Systems of equations word problems
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Example 4: Solving a word problem with substitution
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Mixture problems 1
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Mixture problems 2
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Mixture problems 3
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Systems and rate problems
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Systems and rate problems 2
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Systems and rate problems 3
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Officer on Horseback
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Two Passing Bicycles Word Problem
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Passed Bike Word Problem
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Passing Trains
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Overtaking Word Problem
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Problem Solving Word Problems 2
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Systems of equations word problems
Problem Solving Word Problems 2
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- Let's do a couple of word problems. This tells us that
- Nadia's at home and Peter is at school, which is 6 miles
- away from home.
- So let me draw a little diagram here.
- So Nadia is there.
- Peter is over there, and they're 6 miles apart.
- So this is the distance between home and school.
- That is 6 miles.
- They start traveling to each other at the same time.
- So they both leave at the same time, traveling
- towards each other.
- Nadia is walking at 3 and 1/2 miles per hour.
- So she's travelling at 3 and 1/2 miles per
- hour in that direction.
- And Peter is skateboarding at 6 miles per
- hour in that direction.
- So Peter is going in that direction at 6 miles per hour.
- When will they meet and how far from home is
- their meeting place?
- So let's say that their meeting place is right-- I
- don't know.
- It's going to be closer to home, because this guy is
- going faster.
- Peter is going faster.
- So let's say, let x be equal to the distance
- from home they meet.
- So this distance right here is going to be x.
- And then what's this distance going to be?
- Well, if that's x, and this whole thing is 6 miles, then
- this distance right there is going to be 6 minus x.
- So Nadia will travel x miles and Peter will
- travel 6 minus x miles.
- And they're going to travel the same amount of time.
- Remember, they both leave at time 0, and after time t, they
- will both meet right there.
- So they're both going to take the same amount of time.
- So Nadia's equation, if we just remember that distance is
- equal to rate times time, we could use that for both of
- them, in Nadia's case, the distance she travels is x.
- x is going to be equal to her rate, which is 3.5 miles per
- hour, times t, which is the time that she travels.
- And then Peter's equation is going to be his distance.
- Well, he travels 6 minus x miles, so it's 6 minus x is
- going to be equal to his rate, which is 6 miles per hour
- times the time he travels.
- Well, he also travels t hours, so times t.
- So we have two equations and two unknowns, so we
- can solve for them.
- We've already solved for x here.
- So let's substitute this for x right there.
- So we can rewrite this equation as 6 minus, but
- instead of an x, we can put 3.5t, 6 minus 3.5t, because we
- know that x is equal to 3.5t, is equal to 6t.
- And then let's see, if we add 3.5t to both sides of this
- equation, we get 6 is equal to 3.5t plus 6t is 9.5t, or if we
- divide both sides by 9.5, you get t is equal to 6 over 9.5
- hours, which is a bit of a bizarre number.
- But 9.5, we can rewrite that.
- That is the same thing as 6 over-- see, 9.5, that's the
- same thing as 19/2, right?
- 19/2, you divide that and you would get 9.5.
- So this is equal to 6 times 2 over 19.
- So this is equal to 12/19 of an hour.
- That's how long it'll take them to meet.
- And then if you ask the question how far from home
- will they meet?
- If you want to figure out the x value, x is going to be
- equal to 3.5-- actually, instead of writing 3.5, I can
- write 3.5 as 7/2; that's the same thing as 3.5-- times our
- time, times 12/19.
- So you divide the numerator and the denominator by 2, and
- we get 7 times 6 is 42 over 19 miles.
- And this right here, that is in hours.
- So in 12/19 of an hour-- it's a weird fraction-- they will
- meet exactly 42/19 miles from home.
- So it's a little over 2 miles from home.
- Next problem.
- Peter bought several notebooks at Staples for $2.25.
- And he bought a few more notebooks at
- Rite Aid for $2 each.
- He spent the same amount of money in both places and he
- bought 17 notebooks in total.
- How many notebooks did he buy in each store?
- So let's define some variables.
- Let's let S equal number bought at Staples.
- And then we could say that R is equal to the number bought
- at Rite Aid.
- So what do we know?
- So he bought a total of 17 notebooks.
- Let me do that in a different color.
- He bought a total of 17 notebooks.
- So that tells us that S plus R is equal to 17.
- And we also know that he spent the same amount of money in
- both places.
- So how much did he spend at Rite Aid?
- Let me do this in orange.
- Spent the same amount of money in both places.
- So at Staples, how much did he spend?
- He bought S notebooks for $2.25 each, so 2.25S, that's
- how much he spent at Staples.
- That's going to be equal to the amount he
- spent at Rite Aid.
- He spent $2 per notebook at Rite Aid.
- So $2 times the number of notebooks from Rite Aid, so
- this is spent at Rite Aid.
- So once again, two equations with two unknowns.
- We can do a little bit of substitution maybe.
- So what's the best way to substitute here?
- Well, let's divide both sides of this
- equation by 2 right here.
- So you have-- I'm going to rewrite 2.2-- well, I'll just
- do it like this.
- So you have R.
- If you divide both sides of this by 2, you have 2.25 over
- 2S is equal to R, right?
- I just divided both sides of this by 2.
- So if you take this value and you substitute it in for R
- right there, this equation becomes S plus-- instead of an
- R we have 2.25 over 2S is equal to 17.
- And let's just simplify.
- So this is 1-- we could view this as 2 over 2S, right?
- The coefficient there is just 1.
- So we can view this as we have a common denominator.
- This is the same thing is 4.25 over 2S is equal to 17.
- I'm avoiding doing any hard math just yet.
- Let me multiply both sides by the inverse.
- 2/4.25.
- It's a little bizarre to have an expression with both a
- fraction and a decimal, but it's not illegal.
- Let's multiply both times 2/4.25.
- These cancel out, and so you get S is equal to 17 times 2
- is 34 over 4.25.
- And actually, I can eyeball that.
- That looks like that should be equal to 8, right?
- 4.25 is 4 and 1/4, which is the same thing as 17/4 over 4.
- So this is the same thing as 34 over 17/4, which is the
- same thing as 34 times 4/17.
- Put a 1 there, divide by 17, you get a 2, divide by 17, you
- get a 1, 2 times 4 is 8.
- So he bought 8 notebooks at Staples, this is 8, and then
- at Rite Aid, he bought-- well, 8 plus R is equal to 17.
- Subtract 8 from both sides.
- He must have bought 9 notebooks at Rite Aid.
- Let's do one more.
- This one is especially fun-looking.
- Peter is outside, looking at the pigs and
- chickens in the yard.
- Nadia is indoors and cannot see the animals.
- Peter gives her a puzzle.
- He tells her that he counts 13 heads and 36 feet and asks her
- how many pigs and how many chickens are in the yard.
- So let's once again define our variables.
- P is equal to the number of pigs.
- And let C is equal to the number of chickens.
- So the number of heads will essentially be the number of
- pigs and chickens, assuming they each have one head.
- So the number of pigs plus the number of chickens will be
- equal to the number of heads, so that is right here.
- That is 13 heads.
- And then 4 times the number of pigs, right?
- Each pig has 4 legs, so 4 times the number of pigs plus
- 2 times the number of chickens, assuming we're
- dealing with two-legged chickens, is going to be equal
- to the number of feet, is equal to 36.
- So once again, we have two equations with two unknowns.
- Instead of solving it with substitution, I'm going to do
- it by adding and subtracting the two equations.
- So what we can do, this equation, we can
- multiply it times 2.
- And when I say multiply it times 2, we have to multiply
- the entire equation times 2.
- And so it's still true.
- So we multiply the entire equation times 2.
- This becomes 2P plus 2C is equal to 26.
- And now what I'm going to do is I'm going to subtract this
- equation from that equation, which you can do.
- Because that is equal to that, that is equal to that, and so
- we're not doing anything that violates the laws of algebra.
- So if you subtract the bottom equation from the top
- equation, or I could just put a negative sign everywhere
- here, just so you know we're subtracting,
- 4P minus 2P is 2P.
- 2C minus 2C is 0.
- That's why I did this, so that these would cancel out.
- And then 36 minus 26 is equal to 10.
- So 2P is equal to 10.
- Divide both sides by 2.
- We have 5 pigs.
- And then 5 plus the number of chickens is equal to 13, so we
- must have-- subtract 5 from both sides-- 8 chickens.
- Now if this confuses you, you might want to try it with
- substitution, but I have many videos on solving systems of
- equations where I go a little bit slower and explain a
- little bit more of the logic of it.
- So whatever floats your boat.
- But either way, Nadia should hopefully guess or hopefully
- solve for 5 pigs and 8 chickens.
- And it should work out, right?
- You have 13 animals.
- You have 13 heads.
- And if you multiply 5 times 4, that's 20 pig feet plus 16
- chicken feet.
- 20 plus 16 is 36 total feet.
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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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