Practical case for solving with system of 2 equations

When I teach basic math I want to emphasize on it's power (algebraic part for starters) as a tool for solving certain problems you cannot solve with naked brain, so that one models a problem with mathematical notation and then applies simple math (algebra) to solve it. I noticed that most of the tasks learners tend to crack up with just their brain, thus losing potential motivation to learn more abstract math which has to power you up for future more complicated problems.

Are there some cool/fun/practical tasks to be used for modeling a system of 2 (maybe 3 at most) equations?

P.S. For now I have something like this:
Allegra has a secret cake recipe everyone likes. It's made of just 2 ingredients and weights 600 grams. Alice wants to figure out the recipe so she sneaks into Allegra's granary and discovers that ingredient x gets used up 4 times faster than ingredient y. How can Alice figure out exact weights for ingredients in the recipe?

which some still manage to brute force (without algebra usage) x = 480; y = 120

• – Dave L Renfro Sep 16 '20 at 8:55
• Note that this is a reasonable problem to give without algebra. The total is 600 grams, and it is split into 4+1=5 equal sized pieces. So each piece is 600/5 = 120 grams. Hence 120g and 480g for each ingredient. This is not brute-force: it just required understanding of the 4 operations and their conceptual meaning. – Steven Gubkin Sep 16 '20 at 14:08

A bottle and a cork cost 1 dollar and ten cents. The bottle costs $1.00 more than the cork. How much does each cost? Student often think that it is one dollar for the bottle and 10 cents for the cork. That's incorrect because \$1.00 isn't \$1.00 more than 10 cents but the answer is easily discovered with algebra. You might want to update it for items that are more realistically priced and more relevant to students' lives. Even if you do, I am not sure you will consider it a practical problem. • I like this one! I could imagine a solution path which looks like "Well, I will guess \$1 and \$0.10. Oh, that isn't right. I need to increase the distance between these two prices by 10 cents. When one price goes up, the other must go down by the same amount. Ah, so I can just raise it to \$1.05 and lower the cork to \$0.05". – Steven Gubkin Sep 17 '20 at 1:04 • Awesome practical problem, thank you! @StevenGubkin indeed shown how to crack it by simple substitution (that's what I'm feared of because the whole lesson turns to be: "you see, your algebra isn't needed; won't learn it; bye" ^^), but I guess in this case it's harder to do because numbers differ too much (x is ~100 times bigger than y). – Slaus Sep 17 '20 at 8:05 • @StevenGubkin Guessing should not be allowed. It can be arithmetic, but no guessing and no writing programs on scientific calculator to do it in steps. Here: if there were two corks, than the total price would be \$1 less, which is 10 cents. Hence, once cork is 5 cents, and the bottle is \$1 and 5 cents. I guess one can use problems like this in two different ways: one can either say that algebra dumbs people down, just make an equation and find x. Or one can say that algebra always works. Algebra is like stacked addition or multiplication: don't think much, just follow an algorithm. – Rusty Core Sep 17 '20 at 17:58 • @Rusty Core: By guessing I suppose is meant ad-hoc/improvisational or "iterative" (in the sense of multi-step)? Thus forbidding a line such as "If all of them were cars then there would be an excess of$n$wheels originating from$n/2$motorcycles" in response to "Find numbers of 4-wheeled cars and 2-wheeled motorcycles, given total numbers of vehicles and of wheels"? I remember because I was rarely one to do things "systematically". – Vandermonde Sep 18 '20 at 9:30 • @RustyCore I don't think any sort of thinking should be forbidden. If a student has an incorrect thought (such as the initial thought that the bottle was \$1 and the cork \0.10), I would rather see them recognize the error and correct it than just abandon their original strategy and start from scratch. Especially it is hard to have an external authority say "No, you are wrong", and then present a completely new way rather than riffing on your method. Riffing feels like play and collaboration, which is inherently reinforcing. Then alternatives can be presented as a gift, not a punishment. – Steven Gubkin Sep 18 '20 at 11:58 Algebra is a like a hammer that always works unlike arithmetics that may require inventive tricks, different for each problem. You may want to read a pertinent short story Tutor by Anton Chekhov. The problem posed in the story can be solved arithmetically, algebraically and arithmetically with tools like abacus. Do you do word problems with your students? There are tons of word problems like a pool with pipes or two railroad trains or a guy rafting down or across a river that are solved with a system of two or three linear equations. These problems prepare elementary and middle-schoolers to algebra and physics. I think you using x and y in your problem is too on the nose, can you use actual names of components? I wonder what cake can be made just of two components unless it is a ready-made cake for which you need cake mix and water. Why not just calling them cake mix and water? Also, the word "faster" implies that there is a time component to this, which complicates the matters. Here is a basic word problem: Two people left their respective towns simultaneously and started walking over the same road that connected the towns towards each other. The first pedestrian walked 24 miles until they met, walking at 4 miles per hour. The other pedestrian walked at 5 miles per hour. What distance have the second pedestrian walked before they met? Here is a more fun one. Can be done either arithmetically (need to think a little) or algebraically (no need to think much, just use the hammer): Every day an engineer arrives to a station at 8 a.m. by train. Exactly at the same time a car, sent from a factory, drives up to the station, picks up the engineer and takes him to the factory. One day the engineer arrived at 7 a.m., decided not to wait for the car, and started walking towards the car. When the car met the engineer, it picked him up, turned back and arrived to the factory 20 minutes earlier than usual. For how long did the engineer walk? Consider the speeds of the engineer and the car constant. or A boat travels upstream the Mississippi river at full steam. When it passes Lexington Bridge, a barrel with corn syrup falls overboard into the river. When the loss is noticed 40 minutes later, the boat quickly turns back to chase the barrel at full steam. The boat catches with the barrel at Great Western Bridge four miles downstream. What is the speed of river current? • Thank you for for awesome cool problems! Last two ones, I think, are totally uncrackable without algebra. I'm unaware of terminology; what are word problems? I try to teach math to my friends and college artists/designers, nephews and family. I'm dreaming of following people all the way up to Cantor's infinities (I managed to do it once, but didn't see an awe), so for that they need to get accustomed to working with plain math, but the biggest stumble I observe is ~ "why the hell one would use letters instead of numbers!?", so I guess introduction of algebra is the most troublesome. – Slaus Sep 17 '20 at 8:16 • @Slaus The one with the engineer: the car returned 20 minutes earlier than usual, meaning that it would take 10 minutes for it to get from the point where it met the engineer to the station at 8 o'clock. So, the car met the engineer at 7:50. Hence, the engineer was walking for 50 minutes. – Rusty Core Sep 17 '20 at 17:39 • The one with the barrel: relative to the river the barrel does not move. If it took 40 minutes for the boat to get away from the barrel, it will take another 40 minutes to reach the barrel. Altogether is 80 minutes, during which the barrel floated down 4 miles. Thus the speed of the current is 4 miles / 80 minutes = 1 mile in 20 minutes = 3 miles/hour. – Rusty Core Sep 17 '20 at 17:39 Another common problem often used in school algebra classes are of the form, similar to the following: Maria and Juan are siblings. The sum of their ages now is 16. In four years, Maria will be twice as old as her brother Juan. What are Maria's and Juan's ages now? Algebraically, we'd have the system of equations, with $$j$$ representing Juan's current age, and m representing Maria's current age: \begin{align} j+m &= 16\\ \\ m+ 4 &=2(j+4) \end{align} One of the most interesting word problems of all time, which broadened the human intellect in many ways, is the Archimedes cattle problem. There are many excellent books and articles about this--start with the wikipedia page. Also look for "The Sand Reckoner." Archimedes is trying to explain that "infinity" is (conceptually) much more than just a very large number. He starts with, "There are some, King Gelon, who think that the number of the sand is infinite.." He poses an innocent sounding word problem about the number of cattle in a herd with various colors. "Compute, O friend, the number of the cattle .. He sets up a word problem that leads to seven linear equations in eight unknowns. The smallest solution is about 50 million. Archimedes says that if you can get this far, then "thou art no novice in numbers." So pat yourself on the back. But then he adds two more equations, which are non-linear, but still seem innocent. One is that the sum of two of the eight unknowns is a square, so $$x+y=n^2$$. As it turns out, the smallest solution for the size of the herd is then represented as a base-ten numeral with over 200,000 digits. This is an "incomprehensibly" large number. If you can solve this, he says, then "then exult as a conqueror, for thou hast proved thyself most skilled in numbers." The number is more than the number of grains of sand on the earth, indeed much, much more. So the point here is not that Archimedes is using the equations to solve a practical problem to help a certain person figure out how many cattle are in a herd. Instead, he is setting up a word problem that seems, on the surface, to be rather ordinary and not too ridiculous. But the solution! It is not infinite, but wow..it's a big number. In the Sand Reckoner, Archimedes uses this practical sounding word problem about counting cattle to discuss the size of the universe (3rd century BC!), how many grains of sand would fit into it, how to invent a system to name such an enormous number, and how even enormous numbers are not infinite. It's also a challenge to understand how Archimedes managed to contrive the problem to look innocent, but to have such a large solution. Maybe this mystery just comes down to the fact that he was one of the most brilliant persons to have ever lived. Here's a problem that's so practical, I solved it on a piece of scrap wood in the middle of building a gate for my wooden picket fence. I used it in my intermediate algebra class and it went over really well. I have a section of pre-built picket fence that is irregularly sized (I had to cut an 8 foot panel in half to fit it in my hatchback). It is constructed of two horizontal rails with 9 pickets nailed to them. The rails are (say) 44 inches long, and my gate needs to be 41 3/4 inches wide, so I need to trim 2 1/4" from the section. But the pickets are not attached symmetrically; the rail sticks out beyond the last picket by 3" on the left side and by 4 1/2" on the right. Being a bit compulsive about symmetry, I want to trim some amount from each side so that the final gate is both 41 3/4" wide and so that the rails extend the same amount on each side. To set up a system of two equations: Let $$L$$ and $$R$$ be the amounts I will cut off the left and right sides, respectively. Then we have: $$3 - L = 4.5 - R$$ (the rails should extend the same amount after trimming) and $$L + R = 2.25$$ (I want to trim a total of 2 1/4"). This turned out to be a great problem for two reasons. 1) It can actually be solved with a single equation, by defining $$x$$ to be the length of rail remaining after trimming. This is a valuable lesson about the importance of setting up the problem properly. [NB: I did it the two-variable way in my garage. Another important lesson: even experts don't see 'tricks' right off the bat] And 2), The process of actually solving it with algebra corresponds very very closely with how students inevitably solve it in their heads. They'll say: "I subtracted 1.5 inches from the longer (right) side to make the sides even, and also from the 2.25 amount, leaving 0.75 remaining. I split that in half, getting 3/8" more to be trimmed from each side." These are, of course, the exact same operations done when solving the (single equation) version. • In real life you would measure from the outer edge of the picket on one side of the future gate to the outer edge of the picket on the other side (36\frac{1}{2}$), subtract it from the desired gate width ($5\frac{1}{4}$), divide by two ($2\frac{5}{8}\$), verify that both sides have enough rail, mark it and saw it off. You don't care how much to cut off, you care how much to keep. (Actually, you may want to cut a bit more to make sure the gate swings without jamming, and there is space for hinges. Which means, you need to cut more from one side for the installed gate to look symmetrical.) – Rusty Core Sep 23 '20 at 17:22