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Solve the following application problem using three equations with three unknowns.

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And they tell us the second angle of a triangle is 50 degrees less than 4 times the first angle.

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The third angle is 40 degrees less than the first.

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Find the measures of the three angles.

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Let's draw ourselves a triangle here.

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And let's call the first angle "a",

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the second angle "b",

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and then the third angle "c".

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And before we even look at these constraints,

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one property we know of triangles is that the sum of their angles must be 180 degrees.

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So we know that a + b + c must be equal to 180 degrees.

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Now with that out of the way let's look at these other constraints.

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So they tell us the second angle of a triangle is 50 degrees less than 4 times the first angle.

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So we're saying b is the second angle.

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So they're second the angle of a triangle is 50 degrees less than 4 times the first angle.

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So 4 times the first angle would be 4a (we're calling a the first angle).

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So 4 times the first angle is 4a but its 50 degrees less than that

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so minus 50.

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Now the next constraint they give us:

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the third angle is 40 degrees less than the first.

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So the third angle is 40 degrees less than the first.

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So the first angle is a and it's going to be 40 degrees less than that.

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So we have 3 equations with 3 unknowns

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and so we just have to solve for them.

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Let's see, what's a good first variable to try to eliminate.

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And just to try to visualize that a little bit better,

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I'm going to bring these a's onto the left-hand side of each of these equations over here.

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So I'm going to rewrite the first equation.

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We have

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a + b + c = 180

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and then this equation, if we subtract 4a from both sides of this equation we have

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-4a + b = -50.

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And then this equation right over here,

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if we subtract a from both sides we get

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-a + c = -40.

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I just subtracted a from both sides.

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So we now want to eliminate variables.

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And we already have this third equation here is only in terms of a and c,

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this is only in terms of a and b,

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and this first one is in terms of a, b and c.

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Let's see, this is already in terms of a and c;

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if we could turn these first two equations,

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if we could use the information in these first two equations to end up with an equation that's only in

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terms of a and c, then we could use whatever we end up with along with this third equation right over

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here and we'll have a system of 2 equations with 2 unknowns.

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So let's do that.

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So if we wanted to just end up with an equation

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only in terms of a and c using only these first 2, we would want to eliminate the b's...

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so we could multiply one of these equations time negative 1

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and one of these positives b's would turn into a negative b.

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So let's do that. Let's multiply this first equation over here times -1.

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So it will become

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-a - b - c = -180,

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and then we have this green equation right over here

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which is really just this equation, just rearranged.

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So we have

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-4a + b = -50

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and now we can add these two equations.

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Actually let me do that in the other color just so you see where that's coming from.

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This is

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-4a + b = -50.

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We can add these two up now

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and we get

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-a - 4a = - 5a,

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the b's cancel out,

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we have a minus c,

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is equal to

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-180 - 50 = -230

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So now using these top two equations we have an equation only in terms of a and c,

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we have another equation only in terms of a and c,

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and it looks like if we add them together the c's will cancel out.

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So let me just rewrite this equation over here.

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And you have to be careful that you're using all of the equations

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otherwise you'll kind of do a circular argument.

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You have to be careful that over here,

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this first equation came from

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these two over here

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Now I want to combine that with this third constraint,

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a constraint that's not already baked into this equation right over here.

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So we have

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-a + c = -40

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We add these two equations:

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-5a - a = -6a,

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the c's cancel out,

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and then you have -230 - 40, this is equal to -270,

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we can divide both sides by -6,

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and we get a is equal to -270 over -6.

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270 is divisible by both 3 and 2 so it should be divisible by 6,

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so let me just divide it;

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the negative signs obviously will cancel,

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a negative divided by a negative is going to be a positive.

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If we take 6 into 270, 6 goes into 27 four time

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4 x 6 = 24

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we subtract

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we get 3, bring down the zero

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6 goes into 30, 5 times

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So we get a is equal to 45.

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Now let's look at the other ones.

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We can substitute back into to solve for c.

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c is equal to a minus 40 degrees.

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So that is equal to, in yellow,

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so c is equal to 45 minus 40 which is equal to 5 degrees.

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So, so far we have a = 45 degrees,

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c = 5 degrees,

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and then you can substitute into either one of these other ones to figure out b.

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We can use this one right over here in green:

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b = 4a - 50

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So b is going to be equal to 4 times 45...

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let's see, 2 x 45 is 90, so 4 x 45 is 180

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so it's going to 180 minus 50 by this equation right over here

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which is equal to 130 degrees.

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So we get b is equal to 130 degrees.

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So let me write it right over here.

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So a is equal to 45.

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If I wanted to draw this triangle it would actually look something like this:

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a is a 45 degree angle,

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b is a 130 degree angle,

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and c is 5.

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So it'll look something like this

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where this is a at 45 degrees,

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b is 135 degrees [oops],

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and then c is 5 degrees.

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And you can verify that it works.

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One, you could just add up the angles

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45 + 5 is 50.

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Oh, sorry, this isn't 135, it's 130.

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We solved it right over here

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and this is 5.

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So when you add them all up

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45 + 130 + 5

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that does indeed equal 180 degrees;

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45 + 5 is 50

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plus 130 so this does definitely equal 180.

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So it meets our first constraint.

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Then on our second constraint

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b needs to be equal to 4a - 50

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well 4 x a = 180

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180 - 50 = 130 degrees

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so it meets our second constraint.

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And then our third constraint

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c = a - 40 degrees

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Well a is 45, c is 5, so if subtract 40 from 45 you get 5 which is c

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so it meets all of our constraints and we are done.