Algebra I
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CA Algebra I: Number Properties and Absolute Value
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CA Algebra I: Simplifying Expressions
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CA Algebra I: Simple Logical Arguments
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CA Algebra I: Graphing Inequalities
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CA Algebra I: Slope and Y-intercept
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CA Algebra I: Systems of Inequalities
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CA Algebra I: Simplifying Expressions
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CA Algebra I: Factoring Quadratics
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CA Algebra I: Completing the Square
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CA Algebra I: Quadratic Equation
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CA Algebra I: Quadratic Roots
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CA Algebra I: Rational Expressions 1
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CA Algebra I: Rational Expressions 2
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CA Algebra I: Word Problems
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CA Algebra I: More Word Problems
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CA Algebra I: Functions
CA Algebra I: Systems of Inequalities 33-37, systems of equations and inequalities
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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 5/4 x plus 2?
- So a line that's parallel will have the same slope.
- And we can just look at this, we could inspect this and say,
- OK, the slope of this line is the coefficient on the x term.
- So slope on this line is right there.
- Because the slope is equal to minus 5/4.
- And so we just have to look at the other choices and say,
- which one has the same slope of minus 5/4?
- And choice A actually is, choice A is y is equal to
- minus 5/4 x plus 1.
- So it has the exact same slope and it's just
- shifted down by 1.
- So A is our choice.
- Now problem 34.
- Let me copy and paste this one.
- Let's see, I can paste it.
- All right, and they ask us, which graph represents the
- solution to this system of inequalities?
- OK, so we could take each of these inequalities and then,
- what 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 it in kind of y is equal to mx
- plus b form.
- So let's do that.
- So the first one says 2x is greater than or
- equal to y minus 1.
- If we add 1 to both sides, we get 2x plus 1 is greater than
- or equal to y.
- And if we just want to write that in a way that we're used
- to seeing it, y-- I'm just switching it around.
- y is less than or equal to 2x plus 1.
- So let's see which one this represents.
- We have a line, 2x plus 1, so the y-intercept is 1 and the
- slope is 2.
- So let's see.
- So that's this line here.
- y-intercept 1 and if we go over to the
- right 2, we go up 4.
- However much we increase in x, we increase twice
- as much with y.
- That's what the slope tells us.
- And that's this line in this chart like that.
- Fair enough.
- So that's that line.
- And they're saying y is less than that.
- So we want the area below that graph.
- So it's not this one.
- This one they're shading in the area above that line.
- For any x and y that's on the line, they're shading in y is
- greater than that point.
- No, but we what y is less than that point.
- For any y on the point, all the y's less
- than it satisfy it.
- So based on the information now, this one 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 than the second line and this first
- line, then we're going to be in choice D.
- Let's see what the second piece of information tells us.
- OK, well they actually have a typo and they don't give us
- enough information to actually-- let
- me look up the answer.
- Because they didn't even write a greater than or equal or
- nothing here.
- So I have to see what they probably intended.
- OK, so I just looked up the answer and they said that the
- answer is C.
- So they must want us to be greater than this second line.
- So let's figure out what this should have been.
- What should have this-- should there have been an equals?
- Well, definitely not an equal sign.
- But should there have been a greater than or
- less than or whatever?
- We'll figure that out.
- So let's see.
- 2x minus 5y circle 10.
- We don't know if there's an equal there or less-- this is
- almost a better exercise than what they intended.
- Let's see.
- If we were to add 5y to both sides, I just want to get y on
- the other side.
- And I don't want to have to say flip signs and
- divide by minus 5.
- So let me get 5y on the other side.
- So I'll add 5y to both sides.
- So we get 2x.
- Whatever inequality sign, so that could
- be an equal or whatever.
- 10 plus 5y.
- Subtract 10 from both sides.
- So you get 2x minus 10 circle 5y.
- Remember that circle could be an equal sign.
- Well, it's not an equal.
- It's either a greater than or equal or it's a
- less than or equal.
- And then we can divide both sides by 5 and you get 2
- divided by 5 x minus 2, some greater than or equal or less
- than or equal sign, y.
- Now they told us C was the answer because I had to look
- it up because they didn't give us any inequality sign there.
- If C is the answer that means we're going to-- we want all
- y's that are below the first line.
- So below this line right here.
- So that's that area.
- And if we're in this grey area, all y's that are above
- this bottom line.
- So if we're above this bottom line, this bottom line
- is 2/5 x minus 2.
- So we want all y's that are greater than that.
- Sorry, I've done it-- we want all the y's that are greater
- than that. y is greater than this thing.
- So that sign in this problem should have
- been a less than sign.
- I'm saying y is greater than this.
- But if you read it left to right, you have 2x minus 2 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?
- Well if you think about it, this area is
- above our first line.
- This area is y is greater than 2x plus 1.
- It's above our first line.
- And it's actually below our second line.
- So it's actually the opposite area.
- Anyway, next problem.
- The answer was C.
- Next problem, 35.
- OK, let's see.
- I'll copy and paste this one.
- I think the choices are interesting.
- What is the solution to this system of equations?
- All right, so let's see if we can get in a form that makes
- it easy to look at.
- 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 2.
- All I did is I did plus 3-- you can't see that.
- I added plus 3x to both sides of this equation.
- And of course, that cancels with that.
- All right, and now let' see.
- That second equation is 6x plus 2 is equal to minus 2.
- Well let me do something else to this equation and I think
- it'll become apparent that these are
- actually the same line.
- So if you take this first equation and multiply both
- sides of it by 2, what do you get?
- Let's see.
- Multiply 3x plus y by 2, you get 2 times 3x is 6x.
- Plus 2 times y.
- You have to distribute the 2.
- Plus 2y is equal to.
- Minus 2 times to 2 is minus 4.
- And they're the same line.
- So when you're solving a system of equations, you're
- figuring out, where do those two equations intersect?
- 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.
- 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.
- OK, so the easiest thing to do is probably to just subtract
- the second equation from the first equation.
- And instead, I'll actually write it out explicitly.
- So the first question is x plus 3y is equal to 7.
- Instead of subtracting this one from that one, let's just
- multiply the second question by negative 1 and then we'll
- add the two equations.
- So if we multiply this bottom equation by negative 1 we get
- minus x minus 2y is equal to minus 10.
- And the whole reason why I'm doing that is I know when I
- add these two left-hand sides, the x and the minus x are
- going to cancel out.
- And then I can just solve for y.
- That's why you immediately see, they have an x and an x.
- If we subtract this from that, they'll cancel out.
- So if we add these two equations, the x' cancel out.
- 3y minus 2y is equal to y.
- And 7 minus 10 is equal to minus 3.
- Fair enough.
- And now we can substitute back in to figure out an x.
- So let's use the first equation.
- x plus 3 times minus 3-- we figured out what y is-- is
- equal to 7.
- And then you get x.
- 3 times-- minus 9 is equal to 7.
- x is equal to-- add 9 to both sides of this equation.
- x is equal to 16.
- So the solution is 16 comma minus 3. x and y.
- And that is choice D.
- All right, problem 37.
- Marcy has a total of 100 dimes and quarters.
- If the total value of the coins is $14.05, how many
- quarters does she have?
- All right, so let's say that d is a number of dimes and q is
- the number of quarters.
- So if you take the number of dimes plus the number of
- quarters she has 100 coins.
- That's this piece of information right there.
- And then the total value of the coins.
- It's going to be 0.10 times the number of dimes plus 0.25
- times the number of quarters.
- And they tell us that that is equal to $14.05.
- And that's this piece of information.
- The total value of the coins is $14.05.
- And they say, well, how many quarters does she have?
- So we just have to solve 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.
- And I'm doing that so it cancels out with
- 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 0.1d-- I could write 1-0 if I want-- d.
- Minus 0.10q is equal to-- what's minus 1/10 of 100?
- Well that's minus 10, right?
- 100 times 0.1 is 10 and we're doing a minus 0.1.
- All right, now we can add these two equations.
- 0.1d minus 0.1d, those cancel out.
- 0.25q minus 0.1q, that's equal to-- let me switch colors.
- That's equal to 0.15q is equal to-- and
- what's 14.05 minus 10?
- That's equal to $4.05.
- Just to get rid of the decimals, we can multiply both
- sides of this equation by 100.
- So we'll get 15q is equal to 405.
- So q.
- We divide both sides of this by 15.
- And so how many times does 15 go into 405?
- 15 goes into 40, what?
- Two times.
- 2 times 15 is 30.
- Get a 10.
- 105.
- 15 goes into 105 seven times I think.
- 7 times 5 is 35.
- 7 times 1 is 7.
- Plus 3 is 10.
- 27.
- So q is equal to 27.
- So Marcy has 27 quarters.
- Anyway, see you in the next video.
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At 5:31, how is the moon large enough to block the sun? Isn't the sun way larger?
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