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# System of equations word problem: walk & ride

Systems of equations can be used to solve many real-world problems. In this video, we solve a problem about distances walking and riding bus to school.

## Want to join the conversation?

• I get confused by this type of solution because of the units. It's funny how you can just have a system of equations despite all their different units. Like he sums one equation that deals with distance (km) and another that deals with time (hours). How is that not conflicting? This cause my brain to shut down a little...
• Maybe it's easier to see this when you make the times, rather than the distances, the unknowns.

If w is time walking and b is time busing, then

1) 5w + 60b = 35
[(5mph * w hrs) + (60mph * b hrs) = 35 miles (that is, speed1*time1 + speed2*time2 = total distance)]

2) 2w + 2b = 3
[from w + b = 1.5. (that is, time1 + time2 = total time)].

You can see here that while equation "1)" is about distance, and equation "2)" about time, that since the rates (5mph and 60mph) are not unknowns, they can be thought of simply as numbers that show another way that the two times are related, which gives you 2 (different) equations in two unknown times, which is enough information to get a solution.
• How did you get 11/12B by adding B and -B/12 together?
I don't get it at all! Please explain in an easier way..
• according to professionals from NASA and millions of researchers all across the globe:
We got B and add -B/12 or
B-B/12
which is equal to
1B-1B/12
look that the Bees have a legitimate exponent corresponding to one(1)
now:
1 is and can be 12/12 or any number over its number like 1/1
so:
12/12B-1B/12 and 1B/12 is 1/12B
so:
12/12B-1/12B
put bees aside and work out the fraction because everything we're working is with bees:
(12/12-1/12)B = 11/12 B
• I like to use substitution when solving problems with multiple equations. Is there a preferred method?
w = time walking, b = time riding the bus
w + b = 1.5, w = 1.5 - b, b = 1.5 - w
5w + 60b = 35 (speed times time equals distance)
5w + 60(1.5 - w)=35
5w + 90 - 60w = 35
5w - 60w + 90 - 90 = 35 - 90
-55w = -55
w = 1, b = 0.5
5(1) + 60(.5) = 35
5 + 30 = 35
The trip to school: 5 kilometers walked in an hour and 30 kilometers traveled on the bus in 0.5 hours.
• this is the simplest and best solution to me by far
• I struggle when doing word problems in how to define my variables. I noticed in this video that Sal went straight to what question the problem was asking and defined his variables based on that. Is that a good idea in general to define your variables based on the question?
• @bhannon039 I struggled with your problem as well in the past, I finally figured out the exact same solution as yours, and it has helped a lot.
• i don't get how you can subtract 1.5hours from 35 kilometers :(
• See, when we convert problems to equations we don't write it with units as we have the same variables on the same side(in this case 'w' and 'b') so we can proceed further without caring about their units.
• SAL moves pretty fast through this video and he does lots of stuff that we've learned before but he doesn't explain in minute detail.

Let's try to explain what SAL did here.

The total distance is the sum of the unknown walking distance W and the unknown bus distance B.

So W + B = 35 km. I will call this the "top equation".

We know that distance divided by speed is time, so we know that the unknown walking distance W divided by speed of 5km/hr = W distance / 5 kmh and the unknown bus distance is B divided by 60 kmh = B/60.

So W+B = 35 and W/5 + B/60 = 1.5. I will call this the "bottom equation".

SAL decides to eliminate the variable "W" and notices that w/5 is 1/5th of W. Remember multiplying fractions? Convert 5 to 5/1, then multiply the top numbers together, then separately multiply the bottom numbers. So W multiplied by 5 is 5W and 5 multiplied by 1 is 5. So W/5 x 5/1 becomes 5W/5. Whenever the top number (numerator) in a fraction and the bottom number are the same number, then the fraction is "1".

2/2 is 1. 3/3 is 1. 4/4 is one. 5/5 is 1.

So 5/5 with a W next to it is 1 x W, which is another way of writing "W".

Sal is trying to eliminate W so what he actually did was multiply by -5, which meant that W/5 x -5/1 is -W.

To keep the equality, SAL then multiplied the rest of the bottom equation by -5.

SAL now tackles multiplying B/60 by -5. Remember -5 as a fraction is -5/1. SAL simplified these fractions. Put the two fractions side by side and you simplify the top left by dividing by the bottom right and the bottom left by dividing by the top right. So... look...

B x -5
60 1

60 / -5 = 12. B / 1 is one. That results in -B/12.

SAL (metaphorically) jumps over the equals sign and multiplies 1.5 by -5 to get -7.5.

So the system of equations now reads

W+B=35
-w-b/12 = -7.5

SAL now adds the bottom equation to the top equation.

W-W = 0, so the W disappears!

B/12 really means "one twelfth of B". 12/12 is a full one e.g. 12/12 of a pie is a full pie, so, just as 12/12 of a pie minus 1/12 of a pie results in 11/12 of a pie, then one full B (or, if you prefer, 12/12 of a B) minus 1/12 of a B = 11/12 of a B i.e. 11/12 x B.

SAL again jumps over the equality. He adds the bottom to the top... 35 plus -7.5 is 27.5.

Because we have a fraction in the calculation, SAL converts the decimal number 27.5 into a fraction.

SAL's pretty clever and spots that 27.5 = 27.5 / 1, and that if he doubles that fraction by 2 he will have the fraction 55 / 2. He does that because he knows that 12 can be divided by 2.

So the bottom equation now reads 11/12B = 55 / 2.

To shove the 11/12 fraction from the left to the right, he multiplies it by its reciprocal which is 12/11. 11/12 x 12/11 is one (the 12s cancel each other and the 11s cancel each other).

That tells him that B = 55/2 x 12/11.

SAL then cancels these. So 2 / 2 is 1. 12 / 2 is 6. 11 / 11 is one and 55 / 11 is 5. That gives him this...

5/1 x 6/1. 5 x 6 = 30. 1 x 1 is 1.

so that means B = 30! That's 30k by bus!

SAL pus 30 into the "B" spot in the equation W+B=35. So it now reads W+30=35.

We know that 35-30 is 5. So we now know that W is 5.

so we know that Yochan walks 5km and takes the bus 30km.

Personally, I think Yochanan should have just taken a taxi.
• Why does he put w over 5 and b over 60 into fractions? Is there other ways you where you don't have to have fractions to solve the linear equation?
• w/5 just means w divided by 5 and same with B/60. You divide because you want to find the rate of how fast he's walking (w) and how long he was on the bus (B).
• Where did the 12 come from? Sounds like it should’ve been
5b/60.

5/60 is NOT 12.
• You are correct - it isn't 12 and Sal doesn't make it 12. 5b/60 when reduced is equal to b/12.
the fraction is reduced by dividing both the numerator and denominator by 5: 5b/5=b and 60/5=12.

Hope this helps.
• This might be dumb, but why did he multiply the equation by -5?
• No serious question is dumb. In order to use elimination on a system of equations, you have to have one of the variables be additive inverses in the two equations. Since you have a w (or 1w) on the first equation, to eliminate the w's, you need a -1w on the other since w - w = 0. Thus, starting with w/5, multiplying by -5 on whole second equation reaches that goal, adds the negative and 5 * 1/ = 1.