Systems with three variables
Three Equation Application Problem Three Equation Application Problem
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- Solve the following application problem using three equations with three unknowns.
- And they tell us the second angle of a triangle is 50 degrees less than 4 times the first angle.
- The third angle is 40 degrees less than the first.
- Find the measures of the three angles.
- Let's draw ourselves a triangle here.
- And let's call the first angle "a",
- the second angle "b",
- and then the third angle "c".
- And before we even look at these constraints,
- one property we know of triangles is that the sum of their angles must be 180 degrees.
- So we know that a + b + c must be equal to 180 degrees.
- Now with that out of the way let's look at these other constraints.
- So they tell us the second angle of a triangle is 50 degrees less than 4 times the first angle.
- So we're saying b is the second angle.
- So they're second the angle of a triangle is 50 degrees less than 4 times the first angle.
- So 4 times the first angle would be 4a (we're calling a the first angle).
- So 4 times the first angle is 4a but its 50 degrees less than that
- so minus 50.
- Now the next constraint they give us:
- the third angle is 40 degrees less than the first.
- So the third angle is 40 degrees less than the first.
- So the first angle is a and it's going to be 40 degrees less than that.
- So we have 3 equations with 3 unknowns
- and so we just have to solve for them.
- Let's see, what's a good first variable to try to eliminate.
- And just to try to visualize that a little bit better,
- I'm going to bring these a's onto the left-hand side of each of these equations over here.
- So I'm going to rewrite the first equation.
- We have
- a + b + c = 180
- and then this equation, if we subtract 4a from both sides of this equation we have
- -4a + b = -50.
- And then this equation right over here,
- if we subtract a from both sides we get
- -a + c = -40.
- I just subtracted a from both sides.
- So we now want to eliminate variables.
- And we already have this third equation here is only in terms of a and c,
- this is only in terms of a and b,
- and this first one is in terms of a, b and c.
- Let's see, this is already in terms of a and c;
- if we could turn these first two equations,
- if we could use the information in these first two equations to end up with an equation that's only in
- terms of a and c, then we could use whatever we end up with along with this third equation right over
- here and we'll have a system of 2 equations with 2 unknowns.
- So let's do that.
- So if we wanted to just end up with an equation
- only in terms of a and c using only these first 2, we would want to eliminate the b's...
- so we could multiply one of these equations time negative 1
- and one of these positives b's would turn into a negative b.
- So let's do that. Let's multiply this first equation over here times -1.
- So it will become
- -a - b - c = -180,
- and then we have this green equation right over here
- which is really just this equation, just rearranged.
- So we have
- -4a + b = -50
- and now we can add these two equations.
- Actually let me do that in the other color just so you see where that's coming from.
- This is
- -4a + b = -50.
- We can add these two up now
- and we get
- -a - 4a = - 5a,
- the b's cancel out,
- we have a minus c,
- is equal to
- -180 - 50 = -230
- So now using these top two equations we have an equation only in terms of a and c,
- we have another equation only in terms of a and c,
- and it looks like if we add them together the c's will cancel out.
- So let me just rewrite this equation over here.
- And you have to be careful that you're using all of the equations
- otherwise you'll kind of do a circular argument.
- You have to be careful that over here,
- this first equation came from
- these two over here
- Now I want to combine that with this third constraint,
- a constraint that's not already baked into this equation right over here.
- So we have
- -a + c = -40
- We add these two equations:
- -5a - a = -6a,
- the c's cancel out,
- and then you have -230 - 40, this is equal to -270,
- we can divide both sides by -6,
- and we get a is equal to -270 over -6.
- 270 is divisible by both 3 and 2 so it should be divisible by 6,
- so let me just divide it;
- the negative signs obviously will cancel,
- a negative divided by a negative is going to be a positive.
- If we take 6 into 270, 6 goes into 27 four time
- 4 x 6 = 24
- we subtract
- we get 3, bring down the zero
- 6 goes into 30, 5 times
- So we get a is equal to 45.
- Now let's look at the other ones.
- We can substitute back into to solve for c.
- c is equal to a minus 40 degrees.
- So that is equal to, in yellow,
- so c is equal to 45 minus 40 which is equal to 5 degrees.
- So, so far we have a = 45 degrees,
- c = 5 degrees,
- and then you can substitute into either one of these other ones to figure out b.
- We can use this one right over here in green:
- b = 4a - 50
- So b is going to be equal to 4 times 45...
- let's see, 2 x 45 is 90, so 4 x 45 is 180
- so it's going to 180 minus 50 by this equation right over here
- which is equal to 130 degrees.
- So we get b is equal to 130 degrees.
- So let me write it right over here.
- So a is equal to 45.
- If I wanted to draw this triangle it would actually look something like this:
- a is a 45 degree angle,
- b is a 130 degree angle,
- and c is 5.
- So it'll look something like this
- where this is a at 45 degrees,
- b is 135 degrees [oops],
- and then c is 5 degrees.
- And you can verify that it works.
- One, you could just add up the angles
- 45 + 5 is 50.
- Oh, sorry, this isn't 135, it's 130.
- We solved it right over here
- and this is 5.
- So when you add them all up
- 45 + 130 + 5
- that does indeed equal 180 degrees;
- 45 + 5 is 50
- plus 130 so this does definitely equal 180.
- So it meets our first constraint.
- Then on our second constraint
- b needs to be equal to 4a - 50
- well 4 x a = 180
- 180 - 50 = 130 degrees
- so it meets our second constraint.
- And then our third constraint
- c = a - 40 degrees
- Well a is 45, c is 5, so if subtract 40 from 45 you get 5 which is c
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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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