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# 3-variable linear system word problem

Sal solves a word problem about the angles of a given triangle by modeling the given information as a system of three equations and variables. Created by Sal Khan and Monterey Institute for Technology and Education.

Video transcript

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