which -8x17 so 17x-8=-136 and now we can add these two equations, 15-8, 7r, 8c-8=0 and that was the whole point behind multiplying this type of equation by -8 is equal to 192-36 through borrowing or regrouping they call now that’s 12, that’s 8 so b=56, 7r is equal to 56, so r=8. So r is equal to 8. So Sharita bought 8 dozens of roses and that is choice C. Problem four, what is the solution to the system of equation shown below? And there are three of them with three variables, always a hilarious problem to see. We have 2x-y+3z=8. We have x-6y-z=0 and then when you get -6x+3y-9z=24. Now there’s a bunch of ways to solve this. You could use matrices and inverses and matrices or elimination or you can just kind of use your traditional multiplying equations by each other, and adding them so that you can kind of constrain them or kick out variables, and we will do the last one because I do not want to introduce in the other kind of matrix manipulation here. So what I want to do is let use this equation with both of this and try to cancel out the x’s and the best way to do that. Let’s think of how we can do that. Well, one thing I like to do just to simplify it this to get all of these equations in what I kind of consider, you know, lowest integer form. I know that’s not formal term but look at all the terms in this equation. They are all divisible by three, right? So let’s divide al of them by three. So this last equation would become, if you divide everything by three you get -2x+y-3z=8. So let’s cancel that out, I just divide it by three. So same thing or whatever you do on the one side of the equation, you could do to the other side of the equation. Anyway, so let’s use this equation on this equation and then this equation in this equation, this equation to get two equations with just y’s and z’s. So let say you want to cancel out the x’s. So the canceled the x’s with this one and this one, we can just add them because 2x+-2y, that’s we’re going to cancel out the x’s. So we can just there and then what we do with this equation is we could multiply this equation times a -2 first, and all cancel out the x’s there. So let’s do it. So if I had this equation to this equation, 2x-2x=0x plus -y+-y=0y, these all zero but I’m just writing it down at certain we’re done and then plus-- are they all canceled out, plus 0z. So these are, is equal to 8+8=16, so you get 0=16 which makes no sense and there’s actually no solution here and if you wanted to think about visually what’s happening is each of these equations represent a plane in 3-dimensions. And the only way you get no solution or no intersection between this equation, and this equation or this equation which are essentially the same equation. The reason why you cannot get the intersection between them is that they must be parallel equations. You could get an infinite number of solutions if they were the actually the same plane but these are parallel planes. So they will never intersect. So we saw it and then it kind of makes sense because well, I’m not going to go to the details, we’re just focused on solving problems right now. But it’s a fair enough to say that when you add this equation to this equation to cancel out the x’s, everything else cancels out and then you’ll get a non-sense statement 0=16. So there’s no solution and that’s choice C. Those are parallel planes in 3-dimensions. Problem 5, a restaurant manager bought 20 packages of bagels. Some packages contain 6 bagels each and the rest contain 12 bagels each. There are 168 bagels in all. How many packages of 12 bagels had the manager bought? So the total numbers that he bought some packages of six. So the packages of six plus the packages of 12, that is just my random notation that I invented. He bought a total of 20 packages, so that equals 20 and then he bought total of 168 bagels in all. So how many bagels could he get from the six packs? Well, each of them had six bagels, right? So he got six times the number of packs. So you got six times this. This is the number of bagels we got from the six packs and then he got this type, this is a number of bagels he got from the 12 packs, right? The number of packs times 12 and that is equal to 168, two equations, two unknowns, they’re linear equations. We should be able to solve this. So let’s multiply. They want to know how many packages of the 12 bagels they bought. So they want to know P12 or whatever you want to call that is. So if we want to know, let’s cancel out the P6. So if you’ll multiply the top equation by -6, you get -6x6 packs, -6x12 packs(P12), just multiplying this top thing by -6 is equal to -120. Now you can add the two equations. These canceled out that was a whole point, 12x12 packs-6x12 packs is equal to 6x12 packs is equal to 168-120=48 and so we got the 12 packs. There were 48/6=8, so there were eight 12 packs. The manager bought P12=8. That is choice B. And I’ll see you in the next video because I’m all out of time.

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So we’re on problem number one on the California’s Standard Test Problems for Algebra II and so the theory being is if you were able to do all or at least understand all 80 of these question that I think you have a pretty good understanding of Algebra II at least from the perspective of California.

So in problem number one, what is a complete solution to the equation of the absolute value of 3-6x=15? So let’s think about it, 3-6x could clearly be 15 because the absolute value of 15 is 15. So we could say that 3-6x=15 but 3-6x can also be equal to minus 15 because the absolute value of -15 is also 15. So let us write that down so or 3-6x=-15, right? If this is equal to -15, you take the absolute value you get positive 15 so either of this could be true.

So let solve for x. Subtract three from both sides, you get -6x=12, divide both sides by -6, you get x=-2 and so here if we subtract three from both sides you get -6x=-15-3 that’s -18, divide both sides by -6, you get x=3, right? Negative divide by negative so x=3. So that’s actually could be -2 or 3 and that’s choice B.

Problem two, I’ll switch colors to make things interesting and there’s a similar type problem where the possible values of x in the absolute value of 12-4x=2. Same logic, 12-4x could be two, right? The absolute value of two is two so you get 12-4x=2 or 12-4x could be -2 because the value absolute value of -2 is 2. So then you get 12-4x could be equal to -2. So this is an oral proposition. Subtract 12 from both sides, you get -4x, 2-12 is -10 but when you get x divide both sides by -4 is -10/-4. Here the negatives canceled out and that’s equal to 5 halves or two and a half or 2.5 and just look at the choices they have everything is decimals.

So let’s look at the other possibilities. Subtract 12 from both sides, you get -4x is equal to -2-12 is -14, x is equal to -14/-4. Negatives was cancel out which is equal to 7 halves which is equal to 3-1/2, which is equal to 3.5. So x could be 2.5 or 3.5 and that is choice D.

Problem 3, for a wedding, Sharita bought several dozen of roses and several dozen carnations. The roses cost $15 per dozen. The carnations cost $8 per dozen. If Sharita bought a total of 17 dozen flowers and paid a total of $192.00. How many roses did she buy?

Alright, let’s set some variables that are equal dozens of roses and let’s see, equal to dozens of carnations. Alright, so they said that she bought a total of 17 dozen flowers. So the dozens of roses she bought plus the dozens of carnations she bought must be equal to 17 because she bought a total of 17 dozens. It also tells that she spend a $192.00. So how do we figure out how much money she spends? So she bought r dozen of roses, how much she spend on roses? Well, each dozen of rose is $15.00 so if she spends 15 times the number of dozens, that how much she spend on roses. How much she spend on carnations? Well, she bought c dozen of carnation, each dozen was $8.00. She spends 8 c dollars on carnations. So her total amount of money that she spends was the amount she spends on roses plus the amount she spent on carnations, and that is equal to 192.

Now we have two linear equations and two unknowns, so we can just solve them. Let’s say we want to solve for a number of roses. That was the question. So let’s try to cancel out the number of carnations. So if we multiply this top equation by -8, we got a 8-c that’ will cancel out with that. So let us do that. So let’s multiply that top equation by -8, you get -8r-8c which -8x17 so 17x-8=-136 and now we can add these two equations, 15-8, 7r, 8c-8=0 and that was the whole point behind multiplying this type of equation by -8 is equal to 192-36 through borrowing or regrouping they call now that’s 12, that’s 8 so b=56, 7r is equal to 56, so r=8. So r is equal to 8. So Sharita bought 8 dozens of roses and that is choice C.

Problem four, what is the solution to the system of equation shown below? And there are three of them with three variables, always a hilarious problem to see. We have 2x-y+3z=8. We have x-6y-z=0 and then when you get -6x+3y-9z=24. Now there’s a bunch of ways to solve this. You could use matrices and inverses and matrices or elimination or you can just kind of use your traditional multiplying equations by each other, and adding them so that you can kind of constrain them or kick out variables, and we will do the last one because I do not want to introduce in the other kind of matrix manipulation here.

So what I want to do is let use this equation with both of this and try to cancel out the x’s and the best way to do that. Let’s think of how we can do that. Well, one thing I like to do just to simplify it this to get all of these equations in what I kind of consider, you know, lowest integer form. I know that’s not formal term but look at all the terms in this equation. They are all divisible by three, right?

So let’s divide al of them by three. So this last equation would become, if you divide everything by three you get -2x+y-3z=8. So let’s cancel that out, I just divide it by three. So same thing or whatever you do on the one side of the equation, you could do to the other side of the equation.

Anyway, so let’s use this equation on this equation and then this equation in this equation, this equation to get two equations with just y’s and z’s. So let say you want to cancel out the x’s. So the canceled the x’s with this one and this one, we can just add them because 2x+-2y, that’s we’re going to cancel out the x’s. So we can just there and then what we do with this equation is we could multiply this equation times a -2 first, and all cancel out the x’s there.

So let’s do it. So if I had this equation to this equation, 2x-2x=0x plus -y+-y=0y, these all zero but I’m just writing it down at certain we’re done and then plus-- are they all canceled out, plus 0z. So these are, is equal to 8+8=16, so you get 0=16 which makes no sense and there’s actually no solution here and if you wanted to think about visually what’s happening is each of these equations represent a plane in 3-dimensions. And the only way you get no solution or no intersection between this equation, and this equation or this equation which are essentially the same equation.

The reason why you cannot get the intersection between them is that they must be parallel equations. You could get an infinite number of solutions if they were the actually the same plane but these are parallel planes. So they will never intersect. So we saw it and then it kind of makes sense because well, I’m not going to go to the details, we’re just focused on solving problems right now. But it’s a fair enough to say that when you add this equation to this equation to cancel out the x’s, everything else cancels out and then you’ll get a non-sense statement 0=16. So there’s no solution and that’s choice C. Those are parallel planes in 3-dimensions.

Problem 5, a restaurant manager bought 20 packages of bagels. Some packages contain 6 bagels each and the rest contain 12 bagels each. There are 168 bagels in all. How many packages of 12 bagels had the manager bought?

So the total numbers that he bought some packages of six. So the packages of six plus the packages of 12, that is just my random notation that I invented. He bought a total of 20 packages, so that equals 20 and then he bought total of 168 bagels in all. So how many bagels could he get from the six packs?

Well, each of them had six bagels, right? So he got six times the number of packs. So you got six times this. This is the number of bagels we got from the six packs and then he got this type, this is a number of bagels he got from the 12 packs, right? The number of packs times 12 and that is equal to 168, two equations, two unknowns, they’re linear equations. We should be able to solve this.

So let’s multiply. They want to know how many packages of the 12 bagels they bought. So they want to know P12 or whatever you want to call that is. So if we want to know, let’s cancel out the P6. So if you’ll multiply the top equation by -6, you get -6x6 packs, -6x12 packs(P12), just multiplying this top thing by -6 is equal to -120. Now you can add the two equations. These canceled out that was a whole point, 12x12 packs-6x12 packs is equal to 6x12 packs is equal to 168-120=48 and so we got the 12 packs. There were 48/6=8, so there were eight 12 packs. The manager bought P12=8. That is choice B.

And I’ll see you in the next video because I’m all out of time.

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