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## Systems of equations word problems

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# System of equations word problem: walk & ride

CCSS.Math: , , , ,

## Video transcript

- [Voiceover] Yochanan walked
from home to the bus stop at an average speed of
five kilometers per hour. He immediately got on his school bus and traveled at an average speed of 60 kilometers per hour
until he got to school. The total distance from his home to school is 35 kilometers, and the entire trip took one and a half hours. How many kilometers did Yochanan cover by walking, and how many kilometers did he cover by travelling
on the bus? It's fascinating. So I encourage you to pause the video and try to think about on your own. Alright, so let's just
define some variables here. How many kilometers did
he cover by walking? Let's call this W and how many kilometers did he cover by travelling on the bus? Let's call that, Let's call that B and so what do we know? We know that the kilometers and I can actually even
draw it just to make sure that we're visualizing this thing, right? So this right over here is
his home, that's his home and then he is going to
travel and see they tell us, they tell us that it is
35 kilometers to school so this is his school right over here. I'll draw a bigger
building, that's his school, and we know that this
distance is 35, 35 kilometers and we also know that it
took one and a half hours. One and a half hours, now he traveled at different rates for
different distances. So he traveled some
distance to the school bus so this is, or to the bus stop, so that's the bus stop, right over there, and we're seeing this
distance to the bus stop, that's how much he covered by walking. So this right over here, this
distance, right over here, that is W and the rest of the distance, he covered by the bus, so
the rest of this distance, all of this distance right over there, that is going to be B. So what do we know? We know the distance covered by walking plus the distance covered by bus is going to be 35 kilometers,
35 kilometers here, this is the entire, that
is the entire distance from home to school, so
we know that W plus B, W plus B, plus B is
equal to 35 kilometers, is equal to 35 kilometers
and just with one equation, we're not going to be able to
figure out what W and B are but we have another constraint. We know the total amount of time. So the total amount of time is going to be one and half hours, so we'll
just write that over here. This is going to be 1.5, so
what's the time traveled by, what's the time he walks? Let me write this over here, time time walking, we'll that's going to be the distance walking
divided by the rate walking. So the distance walking is W kilometers W kilometers divided by his rate, the distance divided by your rate is gonna give you your time, so let's see, his rate is five kilometers per hour, five kilometers per
hour and so you're gonna have kilometers cancel kilometers and if you divide by or
if you have one over hours in the denominator, that's
going to be the same thing, this is gonna be W over five
hours, so the units work out. So his time walking is W over five, W over five and by that same logic, his time on the bus is
going to be the distance on the bus divided by,
divided by the average speed of the school bus, so
this is going to be 60. This is all going to be in hours and now we can solve
this system of equations. We have two linear
equations with two unknowns. We should be able to find W and B that satisfy both of these. Now what's an easy thing to do? Let's see, if I can multiply
this second equation by negative five, and I'm gonna, this is going to be a negative W here so it'll cancel out with this W up there. So let's do that, let's
multiply the second equation by, I'm just gonna
switch to one color here, so this top equation is going
to be W plus B is equal to 35. This bottom equation,
if I multiply both sides by negative five, so both
sides by negative five, I'm going to multiply both
sides by negative five I'm going to get negative
five times W over five is negative W, negative
five times B over 60. Let's see, it's gonna be,
it's going to be negative five over 60, that's negative
1/12, so this is negative B over 12 and then it is
going to be equal to 1.5 times negative five is
negative 7.5, negative 7.5. Now we can just add the left and right hand sides of these two equations. Now let me, I can do
this a little bit neater, let me actually delete, let me make these line up a little bit better
so that we, delete that, make this, so this first
equation was, whoops, this first equation was
W plus B is equal to 35 now they line up better and now we can add the left hand sides of these equations and the right hand sides,
so the left hand sides, the W's cancel out. That's what we wanted. Now B minus B over 12 or
B minus the twelfth of B, well that's going to be 11/12B, 11/12B is equal to, let's see, 35 minus 7.5, see 35 minus seven would be 28 and then another
half this would be 27.5. 27.5 and since I'm dealing
with a fraction here, let me write this as a fraction so this is the same thing as 55 over two. Let's me just write it this way. This is the same thing as 55 over two. Now to solve for B, I
just have to multiply both sides times this reciprocal. I'll switch colors just
to ease the monotony. So multiply both sides by
12 over 11, 12 over 11, what we get, these cancel out, what we get is that B, so I'll do this in this color, B is equal to, let's see, I
have a 12 in the numerator two in the denominator so I can make that a six and a one,
then I have a 55 and 11, I can divide both by 11, so
it's gonna be five and a one. So it's five times six, B is equal to 30 and B was in kilometers so
if he travels 30 kilometers, 30 kilometers by bus, B
is equal to 30 kilometers, and the amount that he walks,
well, we can figure that out. If this is 30, well the amount that he walks is going to be five. This is going to be five kilometers, W, let me write this, W is
equal to five kilometers. He walks five kilometers and then he goes by the bus 30 kilometers.