Ex 3: Distributive property to simplify Equations with Variables on Both Sides
Ex 3: Distributive property to simplify
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- Like the last video, I want to start with two warm up
- problems. And then we'll do an actual word.
- And you're going to see these are going to be a little bit
- more involved than the equations in the last video.
- But we're still going to be doing the exact same
- operations, or what we could consider legitimate
- operations, to get our answer.
- So here we have 3 times x minus 1 is equal to 2
- times x plus 3.
- So let's see what we can do here.
- The first thing I like to do-- and there's no definite right
- way to do it-- there's several ways you could do these
- problems-- but I like to distribute out
- the 3 and the 2.
- So 3 times x minus 1, that's the same thing as 3x minus 3--
- I just distributed the 3-- is equal to--
- distribute out the 2.
- 2 times x, plus 2 times 3, which is 6.
- Now what I like to do is get all of my constant terms on
- the same side of the equation and all my variable terms on
- the same side of the equation.
- So let's see if we can get rid of this 2x term on
- the right hand side.
- So let's subtract 2x.
- I'm going to do a slightly different notation this time
- because you might see it done this way.
- Or you might find it easier to visualize it this way.
- It doesn't matter.
- It's the same thing we did in the last video.
- But I want to subtract 2x from this side of the equation.
- But if I subtract 2x from the side of the equation, I also
- have to subtract 2x from that side of the equation.
- So then when we subtract 2x from both sides of the
- equation, what do we get?
- Here we get 3x minus 2x.
- That's just 1x, or x minus 3.
- 2x minus 2x is no x's, or 0.
- Then you just have the 6.
- So we get x minus 3 is equal to 6.
- That was by getting rid of the 2x from the right hand side;
- subtracting it from both sides of this equation.
- Now we have this negative 3 on the left hand side.
- I don't want it there.
- I just want an x there.
- So to get rid of that we can add 3 to both
- sides of this equation.
- You could imagine this is being adding this equation to
- the equation 3 is equal to 3.
- 3 is, obviously, equal to 3.
- Negative 2x is, obviously, equal to negative 2x.
- You could do it either way.
- But if you add 3 to both sides of this equation the left hand
- side of the equation becomes just an x, because these two
- guys cancel out. x equal, and 6 plus 3 is 9.
- And we are done.
- And we can even check our answer.
- 3 times 9 minus 1 is what?
- This is 3 times 8.
- This is 24.
- So that's what the left hand side equals.
- What does the right hand side equal?
- That is 2 times 9 plus 3.
- That's the 2 times 12, which is also equals 24.
- So it all works out. x is equals to 9.
- Next problem.
- z over 16 is equal to 2 times 3z plus 1, all of that over 9.
- So it's a hairy looking problem.
- Let's multiply both sides of this equation by 9.
- So if you multiply both sides of this equation by
- 9, what do we get?
- We get 9 over 16z is equal to-- this 9 and that 9 will
- cancel out-- 2 times 3z plus 1.
- Now let's distribute this 2.
- So we get 9 over 16z is equal to 2 times 3z is 6z plus 2.
- Now let's get all of z's on the same side of the equation.
- So let's subtract 6z from both sides of the equation.
- So let's subtract minus 6z there, and that, of course,
- equals minus 6z there.
- And what do we get?
- On the left hand side, we get 9/16 minus 6.
- What is 6 if I have 16 as a denominator?
- 6 is equal to what over 16?
- Let me think about it.
- 60 plus 36.
- It's equal to 96/16.
- So it's 9 minus 96 over 16z.
- This is just 6-- I just rewrote minus 6 here-- is
- equal to-- these two cancel out.
- That's why I subtracted 6z in the first place.
- So it's going to equal 2.
- So what does this equal right over here?
- Let me do it in orange.
- 9 minus 96.
- Well let's see.
- The difference between 9 and 96 is going to be 87.
- And, of course, we're subtracting the larger from
- the smaller.
- So it's going to be negative 87 over 16z is equal to 2.
- And we're almost there.
- We just have to multiply both sides of this equation by the
- inverse of this coefficient.
- So multiply both sides of this equation by negative 16/87.
- These cancel out.
- 87, 87, 16 and 16.
- The negative signs plus.
- You're just left with z is equal to 2 times this thing.
- So 2 times negative 16 is negative 32/87.
- And we are done.
- That's not a pretty looking answer, but
- that's what we got.
- And you can try it with multiple methods.
- Maybe you can multiply both of the equation by 16 first.
- Maybe you can distribute out the 2 9's first. All sorts of
- things you can do.
- And you could also verify that this is, indeed, the answer.
- Let's do a word problem now that we're warmed up.
- All right.
- It says Manoj and Tomar are arguing about how a number
- trick they heard goes.
- Tomar tells Andrew to think of a number.
- Let's say that x is the number that Andrew thinks of.
- Think of a number.
- Multiply it by 5.
- So this is what Tomar is saying.
- Ć is telling Andrew to think of a number, multiply it by 5,
- so let me do that.
- So multiply it by 5, and subtract 3 from the result.
- We'll do that in a different color.
- We'll do it in blue.
- Subtract 3 from the result.
- So subtract 3 from the result.
- Then Manoj comes along.
- So here's Manoj.
- Manoj comes along and tells Andrew once again
- to think of a number.
- So once again, think of a number.
- We'll call that x.
- So they both are telling him to think of a number.
- Add five.
- So now he's saying add five to the number.
- So you add 5 to number.
- And then multiply the result by 3.
- So multiply the result by 3.
- Andrew says that which ever way he does the trick, he gets
- the same answer.
- What was Andrew's number?
- So regardless of whether he does 5 times the number, and
- then subtracts 3, or whether he adds 5 and then multiples
- the whole answer by 3, he gets the same number.
- So that must mean that these two things are equal.
- He gets the same answer.
- That means that these two are equals.
- So let's solve this equation.
- We get 5x minus 3 is equal to 3 times x plus 5.
- A good place for me-- I like to distribute out this 3, so
- that is equal to 3x plus 15.
- Always have to remember to multiply the 3 times all of
- the terms in parentheses.
- This is a 15.
- And, of course, 5x minus 3 is equal to that.
- Now, I think you know how I like to operate.
- I like to get all of my ex-coeffients on one side.
- So let's get them all on the left hand side.
- Which means, let's get rid of them on the right hand side.
- So let's subtract 3x from both sides.
- So minus 3x minus 3x and then what do we get?
- Our equation becomes 5x minus 3x is 2x.
- You still have a minus 3 is equal to-- these cancel out--
- equal to 15.
- Equal to that yellow 15 right there.
- Now we want to get rid of this negative 3 on
- the left hand side.
- And the best way to do it is to add 3 to both
- sides of the equation.
- So add 3 to both sides of this equation.
- So what do you get?
- You now get 2x-- and these cancel out-- is equal to 18.
- And then you divide both sides of this by 2.
- And you get x is equal to 18 over 2.
- Or 9.
- So in either situation, Andrew was thinking of the number 9.
- Let's see if it works according to both Tomar and
- Manoj's method.
- If I were to take 5 times-- if this is 9-- you get 5 times 9
- is 45 minus 3.
- So you get 42.
- That's when you do Tomar's method.
- When you do Manoj's method-- this is a 9, right there.
- 9 plus 5 is 14.
- So this thing becomes 14, and it becomes 3 times 14.
- 3 times 14 is 30 plus 12, which is 42.
- So Andrew was right.
- When he randomly picked the number 9, and regardless of
- which method he uses, he gets the same results.
- He gets 42.
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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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