What a coincidence! This same question was asked to me in a university entrance exam I gave today (they just changed up the names and heights of the objects) and I had watched this video like 3 yrs ago, so I didn't exactly remember the method, but after reading the question, the solution just popped into my mind and I solved it in a minute! Now I can know for sure that I did the right thing! :)
I (almost) accidentally found the answer xD Wrote the problem down, and tried to add (x + c - t = 170) to (x - c + t = 130) which got me a 2x = 300 => x = 150. But doing that wasn't my original plan... just something I decided to do after writing it down.
I really like this one. I used a visual method as well but just treated the turtle height as "0". Then the 2 equations are table + cat = 170, table - cat = 130. Got the same answer that way
That's a really clever trick. 'Let [variable] = 0' is a typical mathematical technique, but it's not necessarily intuitive to imagine a turtle of zero height
Oh I solved using systems of equations: (t = table c = cat r = turtle (r is for reptile but I already used t)) r + t - c = 130 c + t - r = 170 Using algebra, we can see that the difference in height caused by switching the turtle and cat is 40 cm, and half that height will come from the added height of the cat, the other half from the subtracted height of the turtle, so the difference in height from the turtle to the cat is 20 cm. Therefore c - r = 20. This can be rearranged to c = r + 20 Substitute: (r + 20) + t - r = 170 r's cancel, 20 + t = 170 --> t = 150 The table is 150 cm tall.
I did this one two also. But my other method was made without any calculations in my head. Just put both pictures on top of each other (x2). You get 2x table - cat + cat - turtle + turtle=300. So 2x table = 300. 1x table = 150. Edit: hahahhaha... and its the first solution in video.
I did it a different visual way and got a different answer. In the first picture it shows 170cm being from the top of the turtle to the top of the cat. That means that the 170cm doesn’t include the turtle’s height and includes the cat’s full height. Then in the second picture it shows the cat on the bottom and the turtle on top with the 130cm not including the cat’s height and fully including the turtle’s height. If you then make the second picture fully include the cat’s height and not include the turtle’s height like is shown in the first picture the 170cm would then go from the bottom of the cat to the bottom of the turtle which is just from the bottom of the table to the top of the table. Therefore the table is 170cm.
I started solving with algebra, then i figured out a simplier and quicker solution with logic: Top turtle to top cat = 170 Top cat to top turtle = 130 The difference is based on the height of the cat: it's taller than the turtle by 40/2 (caused 1x its tallness is subtracted when switching cat with turtle on top of the table; another 1x is subtracted when switching the turtle with the cat on the bottom of the table), so the height from turtle to turtle should be 170-20=150, but it's the same as floor to top table height
It is not 150cm. It is 170cm. The cat has 70cm and the turtle 30cm.170cm - 100cm = 70cm. 130cm - 100cm = 30cm. 200cm - 30cm =170cm. The table is 170cm high.
@@robertzappe9747 that's impossible, 'cause top turtle to top cat is 170 cm and the turtle and the cat are not of the sane height (otherwise top cato to top turtle should be 170 cm too, but instead it is 130 cm).
So I concluded it was 150 but my method was realizing the 40 unit difference between the two values meant the cat was 20 units taller than the turtle (at least that was my hope, thankfully that's a really big turtle) Then I looked at the picture with the turtle on top and added 20 units to imagine it was 2 cats and no turtle existed. From here I had the measurement of the table, 150, it was just offset by one whole cat hahah
This is interesting because we appear to have two equations in three unknowns, which usually we would consider not solvable. When I first glanced at the thumbnail, my mind said, both at the same time, "That's unsolvable. That's easy." ;-) The diagram method is exactly the same as the algebra method. It is like using rulers to add numbers. The animals should not be taken away before moving the bracket, though. It is clearer if the upper leg and the lower leg of the bracket are both moved down by one cat height. It can also be done the other way, with the turtle on the top of the upper diagram and the bottom of the lower diagram, and move the bracket down by one turtle height. This corresponds to summing the equations in the reverse order. It also gives a hint into why this is not two equations in three variables. The height of the cat and turtle do not matter, as long as they differ by 20, they can be anything. We are not trying to solve for three unknowns. Two of them are still variables. If we had additional information about the height of the cat and the height of the turtle, like their sum or product, we could solve for those heights as well.
Let me show one way to visualize why we can know the third variable that you may not have thought about yet. We have 3 equations so lets think of a 3 dimensional room. We have a x,y and z axis. Each of those represents one of our 3 variables. Each solution is a ordered number triple (x,y,z) which means that each point in our room is a possible solution. Now we introduce our first equation. We can think of this equation as a plain. Only points on this plain satisfy the equation. Our second equation is another plain. As long as our two plains are not parallel they will cut each other in a line. Only points on this line satisfy both equations. This line can have the property that it is in a 90 degree angle to one axis. In that case the value of the variable associated with this axis is the same for all points. This means that there is only one solution to this variable. This is exactly the case in our problem.
Visualization is a good idea. We already know what the graphical solution looks like; it is a line perpendicular with the table axis at a value of 150 and which projects into the cat, turtle plane as a line which intercepts the cat axis at 20 and the turtle axis at -20. We can see this as the intersection of two planes; first the plane perpendicular to the table axis at 150 and second, the plane perpendicular to the cat, turtle plane through the projected line in the cat, turtle plane just described. The described solution line has one value for the table height and an infinite number of values for each of the cat and turtle heights. This isn't exactly what you described, you described it with two different planes. Following your original thought, the two intersecting planes may be described more particularly as follows. Both planes contain the solution line described above. The plane corresponding to the first equation passes through the cat, turtle plane, where the height of the table is zero, on the line which has cat intercept 170 and turtle intercept -170. The second plane passes through a line in the cat, turtle plane on the line which has cat intercept -130 and turtle intercept 130. These two lines are on opposite sides of the origin, they are parallel and the corresponding planes lean toward each other to intersect on the solution line described above. Either of these pairs of intersecting planes is a strong visualization of how there is one solution for the height of the table, but an infinite number of solutions for the cat and turtle heights.
@aeromodeller 1 I am also fascinated that "There exist some 3 variable problems with 2 equations, whose exact answer/value can be found for 1 variable." That's something NEW to me! Like you, I previously believed that you can either solve for all 3 variables IF AND ONLY IF you have 3 linearly independent equations, or none at all if you have less than 3 equations. This problem proves it wrong. I can see now after having thought about it why this problem is solvable for Table. Because by chance it is a 2 variable - 2 equations for Table, if you make the equations like this: Table + (Cat - Turtle) = 170 Table - (Cat - Turtle) = 130 If we rename (Cat - Turtle) as DIFF ... the difference in height btw Cat and Turtle Table + DIFF = 170 Table - DIFF = 130 This is a 2 eq - 2 var problem. Both Table and DIFF are solvable to be Table = 150 DIFF = 20 = Cat-Turtle
@Random Person Your explanation is quite useful. I've never thought of it as "the answer line is in 90 degree angle to one axis.", though. I would just describe the answer line as a line in 3 dimensional (x,y,z) space, whose coordinates satisfy the following line equation: y - x = 20, z = 150 Essentially and visually, it is a line y = x + 20 in 2 dimensions, flat x-y plane, floating at the z - height held constant at 150.
Table = 150 DIFF = 20 = Cat-Turtle and y - x = 20, z = 150 are different forms of the two equations that correspond to one of the pairs of intersecting planes that determine the solution line; first the plane perpendicular to the table axis at 150 and second, the plane perpendicular to the cat, turtle plane through the projected line in the cat, turtle plane. Both of these equations in this problem contain all three variables, sometimes hidden. Cat - Turtle + 0 x Table = 20 contains Table multiplied by zero, so all values of Table satisfy this equation. When Table = 0, this plots in the Cat, Turtle plane as the line with Cat intercept 20 and Turtle intercept -20. This same line is valid at every value of Table, so this equation represents the plane which is perpendicular to the Cat, Turtle plane through the projected line in the Cat, Turtle plane. Similarly, Table + 0 x Cat + 0 x Turtle = 150 is satisfied by all values of Cat, Turtle, but only one value of Table. This is the plane that is perpendicular to the Table axis at 150. The intersection of these two planes is our solution line, in which Table = 150 and Cat - Turtle = 20. It is interesting that we have two sets of equations and two corresponding sets of intersecting planes producing the same solution line.
I used the algebra method, but decided to add 40 to the table with the turtle on top, to equate the two figures. I then defined cats in terms of turtles, and decided that the cat was 20 taller than the turtle. Using this knowledge, I replaced all cats with turtles, and then all turtles cancelled out. This is why math is fun. Many ways to get an answer.
I wrote out both equations, but I ddin't add them. I disregarded the table as the common element in each equation. That showed me that the difference between cat-minus-tortoise and tortoise-minus-cat was 40. Given that the differences had to be symmetrical they had to be 20 and -20. So the height of the table was 170 minus 20 and also 130 plus 20. (Cleverer or more practised people than me would have been able to think through this almost instantly of course.)
Actually, the height of either animal is not able to be calculated. All that is known once the solution is found is that the cat is 20cm taller than the turtle. They cat could be 160cm tall and the turtle 140cm tall. You can't assume diagrams are to scale unless stated.
Yes but it works nonetheless. The cat must be 20cm than the turtle because from the left picture to the right one it loses 40cm because of the height difference between cat and turtle 20 twice. So the cat must be 20cm taller. Now replacing the turtle on the left picture with a cat, we have a height of 150cm, since we have to subtract the 20cm from the height difference. Now 150 is the height of the table, since both cats are equally tall and therefore can be subtracted from the equations. At least that's how my solution was
@@thebigmacd It actually is! :) The height of the cat equals to 24,4.. and the height of the turtle equals to 4,4... To confirm: 24,4-4,4=20 170-20=150 and 130+20=150. If anybody still wants this worked out I can post it here, worked it out on paper.
Put these two structures on top of each other. On its left side, that whole structure is: table, cat, table, turtle On its right side: turtle, 170, cat, 130 So 2 tables equal 300 and the table is 150.
I used algebra but somehow took atleast 15 minutes doing the complex way Here's how I did it: Distance from top of Tortoise and top of Cat in the first picture is 130cm and in 2nd picture it is 170cm then height of cat is 40cm. Then from first picture height of table is 130cm + tortoise height. You can confirm this by compering the same with second picture. In the second picture height of table is also 130cm + tortoise height as the cat's height is excluded. But height of cat is 40cm, then in the 2nd pic distance from tabletop to cat top is 130cm - 40cm + tortoise height = 90cm + tortoise height. Also in the 2ne pic, given distance is 130cm. Therefore, (90cm + tortoise height) + tortoise height = 130cm which means height of tortoise is 20 cm. Now from either pictures, adding the remaining height of the tortoise height of table is (170-40+20=)150cm in the 1st picture, and (130-20+40=)150cm..... THANK YOU
Just stack one arrangement above the other; the given dimensions will add up to give the height of two such tables stacked, giving 150 has the height of one table. Approaching a problem visually is more satisfying than applying equations!!! 😊 EDIT: Oh, you mentioned the visual method as well in your video!!! I was not expecting this to be there!!!
Luckily this question only ask the height of the table, not cat or turtle. Otherwise, it would be unsolvable. You have 3 unknowns but only 2 equations. You won't solve the height of the other two
Two animals interchange positions depending if they were at table 1 or table 2, therefore any difference in distance between the two dimensions is obviously due to just swapping their positions therefore making the 40 cm difference between the two dimensions divisible by two (since there are two tables) in order to effectively eliminate the animals from the picture altogether. This invariably means that 20cm must be added to the shorter dimension and 20cm must be subtracted from the bigger dimension in order to arrive at the same measurement for both tables since both tables must be the same height.
The expression minus C plus T is common to each equation so it is just a case of substituting -C+T to get the answer without even needing to know what C and T are. The question doesn't ask what C and T are anyway. 150 cm.
I figured it out through pure intuition because you said it was an elementary school problem. 😆 At that point I just had to look at the elements and notice the pattern. It's asking for a specific height so the numbers are obviously the starting point. The cat and turtle swapping positions leads to the difference between the two numbers, and that is the only relationship between the numbers. That difference is 40 so add 20 to the lower number, 150.
because you only need two: x - y = 170 cm y - x = 130 cm y = 130 + x x - 130 + x = 170 x + x = 300 2x = 300 x = 150 somehow as long as 130 and 170 are added together on one side of the equation, then it will work out.
@@paulke1859 it does add up, 170 should be -170, but it isn't a matter of algebra but of relations. It is a problem of logic. There would be no need to show the right equation, all I am showing here is that the relation between 130 and 170 is such that it naturally suggests 150 as the answer.
I solved it with a third method. The difference in the two heights was 40 cm. The turtle was shorter than the cat so the difference in their heights was added to the table in one case (+ Cat - turtle) and subtracted from the other case (- cat + turtle) so you could add half the difference to the shorter or subtract it from the taller either gave 150cm
One of the few problems of yours i solved My sol was setting it up as an algebraic problem using variables x,y,z to represent the cat,turtle and table respectively x+z-y=170 y+z-x=130 adding the 2 gives 2z=300 dividing 2 from both sides gives z=150 which is the height of table
please, help me in understanding what's wrong in my reasonement: turtle is X, cat is Y, table is Z: so 2Z+x+y=300 That means that, if the solution of the video is right (table is 150), cat and turtle are 0 cm high. I can't understand what's is wrong in my idea.
please, help me in understanding what's wrong in my reasonement: turtle is X, cat is Y, table is Z: so 2Z+x+y=300 That means that, if the solution of the video is right (table is 150), cat and turtle are 0 cm high. I can't understand what's is wrong in my idea.
please, help me in understanding what's wrong in my reasonement: cat is x, table is y. So x+y=170 table is y, turtle is z. So y+z = 130 so 2y+x+y is 300 But, if the solution of the video is right (table is 150), cat and turtle are both 0 cm high. I can't understand what's is wrong in my idea. Please, help me!
Put the second picture on top of the first one with the cat overlapping. It's 300cm from top of first turtle to top of second turtle. Half of it is one table
1st noticable thing is that there is 2 numbers with 3 unknowns so if any certain solution exists then one of unknown is free to select whatever you want. Thus we can just assume turtle is zero height and solution comes obvious.. table + cat = 170, table - cat = 130 => table is in the middle of those which is 150..
The visual and algebraic methods are identical, the visual method is just a visual representation of the maths that was done in the algebraic method when the equations were added together and vice versa
Something that is maybe interesting (or maybe I did it wrong). I used Algebra, but I formulated different equations than you did in the answer. I wrote (for the left-hand drawing) that the Table Height = Turtle Height + 170 - Cat Height. And then for the right-hand drawing, I wrote Table Height = Cat Height + 130 - Turtle Height. Then I solved the equation: Turtle Height + 170 - Cat Height = Cat Height + 130 - Turtle Height. The answer is Turtle Height = 2 (Cat Height) - 40. Then I substituted that answer for Turtle Height in the earlier equation, so I had: 2 (Cat Height) - 40 + 170 - Cat Height = Cat Height + 130 - (2 Cat Height - 40). That works out to Cat Height = 20. Then I plugged that value into the earlier result that Turtle Height = 2 (Cat Height) - 40. Although I got the seemingly incorrect result that the Turtle Height = 0, when I plugged the values for Cat Height, which was 20, and Turtle Height, which was 0, into either of my original equations for the Table Height, I ended up wit the correct result of the Table Height being 150 (seen here: Table Height = Turtle Height (expressed as 0) + 170 - Cat Height (expressed as 20)... or TH = 0 + 170 - 20...or TH = 150). So my question: Did I do something wrong or is the turtle's height actually 0?
well, I had the same issue, tried to see the height of the cat or the turtle. Bad at math so obviously the only thing I know is that the cat is 20 cm taller than the turtle. I have no idea how to formulate the fact that we cannot know the exact numbers using algebra, but yes 0 cm turtle height works fine :) I wish I had a mathematician on a shelf here so that I could ask for explanation on this.
@@wundermax1993 Is a free variable, you need more information to know the height of the cat and the turtle. You can as well have a turtle with negative height or even multidimensional and the math will still work.
I could only solve it on paper with some slightly complicated algebra. If h = table height, c = cat height, and t = turtle height, the h = 170 - c + t and h = 130 - t + c. That means that 130 - t + c = 170 - c + t. -2t + 2c = 40, then (-2t + 2c)/2 = 40/2, then -t + c = 20, then h = 130 + 20, therefore h = 150.
in a minute I solved it simultaneously with x+c-t = 170 and x+t-c = 130, seems easy but...explaining it to your child at primary school in a lucid way without using any algebra is, however, no easy task at all
YES, you see the problems. I am Hongkonger, such math problem is common in Hong Kong. It is just a P2 or P3 math problems. you cannot use algebra. you just can use simple method to solve it and explain it to child. I know algebra is easy to solve. Nearly all of the solution here can't see the major problem.
I figured it by looking at the difference between the two. The difference between the cat and turtle is 40/2 (because the pictures have them switching places), which is 20. Adding 20 to the 130 in the picture with the turtle gives me 150, which is what the distance would be from the top of a turtle on the floor to the top of a turtle on the table. That would also be the distance from the top of a cat on the floor to the top of a cat on the table, or just the height of the table without animals.
I just wrote cat, turtle and table height as x,y and z respectively. Added the total height for both cases, then added the 2 equations in a system, thus removing the cat and turtle. Easy.
"you can rearrange it" as if you can rearrange a printed picture. Imagine solving this question that already printed out and using the "stacking and rearrange" Love using algebra.
This is an easy problem, but there's more to analyse here. With the height of the table being 150 cm, the only there constraint is (heigh of cat) - (height of turtle) = 20 cm. Purely algebraically, this allows for unrealistic and even negative values of heights of both cat and turtle for eg. (-40, -60). Hence, we could set the height of the turtle unrealistically high (higher than the table) and still have the table height to be 150cm, just making sure that the cat's height is 20cm more than the turtle's.
c + x - t = 170 t + x - c = 130 This is two-equation algebra. c + x - t + t + x - c = 300 c - c + x + x - t + t = 300 2x = 300 *x = 150* _Averaging the measures would also get one here._
i found option 3 by imagining an generic y-axis. my table has a height of 5 units. on the ground (y=0) i have a box with height of 1 unit. on the table i have a box with height of 2 units. y1 y7 = 6. now i switch the boxes and it's y2 y6 = 4. the average of these distances is 5, the height of my table. tested with bigger boxes and other heights of the tablet and results were always the same. so it's (130 + 170) : 2 = 150 for this logic test.
I got the correct answer but I did it the hard way. I basically kept guessing the table and the cat height until it fits with the equation. I started with 100 centimeters for the table and kept working my way up
I did it a bit diffrently, but algebraic. I solved both equations for the table and equated them. Then I solved for one of the animals as a function of the hight of another animal. Then I put the other animal in as function of the other to solve for the table. T+c-t=170 -> T=170-c+t T+t-c=130 -> T=130-t+c 130-t+c=170-c+t 2c=170+t-130+t=40+2t -> c=20+t -> T=170-(20+t)+t=170-20-t+t= 150
Denote the height difference between the cat and turtle as x. In the image on the left, imagine replacing the turtle with a duplicate cat. The measurement shortens from 170, by an amount x, to the table's true height h. 170-x=h In the image on the right, imagine again replacing the turtle with a duplicate cat. The measurement lengthens from 130, by an amount x, to the table's true height h. 130+x=h Add these equations together. 300=2h h=150 cm
Before watching and only from the thumbnail. By drawing up some equations, you can tell that the cat is 20 units taller than the turtle. With that you arrive that the height of the table is 150.
I’m an engineer, so I solved it empirically. I built a table with adjustable height. Placed my kid’s turtle at the bottom. Got my tape measure all ready. Grabbed the cat from outside. She started hissing and scratching me. Tried to place her near the table, but she attacked the turtle. My kid started screaming. My wife came in and started yelling at me. Technically, I never actually figured this out. Your solution is much more elegant.
What I did: Initially one of the first things I did was look for the difference. 170-130=40. Good. Now where does this difference come from? 1. Table 1. you are taking away a turtle. (So you are taking away less). At the same time you are adding more since you are adding a cat. 2. Table 2. you are talking away a cat (so your are taking away more this time). And at the same time you are adding a turtle, which is shorter than a cat (so you are adding less). We know these two fact have to equate for a difference of 40cm. Since: -Turtle < -Cat And Turtle < Cat The difference between a negative cat and turtle will be the same as a positive one. So both differences have to be 20 (half of 40). Now looking at the first table…170cm -20cm = 150 (we are substituting the bottom turtle with a cat) if you have two cats one on the floor and one on top of the table they will the difference between their heads will be the height of the table. So 150 = table height. Yes a super over complicated method and not the most efficient but it’s what I came up with so I thought I’d share it as well.
I've done it diferently, but also artmeticly. For clarity - T is table, c is cat, t is turtle I started with basic equasions you had, but I change them to be about table: T = 170 - t + c T = 130 + t - c Than i can merge them together, and than bring all the unknown to one side 170 - t + c = 130 + t - c 170 = 130 +2t - 2c 40 = 2(t-c) t - c = 20 And with (t-c) calculated, I can just drop it into the one of the main equasion, and have a solution. T = 170 - 20 = 130 + 20 Simple and basic method, I probably forgot all the more advanced one since I graduate :D
Ha ha, I'm usually no good at math problems, but this one I solved by coming up with method number one on my own. The REAL trick to this problem is to not get distracted by trying to figure out the height of the cat and turtle first, those don't matter at all.
The visual way of solving it is so interesting though, like it feels better than the algebraic way because anyone can use that method without knowing formal math.
I arrived at the same solution but took a different approach. From the first two images we can see that the difference between the two situations is 20cm. That’s the combined height of the turtle and the cat. Take that away from the first situation, the max height of table, cat and turtle you get 150cm. Thus, the height of the table is 150cm. Did anyone else do this?
a used a 3rd metod. visual. table with cat above = 170. table with turtle above, 130. difference = 40. diference between cat and turtle = 40/2 = 20. table with car above = 170. minus differente between cat and turtle, 150. turtle - turtle = 0. table = 150.
With T = height of table, C = height of cat, and S = height of tortoise : 1) T + C - S = 170 2) T - C + S = 130 3) T + C - S + T - C + S = 170 + 130 = 300 4) T + C - S + T - C + S = 2T = 300 5) T = 150
Fun fact, you cannot figure out just how tall the cat and the turtle are respectively. You only know for sure that the turtle is taller than zero and that the cat will always be 20 cm taller than the turtle but shorter than the table - but we do not know how thick the table's plate is.
Intuitively, it's the average of the two heights. My first thought was to enter the height of each animal on the Y axis using Cartesian coordinates, but when I realized neither height was labelled and there were two drawings, it's like "Half way between 130 and 170 is 150."
Ohh I was so proud of myself for figuring it out, then you explained it and I am once again learning just how bad I am at intuiting math... I did a stupider version of the algebra. I first figured out how much taller the cat is than the turtle, by calculating the difference between the measurements and halving it. The cat is 20cm taller. Then I plugged that into the measurement of the first image: 170 = Table+Cat-Turtle = Table+(Turtle+20)-Turtle = Table+20 = 150+20 Then I checked it by doing the reverse.
It's fun to find the "visual" solution to these types of problems and avoid using algebra whenever possible. Visually the answer came to me when I tried a trick from the solution to one of Presh's other puzzles (I won't say which one).
You can just say that once the length is too big and once it too small and the differences is suppose to be equal so 170-130=40 40÷2=20 and now or you do 130 + 20 =table or 170-150=table
My thought process, let x be cat, y be table and z be turtle From the picture we have: x+y-z=170, by rearrangement we have: y=170-x+z Then we have -x+y+z=130, again by rearrangement we have y=130+x-z Therefore we know that 170-x+z = 130+x-z 170-130 = 2x-2z... 20 = x-z since y=130+x-z, y= 130+20= 150. The table is 150 cm in height edit: Hell yeah, algebra rules
spent 10 mins trying to figure out turtle height always ended up at 0 so I thought I must be wrong somewhere. Turns out the task required only to find table.
never went to the visual one cause i don't want to assume the visual representation is accurate to the measurement. also, i was so close yet so far lol. i went with too complicated solution by including the space between the cat and turtle, along with the cat, turtle and table. and i was too focused in determining the heights of the cat and the turtle, thinking it's part of solving it. like i tried playing the video up to the point where we are adding the two equations and got the 150, but i still insisted in trying to find what the height of the cat and turtle. then i stepped back, and looked again, and thought to myself how the question is only asking for the table XD also, i was so close with adding the two equations, but my equations were two versions for the same answer. like i made an equation in finding the size of the table by using the space between the two animals + the animal below. so like figure 1's equation was cat and turtle's space + turtle = table, while figure 2's was turtle and cat's space + cat = table, then added the two, hoping i'll be able to find each variable's values, which just lead to transpositions and a lengthier equation
I think the key to solving it visually is that even if the proportions of the were super inaccurate, it wouldn't effect the equation, its just a different way of looking at the variables, which are unaffected by how the picture looks
@@lapiras1054 but part of the solution in the visual one is that the sum is same as double the table, which is what i mean by not sure how accurate that is. maybe it looks like it's double, but maybe it's actually more or less of a centimeter. i did try it in pixels to test it in smaller scale with visuals and i think it does work so that's neat
When I see the thumbnail: X is height table, Y is height of turtle, Z is height of cat. X - Y + Z = 170 X + Y - Z = 130 -> X - Y + Z + (X + Y - Z) = 170 + 130 = 300 -> 2X = 300 -> X is 150 cm. But, what is hell kind of 150 cm height table?
I solved this problem before watching this video. Here was my logic. STEP 1) The cat looked to be twice the height of the turtle, so I was going to start with that guess. STEP 2) From the left image, the table was obviously under 170 cm. STEP 3) From the right image, the table was clearly above 130 cm. STEP 4) I decided to start with a will guess that the table would be halfway between 130 and 170. STEP 5) Based on step 1, I made a wild guess that the turtle and cat would be 20 cm and 40 cm because those values seemed likely if I were creating a puzzle like this. STEP 6) I tested my values and found out that all my values were correct. STEP 7) I wrote this comment before watching the video or reading the comments.
The one and only case where a system of two equations in 3 unknowns has a nontrivial solution. When two of the unknowns are equal. Don't let the image fool you.
Difference between 170 and 130 is 40 then divide by 2 to equal height of turtle and cat, then add to the lower value = 150, or minus from the higher value = 150. 2 seconds in my mind
The visually one is such a no-go in math. You can't simply say: "oh, it visually looks like they're the same", if that's how math works, then there would be so many wrong answers and formulae in life
My was was different but I did get the same answer, for me is to substract the 70 with 30 because 100 is in both the pictures, and add the result up and got a 50 (from 70-30) + 100 = 150
