Why are we still teaching algebra?

[Charlie Velasquez]

t = d/v

Just because you don't show your work doesn't mean it isn't algebra.

Yes! Another example of algebra simplifying a problem! Done by a fifth grader...

The problem was the classic fly on the trains.

There is a train on a track in Muscatine, IA, going west at 2mph. There is another train on the same track in Des Moines, IA, going east at 3 mph. There is a fly on the front of the train in Muscatine, it flies at 7 mph straight to the Des Moines train then immediately turns around, flies back to the Muscatine train, then back to the Des Moines train, and so on.. till the trains collide!
The fly begins its journey when the trains are at speed and 150 miles apart.
How far does the fly fly?

Most of my class struggled with this. Then one student answered correctly.
Me: "Outstanding! how did you get this answer?"
Lori: "Well, you said we don't do hard problems, so I changed it to a simple problem.
If I was driving 10 mph and I wanted to know how far I have gone I need to know how long I was driving, then multiply. So if I wanted to know how far the fly went I need to know how long it was flying. So the real problem is how long before they crashed. Then you just multiply"


She came up with a general formula, inputted the data specific to this problem, then it was just arithmetic. The essence of teaching math.

[Charlie Velasquez]

We were always told, to get credit, you must not only have the right answer but also show your work.

You would probably not have liked some of my tests. Sometimes they already had all the answers on them. They only got credit if they showed how to get it. And just to make it more fun, sometimes I put the wrong answer on 2 or 3 problems and they had to find them.

[Matt Mattingley]

I personally use algebra all the time. Algebra is just a way of thinking through a word problem. Usually a alphabetical symbol or symbols can be used. a² x b² =c² Sure you could use trig.

Here’s one I give my students. I actually get them to work through it. Solve for X. They already have the training in other areas of math. They need to give me an algebraic equation that will always solve for X. There could be more or less sides or angles but always one angle is missing. How do you solve for X?

Algebra is usually used when trying to solve an unknown with enough known. I’ll leave this one with you guys for fun. 0A107D10-0BCA-4C56-857E-CEA8120D2E7D.jpg

[Bill Boehme]

Knowing and using higher math on a daily basis saved me from a career of asking "would you like fries with that"?

[Tom Stenzel]

Knowing and using higher math on a daily basis saved me from a career of asking "would you like fries with that"?

Algebra is easy.
Board-feet is hard.

By the way Bill, you DID ask for fries. I remember that when I took your order...

-Tom

[Bob Glenn]

Charlie, that was great, I don't think I could have sorted that one out on my own. Maybe you can help me out with this classic puzzle:

Three guys go into a hotel to spend the night, but there is only one room left, so they decided to share the last room. The hotel manager said that will be thirty dollars for the room, so each man paid ten dollars to make up the thirty and they went up to the room. Soon after, the hotel manager realized that he had charged the men five dollars too much, so he had the bell boy refund the five dollars. However, the bell boy didn't know how to split the five dollars three ways, so he just gave each man a one dollar refund and kept the two dollars left over as a tip.

So now, each man paid ten dollars then got a dollar back, so now each man paid nine dollars. So the total the men paid was twenty seven, plus the two dollars the bell boy kept, make it twenty nine dollars. What happened to the other dollar?

[Brett Luna]

Charlie, that was great, I don't think I could have sorted that one out on my own. Maybe you can help me out with this classic puzzle:

Three guys go into a hotel to spend the night, but there is only one room left, so they decided to share the last room. The hotel manager said that will be thirty dollars for the room, so each man paid ten dollars to make up the thirty and they went up to the room. Soon after, the hotel manager realized that he had charged the men five dollars too much, so he had the bell boy refund the five dollars. However, the bell boy didn't know how to split the five dollars three ways, so he just gave each man a one dollar refund and kept the two dollars left over as a tip.

So now, each man paid ten dollars then got a dollar back, so now each man paid nine dollars. So the total the men paid was twenty seven, plus the two dollars the bell boy kept, make it twenty nine dollars. What happened to the other dollar?

The 'other dollar' is a red herring. You don't add the two dollars, you subtract it.

The math should be: Actual room cost ($25) = amount paid ($30) – partial refund for overcharge ($3) – amount kept by bellboy ($2)

[Matt Mattingley]

This is why the saying BEDMAS needs to be applied. Brackets Exponents Division Multiplication Addition subtraction. This is called order of operations. 10x3=1+1+1+2+25=25+3+2= (25÷3)(1x3)+2



The three men are out $28. 28÷3=$9.33 they each gave $.66 tip to the bellboy. $.66x3=$2.00

[Charlie Velasquez]

The 'other dollar' is a red herring. You don't add the two dollars, you subtract it.

The math should be: Actual room cost ($25) = amount paid ($30) – partial refund for overcharge ($3) – amount kept by bellboy ($2)

Yes, But I will use your equation to form the equation I would have used to explain it.... by using the process Kev Williams used in post #46 to get the $30 by itself on one side.

Actual room cost ($25) = amount paid ($30) – partial refund for overcharge ($3) – amount kept by bellboy ($2)
First move the Bellboy'"tip" to the other side by adding the same to each side to zero it on the right.
Actual room cost ($25) + amount kept by bellboy ($2) = amount paid ($30) – partial refund for overcharge ($3) – amount kept by bellboy ($2) + amount kept by bellboy ($2) =>
Actual room cost ($25) + amount kept by bellboy ($2) = amount paid ($30) – partial refund for overcharge ($3) – 0


Next move the amount paid to the left by adding the partial refund for overcharge ($3) to both sides
Actual room cost ($25)+amount kept by bellboy($2)+partial refund for overcharge($3)=amount paid ($30)–partial refund for overcharge($3)+partial refund for overcharge($3) –0 =>
Actual room cost ($25)+amount kept by bellboy($2)+partial refund for overcharge($3)=amount paid ($30) -0 -0 = $30

But to address the apparent anomaly,
The two dollars is the red herring as Brett said.
The $27 dollars that the men paid included the $25 charge AND the the two dollars the bellboy kept, so you don't add that again. You only add the $3 change they got back to get the $30. You can see in the last equation that you already added that $2, so you know you don't do it again.

[Bob Glenn]

This is why the saying BEDMAS needs to be applied. Brackets Exponents Division Multiplication Addition subtraction. This is called order of operations. 10x3=1+1+1+2+25=25+3+2= (25÷3)(1x3)+2

Matt, I always heard, Pretty Please my dear aunt Sally. Parenthese, Powers, Multiplication, Division, Addition, Subtraction.

[Bob Glenn]

Ding, Ding, Ding, Ding.........Exactly! You guys are too smart! Good on ya. Bob

[Brett Luna]

Just to illustrate it another way, since I'm a Government budget professional:

Guest Room	$ 30.00
Room Rate Adjustment	$ (5.00)
Bellhop service charge	$ 2.00
Subtotal	$ 27.00
CASH PAID	$ (27.00)
AMOUNT DUE	$ 0.00

[Ben Rivel]

lol "BEDMAS"! Or in the US we learn "PEMDAS" to remember the order of operations (LINK), see Mnemonics section.

[Pat Barry]

OMG - This example is just sleight of hand, not algebra.

[Matt Mattingley]

lol "BEDMAS"! Or in the US we learn "PEMDAS" to remember the order of operations (LINK), see Mnemonics section.

They’re both right. Which one seems easier to remember???