Learn about CA Algebra I: Systems of Inequalities Video

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Learn about CA Algebra I: Systems of Inequalities

We’re on problem 33 and it asks which equation represents a line that is parallel to Y is equal to minus 5/4X plus two. So a line that’s parallel will have the same slope and we can just look at this we can inspect this and say okay the slope of this line is the coefficient on the X term. So slope on this line is right there, this is the slope is equal to minus 5/4 and. So we’re just going to look at the other choices and say which one has the same slope of minus 5/4 and choice A is actually is, choice A is Y is equal to minus 5/4X plus one. So it has the exact same slope and it just shifted down by one so A is our choice in problem 34. Let me copy and paste this one okay let see I can paste it. Alright and they ask us which graph best represents the solution to the system of inequalities okay. So we could take each of these inequalities and then we want to do is, we want the area that satisfies both of these inequalities. So let’s take a look at the first one. And I like to always have an in kind of Y is equal to MX plus B form. So let’s do that. So we have the fist one says 2X is greater than or equal to Y minus one if we add one to both sides we get 2X plus one is greater than or equal to Y and we just want to write that in the way we are used to seeing it. Y, we come to switching to around switching it around Y is less than or equal to 2X plus one. So let’s see which one does represents, we have a line 2X plus one. So the Y intercept is one and the slope is two. So let’s see. So that’s this line here right. Y intercept one and if we go over to the right two we go up four right. However much we got to the, however much we increase in X we increase twice as much with Y that’s what the slope tells us right. And this is this line in this chart like that. Fair enough. So that fat line and they’re saying that Y is less than that. So we want the area below that graph right. So it’s not this one right, this one they’re shading in the area above that line right. For any X and Y that’s on the line they’re shading in Y is greater than that point—we want Y is less than that point. So for any, for any Y on the point, all the Y’s less than it satisfy it. So based on the information now this on seems to be a pretty good candidate. I haven’t looked at the other line yet. And frankly neither of, I guess, no neither, maybe this one might be a candidate because this is also less than. Maybe we have to be less than both of these lines right; this is the line that we just figured out. We’re definitely less than it; maybe we have to be less than that second line as well. So if we have to be greater than that second line, we haven’t looked at that yet it’s going to be choice C. If we have to be less then the second line and the first line then we’re going to be in choice D. Let’s see what the second piece of information tells us, well okay they actually have a typo and they don’t give us enough information to actually—let me see if I can—let me look up the answer because they don’t even they didn’t even write a greater than or equal or nothing here so I have to see what they probably intended. Okay so I’m I just looked up the answer and they said the answer is C. So they must want us to be greater than the second line greater than the second line. So let’s figure out what this should have been. So if we you know what, what should’ve this, should there been an equals? Well definitely an equal sign but should’ve been a greater than or less than or whatever we’ll figure that out. So let’s see 2X minus 5Y circle 10. Right, we don’t know if there’s an equal there or less this is almost a better exercise than, than what they intended, okay. So then if we, let’s see if we were to add 5Y to both sides I just want to get Y in the other side and I don’t want to have to say flip signs and divide by minus 5. So we get 5Y and. So just add and. So add 5Y to both sides we get 2X whatever inequality sign is, So that can be an equal or whatever 10 plus 5Y, we could have subtract 10 from both sides. So you get 2X minus 10 circle 5Y remember that circle could be an equal sign or not equal to, it’s either greater than or equal or it’s a less than or equal. And then we can, and then we can divide both sides by five and you get two divided by 5X minus two. Some greater than or equal or less than equal sign Y then they told us C was the answer because I had to look it up because they didn’t give us an inequality sign there if C is the answer that means we’re going to, we want all Y’s that are below this below the first line. So below this line right here. So that’s that area and if we’re on the square them, all Y’s that are above this, this bottom line right. So if we’re above this bottom line this bottom line is 2/5X minus two. So we want all Y’s that are greater than that, greater than that oh. Sorry I’ve done it I just, we want all the Y’s that are greater than that Y is greater than this thing. So that sign in this problem should’ve been should’ve been a less than sign. Right I’m saying Y is greater than this but if you read it left to right you have 2X minus two is less than Y and. So we figured out the sign and you might be saying “hey wait how come this area over here doesn’t work?” right. Well if you think about it this area is above our first line. This area is Y is greater than 2X plus one it’s above our first line and it’s actually below our second line. So it’s actually the opposite area, anyway next problem. So the answer was C. Next problem 35, okay let’s see I’ll copy and paste this one because I think the choices are interesting. What is the solution to this system of equations? Alright. So let’s see if we can get it in a in a form that makes it easy to look at. So if we, I’ll just, let’s take that first equation and let’s add 3X to both sides if you add 3X to both sides that first equation becomes positive 3X plus Y is equal to minus two right all I did is I did plus, well you can’t see that. I added plus 3X to both sides of these equation and of course that cancels with that. Alright and now let’s see that second equation 6X plus two is equal to minus two. Well, you can already see let me, let me do. Something else to this equation I think it will become apparent that these are actually the same line. So if you take this first equation and multiply both sides of it by two what do you get? So you multiply 3X plus Y by two you get two times 3X is 6X plus two times Y alright you have to distribute the two plus 2Y is equal to minus two times two is minus four and then the same line, same line. So if when you’re resolving a system of equations you’re figuring out where those two equations intersected. If they’re the same lines they intersect everywhere. So they have an infinite number of solutions. So the answer is choice D. Next problem okay they give us another one like that, they want to know the ordered pair that’s a solution to this equation. I’ll just copy and paste the equation there okay. So the easiest thing to do is probably just subtract the second equation from the first equation and instead of I actually write it out of explicitly. So the first equation is X plus 3Y is equal to seven instead of subtracting this one from that one let’s just multiply the second equation by negative one and then we’ll add the two equation. So if we multiply these bottom equation by negative one, we get minus X minus 2Y is equal to minus 10 and the whole reason why I’m doing that is that I know when I add these two left hand sides, the X and the minus X are going to cancel out and can solve for Y that’s why you immediately see if they have an X and X of we subtract this from that they’ll cancel out. So if we add these two equations the X is canceled out 3Y minus 2Y is equal to Y and seven minus 10, seven minus 10 is equal to minus 3. Fair enough and now we can substitute back in if you got an X. So let’s use the first equation X plus 3 times minus 3 we figured out what Y is, is equal to seven and then you get X—3 times minus 9 is equal to seven X is equal to add 9 the both side of this equation X is equal to 16. So the solution is 16 comma minus 3 X and Y, and that is choice D. Alright problem 37, 37. Marcy has a total of a hundred dimes and quarters, if the total value of the coins is 14.05 how many quarters does she have? Alright. So let’s, let’s say that D is a number of dimes and Q is the number of quarters. So if you take the numbers of dimes plus the number of quarters she has a hundred coins right. That’s this piece of information right there. And then the total value of the coins is going to be that’s going to be 0.10 times the number of dimes plus point 0.25 time a number of quarters and they tell as that is equal to 14.05 and so and that’s a piece of information. The total value of the coin is 14.05 and they say how many quarters does she have? So we just have for Q. So let’s do that. So if we want to cancel out the D’s what we could do is we can multiply this top equation by let’s say minus 0.1 right and I’m doing that so it cancels out with this, this D right here. Let me do it in a different color. So if I multiply that top equation times minus 0.1, I get minus 0.1 D minus 0.1. We can write one zero if I want D minus 0.10 Q is equal to—what’s minus 1/10 of a hundred? Well it’s minus 10 right minus10. A hundred times 0.1 is 10 and then we’re doing minus 0.1. Alright now we can add these two equations. 0.1 D minus 0.1 D those cancel out. 0.25 Q minus 0.1 Q that’s equal to let me switch colors that’s equal to 0.1 5Q is equal to and what 14.05 minus 10 well that’s equal to four dollars and five cents just to get rid of the decimals, we can multiply both sides of these equation by a hundred right? So we get 15Q, 15Q is equal to 405 and then we just say –. So Q we divide both sides of this by 15 and so what is, how many times is 15 going to 405, 15 goes into a 40 with two times two times 15 is 30 and a 10, 105, 15 goes into a 105 seven times I think, seven times five is 35 seven times one is seven plus 3 is 10 right 27. So Q is equal to 27. So Marcy has 27 quarters, anyway see you in the video.

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We’re on problem 33 and it asks which equation represents a line that is parallel to Y is equal to minus 5/4X plus two. So a line that’s parallel will have the same slope and we can just look at this we can inspect this and say okay the slope of this line is the coefficient on the X term.

So slope on this line is right there, this is the slope is equal to minus 5/4 and. So we’re just going to look at the other choices and say which one has the same slope of minus 5/4 and choice A is actually is, choice A is Y is equal to minus 5/4X plus one. So it has the exact same slope and it just shifted down by one so A is our choice in problem 34.

Let me copy and paste this one okay let see I can paste it. Alright and they ask us which graph best represents the solution to the system of inequalities okay. So we could take each of these inequalities and then we want to do is, we want the area that satisfies both of these inequalities. So let’s take a look at the first one.

And I like to always have an in kind of Y is equal to MX plus B form. So let’s do that. So we have the fist one says 2X is greater than or equal to Y minus one if we add one to both sides we get 2X plus one is greater than or equal to Y and we just want to write that in the way we are used to seeing it. Y, we come to switching to around switching it around Y is less than or equal to 2X plus one. So let’s see which one does represents, we have a line 2X plus one. So the Y intercept is one and the slope is two.

So let’s see. So that’s this line here right. Y intercept one and if we go over to the right two we go up four right. However much we got to the, however much we increase in X we increase twice as much with Y that’s what the slope tells us right.

And this is this line in this chart like that. Fair enough. So that fat line and they’re saying that Y is less than that. So we want the area below that graph right. So it’s not this one right, this one they’re shading in the area above that line right. For any X and Y that’s on the line they’re shading in Y is greater than that point—we want Y is less than that point. So for any, for any Y on the point, all the Y’s less than it satisfy it. So based on the information now this on seems to be a pretty good candidate. I haven’t looked at the other line yet.

And frankly neither of, I guess, no neither, maybe this one might be a candidate because this is also less than. Maybe we have to be less than both of these lines right; this is the line that we just figured out. We’re definitely less than it; maybe we have to be less than that second line as well. So if we have to be greater than that second line, we haven’t looked at that yet it’s going to be choice C. If we have to be less then the second line and the first line then we’re going to be in choice D.

Let’s see what the second piece of information tells us, well okay they actually have a typo and they don’t give us enough information to actually—let me see if I can—let me look up the answer because they don’t even they didn’t even write a greater than or equal or nothing here so I have to see what they probably intended. Okay so I’m I just looked up the answer and they said the answer is C. So they must want us to be greater than the second line greater than the second line. So let’s figure out what this should have been.

So if we you know what, what should’ve this, should there been an equals? Well definitely an equal sign but should’ve been a greater than or less than or whatever we’ll figure that out. So let’s see 2X minus 5Y circle 10. Right, we don’t know if there’s an equal there or less this is almost a better exercise than, than what they intended, okay. So then if we, let’s see if we were to add 5Y to both sides I just want to get Y in the other side and I don’t want to have to say flip signs and divide by minus 5.

So we get 5Y and. So just add and. So add 5Y to both sides we get 2X whatever inequality sign is, So that can be an equal or whatever 10 plus 5Y, we could have subtract 10 from both sides. So you get 2X minus 10 circle 5Y remember that circle could be an equal sign or not equal to, it’s either greater than or equal or it’s a less than or equal.

And then we can, and then we can divide both sides by five and you get two divided by 5X minus two. Some greater than or equal or less than equal sign Y then they told us C was the answer because I had to look it up because they didn’t give us an inequality sign there if C is the answer that means we’re going to, we want all Y’s that are below this below the first line. So below this line right here. So that’s that area and if we’re on the square them, all Y’s that are above this, this bottom line right.

So if we’re above this bottom line this bottom line is 2/5X minus two. So we want all Y’s that are greater than that, greater than that oh. Sorry I’ve done it I just, we want all the Y’s that are greater than that Y is greater than this thing. So that sign in this problem should’ve been should’ve been a less than sign.

Right I’m saying Y is greater than this but if you read it left to right you have 2X minus two is less than Y and. So we figured out the sign and you might be saying “hey wait how come this area over here doesn’t work?” right. Well if you think about it this area is above our first line. This area is Y is greater than 2X plus one it’s above our first line and it’s actually below our second line. So it’s actually the opposite area, anyway next problem. So the answer was C.

Next problem 35, okay let’s see I’ll copy and paste this one because I think the choices are interesting. What is the solution to this system of equations? Alright. So let’s see if we can get it in a in a form that makes it easy to look at. So if we, I’ll just, let’s take that first equation and let’s add 3X to both sides if you add 3X to both sides that first equation becomes positive 3X plus Y is equal to minus two right all I did is I did plus, well you can’t see that. I added plus 3X to both sides of these equation and of course that cancels with that.

Alright and now let’s see that second equation 6X plus two is equal to minus two. Well, you can already see let me, let me do. Something else to this equation I think it will become apparent that these are actually the same line. So if you take this first equation and multiply both sides of it by two what do you get? So you multiply 3X plus Y by two you get two times 3X is 6X plus two times Y alright you have to distribute the two plus 2Y is equal to minus two times two is minus four and then the same line, same line. So if when you’re resolving a system of equations you’re figuring out where those two equations intersected. If they’re the same lines they intersect everywhere. So they have an infinite number of solutions. So the answer is choice D.

Next problem okay they give us another one like that, they want to know the ordered pair that’s a solution to this equation. I’ll just copy and paste the equation there okay. So the easiest thing to do is probably just subtract the second equation from the first equation and instead of I actually write it out of explicitly. So the first equation is X plus 3Y is equal to seven instead of subtracting this one from that one let’s just multiply the second equation by negative one and then we’ll add the two equation.

So if we multiply these bottom equation by negative one, we get minus X minus 2Y is equal to minus 10 and the whole reason why I’m doing that is that I know when I add these two left hand sides, the X and the minus X are going to cancel out and can solve for Y that’s why you immediately see if they have an X and X of we subtract this from that they’ll cancel out.

So if we add these two equations the X is canceled out 3Y minus 2Y is equal to Y and seven minus 10, seven minus 10 is equal to minus 3. Fair enough and now we can substitute back in if you got an X. So let’s use the first equation X plus 3 times minus 3 we figured out what Y is, is equal to seven and then you get X—3 times minus 9 is equal to seven X is equal to add 9 the both side of this equation X is equal to 16. So the solution is 16 comma minus 3 X and Y, and that is choice D.

Alright problem 37, 37. Marcy has a total of a hundred dimes and quarters, if the total value of the coins is 14.05 how many quarters does she have? Alright. So let’s, let’s say that D is a number of dimes and Q is the number of quarters. So if you take the numbers of dimes plus the number of quarters she has a hundred coins right. That’s this piece of information right there. And then the total value of the coins is going to be that’s going to be 0.10 times the number of dimes plus point 0.25 time a number of quarters and they tell as that is equal to 14.05 and so and that’s a piece of information.

The total value of the coin is 14.05 and they say how many quarters does she have? So we just have for Q. So let’s do that. So if we want to cancel out the D’s what we could do is we can multiply this top equation by let’s say minus 0.1 right and I’m doing that so it cancels out with this, this D right here. Let me do it in a different color. So if I multiply that top equation times minus 0.1, I get minus 0.1 D minus 0.1. We can write one zero if I want D minus 0.10 Q is equal to—what’s minus 1/10 of a hundred? Well it’s minus 10 right minus10. A hundred times 0.1 is 10 and then we’re doing minus 0.1. Alright now we can add these two equations.

0.1 D minus 0.1 D those cancel out. 0.25 Q minus 0.1 Q that’s equal to let me switch colors that’s equal to 0.1 5Q is equal to and what 14.05 minus 10 well that’s equal to four dollars and five cents just to get rid of the decimals, we can multiply both sides of these equation by a hundred right? So we get 15Q, 15Q is equal to 405 and then we just say –. So Q we divide both sides of this by 15 and so what is, how many times is 15 going to 405, 15 goes into a 40 with two times two times 15 is 30 and a 10, 105, 15 goes into a 105 seven times I think, seven times five is 35 seven times one is seven plus 3 is 10 right 27. So Q is equal to 27. So Marcy has 27 quarters, anyway see you in the video.

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