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: Simplifying Expressions 38-43, systems, binomial simplification
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- We're on problem 38.
- Which of the following best describes the graph of this
- system of equations?
- OK, so maybe they're the same line.
- Maybe they're parallel.
- Maybe they only intersect in one point-- two lines
- intersecting in only two points.
- Well that's impossible.
- Two lines, I mean that can happen with curves, but that's
- not going to happen with lines.
- So we can already cancel out choice D.
- OK, now let's look at these two.
- See I have a y here and I have a 5y here.
- Let's multiply this top equation times 5 and see what
- it looks like.
- So if you multiply the left-hand side
- by 5, you get 5y.
- I'll do it up here.
- You get 5y is equal to-- 5 times minus 2 is minus 10x,
- plus 5 times 3 is 15.
- So if you multiply the top equation-- both sides of it--
- by 5 and it really doesn't fundamentally change the line,
- the equation might look different, but the equality
- will still hold in the same universe, which is
- essentially that line.
- So if you just multiply both sides by 5, they become the
- same equation.
- 5y is equal to minus 10x plus 15.
- So they are the same lines.
- So that's A, two identical lines.
- Problem 39.
- And they want us to simplify 5x to the third over 10x to
- the seventh.
- So the easiest way to think of this, or at least for me--
- well, there's a lot of ways you can do it and
- we'll do it both ways.
- This is the same thing as 5/10 times x to the third times x
- to the minus 7.
- 1 over x to the 7 is the same thing as x to the minus 7.
- And this is equal to-- 5/10 is 1/2.
- And then here, we have the same base and we're
- multiplying, so we can add the exponents.
- 3 plus minus 7 is minus 4.
- So x to the minus fourth power.
- And we could write that as 1/2 times 1 over x to the fourth
- or 1 over 2x to the fourth.
- And that is choice B.
- Now you could have done it other ways.
- You could have said, OK, let's see.
- Divide the numerator and denominator by 5.
- So this would have been 1.
- This would have been a 2.
- And you say, OK, let's divide the numerator and denominator
- by x to the third.
- So this will become a 1.
- And x to the seventh divided by x to the
- third is x to the fourth.
- You could have done it that way.
- You had 1 over 2x to the fourth.
- Either way.
- Or you could have even said-- you didn't have
- to go to this step.
- You could have said, OK, when I'm dividing with the same
- base, I can just subtract the exponents.
- So 3 minus 7 was minus 4.
- Either way.
- All of them would have valid ways to approach this problem.
- Problem 40.
- This looks like a simplification.
- They write 4x squared minus 2x plus 8, minus x squared plus
- 3x minus 2 is equal to.
- So the key here is just to realize that this is a minus.
- So you could kind of view it as a plus minus 1 times this
- whole thing.
- So we're just going to have to distribute that out.
- So this is equal to 4x squared minus 2x plus 8.
- And now we distribute this minus over this whole
- expression.
- So minus times x squared is minus x squared.
- Minus times 3x, positive 3x.
- So it's minus 3x.
- The minus 1 times negative 2.
- Well now they cancel out and you get a plus 2.
- We switch the sign on everything here because
- they're all being multiplied by this negative 1.
- OK, now we can simplify.
- So let's take the x squared terms first. So we have a 4x
- squared, we have a minus x squared.
- So 4x squared minus x squared is 3x squared.
- 4 minus 1 is 3.
- Then let's do our x terms. We have a minus 2x, we
- have a minus 3x.
- So minus 2 minus 3, that's a minus 5x.
- And then last we have our constants.
- We have 8 plus 2.
- 8 plus 2 is 10.
- So 3x squared minus 5x plus 10.
- And that is choice D.
- Problem 41.
- OK.
- They say the sum of two binomial-- let
- me copy this one.
- It's interesting.
- The sum of two binomials is 5x squared minus 6x.
- So a binomial is just a polynomial with two terms. If
- one of the binomials is 3x squared minus 2x, what is the
- other binomial?
- So this binomial is one of them, so they're saying 3x
- squared minus 2x, and when you add that to some other
- binomial-- and I don't know, let me just write that as A.
- I mean there is no constant term here and there is no
- constant term here, so I'm assuming that my-- and it has
- to be a binomial.
- There's only two terms. So I'm assuming my two terms are an x
- squared term and x term because that's the only terms
- that are involved in both of these.
- So let's say my binomial is Ax squared plus Bx.
- This is the mystery binomial.
- And their sum is equal to this up here.
- Is equal to 5x squared minus 6x.
- Now let's see what we can do.
- Well this is a plus here, so the parenthesis
- really don't matter.
- We can rearrange this as 3x squared plus Ax squared minus
- 2x plus Bx is equal to 5x squared minus 6x.
- 3 plus A.
- 3x squared plus Ax squared, that's the same thing as 3
- plus A, x squared.
- And then, minus 2x plus Bx, or we could switch them around.
- That's the same thing as plus B minus 2-- I just took the
- coefficients and added them together-- x.
- I switched them, but we could have written this in the other
- order to begin with-- is equal to 5x squared minus 6x.
- And now you just compare.
- OK, 3 plus A-- if you just look at the x squared terms--
- 3 plus A has to be equal to 5.
- Because that's the coefficient on the x squared term.
- So 3 plus A is equal to 5.
- Subtract 3 from both sides.
- You get A is equal to 2.
- And then we have B minus 2 has to be the coefficient on x
- here, so it has to be equal to minus 6.
- Add 2 to both sides, you get B.
- Minus 6 plus 2 is 4.
- So the other binomial, just substituting that Ax squared
- plus Bx, is 2x squared plus Bx.
- Oh, sorry.
- This is a minus 4.
- Minus 6 plus 2 is minus 4.
- So plus Bx.
- So minus 4-- that's B-- x.
- And that is choice A.
- Next problem.
- OK, they say, which of the following expressions is equal
- to-- this is problem 42.
- And they write x plus 2, plus x minus 2, times 2x plus 1.
- So we have to simplify this.
- And remember, order of operations, multiplication
- comes first. So we have to multiply these two expressions
- first. So let's do that.
- So this is-- I'll rewrite this one over here.
- x plus 2 plus-- and now let's multiply this.
- When you multiply these two binomials, you're really just
- doing the distributive property twice.
- And let me show that to you.
- We could view this as x minus 2 times 2x plus x
- minus 2 plus 1.
- So I'm just distributing the x minus 2 times each of these
- terms. So I could write this as x minus 2 times 2x, plus x
- minus 2 times the 1.
- All right, and now we can just simplify that by doing the
- distributive property again.
- So this is x plus 2 plus-- let's distribute the 2x times
- each of these.
- 2x times x is 2x squared.
- 2x times minus 2 is minus 4x.
- Plus, well, we're disturbing a 1.
- 1 times anything is just itself.
- So plus x minus 2.
- And let's see what we can do.
- We only have 1x squared terms, so let's write that down.
- 2x squared.
- So 2x squared.
- And then our x terms, we have a plus x, a minus
- 4x, and a plus x.
- So we have 1 minus 4 is minus 3.
- Plus 1 is minus 2.
- So its minus 2x.
- And then, let's see.
- We have a positive 2 and a minus 2.
- They cancel out.
- So we're left with 2x squared minus 2x, and that's choice A.
- Problem 43, I think we can fit in here.
- Let me copy and paste it.
- OK, copy it and now pasting it.
- OK, it says, a volleyball court is
- shaped like a rectangle.
- Let me draw that.
- Well, I didn't want to draw it filled in like
- that, but fair enough.
- Shaped like a rectangle.
- It has a width of x meters and a length of 2x meters.
- So it's width is x.
- Let me write, this could be x and this would be 2x.
- Because this is longer.
- Which expression gives the area of the
- court in square meters?
- Well the area is just the width times the length.
- So it's just x times 2x, which is equal to 2x squared.
- That's the same thing as 2 times, x times x, which is the
- same thing as 2x squared.
- And that's choice B.
- Anyway, see you in the next video.
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