Olympiad Question! Solve this System of Equations if X, Y, & Z are Integers | Step-by-Step Tutorial

[@pranavamali05]

Thanks sir keep rocking and keep giving these questions

[@PreMath]

So nice of you Pranav dear Thank you! Cheers! You are awesome. Keep it up😀

[@arthur_p_dent]

My approach would be to start with the prime factorization of xyz=240 and see where we get from there. 240 = 2^4*3*5. Looking at equations 1 and 2, one can see that x,y,z must all be even. Since we have 4 factors of 2 in xyz, one of the 3 variables must be divisible by 4 and the other 2 only by 2, and looking even further, it becomes clear that x must be the one variable divisible by 4 or equation (2) cannot be solved. So x = 4a, y=2b, z=2c, where abc=3*5. All that is left to do is look where the factors 3 and 5 belong, and upon further inspection it turns out that the only solution that fits is to give the factor 3 to y and the factor 5 to z, ie x=4*1=4, y=2*3=6, z=2*5=10

[@PreMath]

Neat approach Arthur! Thank you for sharing! Cheers! You are awesome. Keep it up😀

[@nayaka-zna]

I'm not pretty good at maths, so may I ask why would one think that x, y, and z must all be even? I can see why x and y must be even, but how about z? Thanks a bunch!

[@arthur_p_dent]

@@nayaka-zna you're right - I was being inaccurate, one cannot conclude that solely from looking at equations 1 and 2. One also needs to consider the prime factorization of 240. Suppose z is odd. Then xy must be a multiple of 2^4. Thus, if both x and y are even, there are two possible cases: 1. both x and y are multiples of 4, or 2. one of them is a multiple of 8, and the other one is even, but not a multiple of 4. Case 1: then xz+y is a multiple of 4, and thus cannot be 46. Thus, equation (1) can't be solved. Case 2: then x+yz is NOT a multiple of 4, and thus cannot be 64. Thus, equation (2) can't be solved. in either case, no solution exists. Therefore, z cannot be odd. ---- However, it has since occurred to me that my solution is incomplete. I do get the correct result, but I have only checked for _natural_ numbers as solution. I have forgotten to check for possible solutions involving negative integers. It would not have been an awful lot of extra work to check all conceivable cases, but it definitely is an omission. A solution involving negative integers does not exist, but implicitly assuming that from the start was a mistake.

[@waheisel]

Very nice! I added the first two equations, factored the left side and got (x+y)(z+1)=110. Then I got 55, 22, 11, 5 and 2 for the factors of 110, so z=54, 21,10, 4 or 1 and I figured it out from there. I like your solution!

[@nayaka-zna]

@@arthur_p_dent thank you very much for your explanation! Now it does make sense

[@luigipirandello5919]

Very nice question and solution. Thank you very much.

[@PreMath]

Most welcome Luis dear You are awesome. Keep it up😀 Love and prayers from the USA!

[@govindashit6524]

You are the beat teacher in my life . Thanks , app ki umar lambi ho krishna se duya mangta huin .

[@PreMath]

Wow, thank you so much Govinda dear. Aap ki bhi umar lambi ho! You are awesome.😀 Love and prayers from the USA!

[@predator1702]

Wonderful solution 👍thank you teacher 🙏.

[@PreMath]

You're welcome 😊 So nice of you dear Thank you! Cheers! You are awesome.

[@iZAPMath]

Your explanation is always diligent! Great work, PreMath!

[@PreMath]

So nice of you my dear friend Thank you! Cheers! You are awesome. Keep it up😀

[@sameerqureshi-kh7cc]

Premath just rock n roll 😊👍🌹love from Pakistan....

[@PreMath]

Thank you Sameer dear! You are awesome. Stay blessed😀 Love and prayers from the USA!

[@242math]

thanks for sharing this system of equations, your step-by-step tutorial makes it easy to understand

[@PreMath]

You are welcome! Thank you my friend! Cheers! You are awesome. Keep it up😀

[@mathfromtheheart]

Great work sir 👍

[@PreMath]

Thank you dear! Cheers! You are awesome. Keep it up😀

[@opytmx]

Having a lot of respect for all the mathematical formulas and procedures, but here it's very obvious that 40 + 6, 60 + 4 and the factor 10 are involved, because of the 0 in 240 (5 x 48 can not work , because of the odd, small 5), and as x, y, and z are integers. Playing a little around with the numbers you get mentally: 4 x 10 + 6 = 46, 4 + 6 x 10 = 64 and 4 x 6 x 10 = 240. PS: But your solution as well is very interesting and, of course, more mathematical approach.

[@PreMath]

Thank you for sharing! Cheers! Keep it up😀

[@nicogehren6566]

very interesting question sir thanks

[@PreMath]

Most welcome Nico dear You are awesome. Keep it up😀 Love and prayers from the USA!

[@devondevon3416]

I haven't work the problem yet, but just by observation I know the values (or one of them) are: 4, 6, 10 such as x=4 z=10 and y=6

[@PreMath]

Awesome job! Thank you my friend! Cheers! You are awesome. Keep it up😀

[@bentels5340]

Nice! 👍

[@PreMath]

Thank you Ben! Cheers! You are awesome. Keep it up😀

[@BubuMarimba]

add 1st to 2nd: xz+y+x+yz=110; (x+y)(z+1)=110=2*5*11; xyz=2*3*4*2*5 => z=10 ; x+y=10 => x=4,y=6 or x=6,y=4 (but the later do not fit to first 2 equations), so 4,6,10

[@kafaichan3371]

(x=4 or 60) & (y=6 or 40), only x=4 & y=6 are valid soln (z=10) as all other pairs are rejected (eg. x=4 & y=40), z will not be an integer.

[@PreMath]

Super Chan! Thank you! Cheers! You are awesome. Keep it up😀

[@millipro1435]

thanks 👍👍❤️

[@PreMath]

Thank you Milli! Cheers! You are awesome. Keep it up😀 Love and prayers from the USA!

[@JohnRandomness105]

I multiplied the first equation by y to get xyz + y² = 46y or y² - 46y + 240 = 0 = (y - 40)(y - 6). But only y = 6 leads to an integer z. I repeated the process with the second equation and got x = 4 as the only value allowing an integer z. The third equation (or any) gives z = 10. The solution (x,y,z) = (4,6,10). Three other solution sets with non-integer z exist.

[@johnbrennan3372]

Worked it out multiplying equation 1 by y and equation 2 by x. That gave me a quadratic equation in y and another equation in x. When I subtracted I got x^2 -y^2 =64x-46y . The quadratic equation in y gave me y=6 and y= 40 . Plugged the values into x^2- y^2= 64x-46y etc

[@PreMath]

Thanks John You are awesome. Keep it up😀

[@alexniklas8777]

The solution and the result is the same as yours. Thank you sir

[@PreMath]

Excellent Alex Thank you! Cheers! You are awesome. Keep it up😀

[@Gargaroolala]

From the last equation, I took x = 240/yz. Then I did plenty of substitution and solving quadratic equations to get the answer. Of cos had to reject the non integers.

[@PreMath]

Super Garrick! Thank you! Cheers! You are awesome. Keep it up😀

[@albertaraujo6304]

That's what I did as well. I went through all the permutations of the quadratic solutions for x and y to find the general solution for x,y,z.

[@mustafizrahman2822]

I faced the problem on the 1st line of what should I do. So, according to this, I failed to solve it basically.

[@PreMath]

No worries! You gave your honest shot. That's what it matters! You are awesome Mustafiz. Keep it up😀

[@mcorruptofficial6579]

Good problem to figure it out, teacher.

[@philipkudrna5643]

Before watching: x=240/yz (from equation 3) yz=64-x (from equation 2) If you plug in, you get x=4. (I did it by trying out, as it was given that it must be an integer.) similarly: y=240/xz (from equation 3) and xz=46-y (from equation 1). Plug in again and y=6. plug both into equation 3 leads to z=10. After watching: Seems like I messed up by not factoring properly, so I missed out on the alternative solutions - but got lucky anyways, since all the other solutions do not lead for an integer for z!

[@PreMath]

Thank you Philip for sharing your experience! Cheers! You are awesome. Keep it up😀

[@graemedurie9094]

I am a hopeless mathematician, but I was able to solve this in my head in a half minute.

[@graemedurie9094]

@Silent Integrals Thanks!

[@armacham]

First, determine that none of them can be zero because the product is 240 Second, determine that none of them can be negative. In order for the product to be 240, either all of them are positive or exactly 2 of them are negative. But you can rule out all possible combinations of 2 negative numbers (xy, xz, or yz) because the first two equations would give a negative result. (e.g. if xy are negative and z is positive, then x + yz must be a negative number, it can't be equal to 64) Third, show that x and y must be even. You know at least one of them has to be even because their product is 240. If x is odd, then at least one of y or z must be even, so yz must be even. But odd + even can't be equal to 64, an even number. Similarly, y must be even. Fourth, add the first two equations together, you get: (x + y)(z+1) = 110 = 2*5*11 Since you know x and y must both be even, you know that their sum must be even, so the "2" factor must be part of x+y, not part of z+1. That means z+1 must be equal to one of the following: {1, 5, 11, 55} if z+1 = 1, then z=0, but we know this is not permissible if z+1 = 5, then z=4 if z+1 = 11, then z=10 if z+1 = 55, then z=54. it is easy to see this is impossible because xz + y = 46, and we know x and y are positive even numbers. since we have cut it down to just 2 possible values for z, break into two cases and substitute those values in. CASE Z=10: xy = 24 y = 24/x 10x + 24/x = 46 it's safe to multiply both sides by x because we know x is nonzero 10xx + 24 = 46x CASE Z=4 4xy = 240 xy = 60 y = 60/x

[@armacham]

It looks like part of my comment is missing. You can solve 10xx - 46x + 24=0 with the quadratic formula. Do that 4 times and you can solve all possible x values when z=10 and when z=4. From this you find that the only solution is 4,6,10

[@davidseed2939]

my approach. add (1)&(2) (x+y)(z+1j = 11*10, or 2*55 or 5*22, possible values of z are 9,10,1,54,4,21 but only 1,4,10 divide xyz 1 is impossible since x+y =110, must be both even xyz=2*5*3*8 , x*y=10*24 or 20*12 or 40*6 . none of the pairs add to 110 z=4, x+y=22 both even, xy= 4*5*3 , so factors can only be 10*6 factors dont add to 22. . z=10, xy=24=3*2*4 x+y=10 both even and 6,4 works. Actually I just guessed this solution since z=10 seemed right.

[@PreMath]

Nice approach David Thank you for sharing! Cheers! Keep it up😀

[@phanimaheswara7492]

Thank you sir

[@PreMath]

So nice of you Phani dear Thank you! Cheers! You are awesome. Keep it up😀

[@satyanarayanmohanty3415]

Awesome

[@PreMath]

So nice of you Mohanty dear Thank you! Cheers! You are awesome. Keep it up😀

[@johnbatchler8551]

To solve that equations u need to use imicit or explicit

[@mahalakshmiganapathy6455]

Very nice problem

[@PreMath]

So nice of you Mahalakshmi dear Thank you! Cheers! You are awesome. Keep it up😀

[@susennath6035]

After getting X =4, we can get other two by inspection from equation 1

[@PreMath]

Super Job Susen! Thank you! Cheers! You are awesome. Keep it up😀

[@bentels5340]

No you can't. With just x = 4, xz + y = 46 has infinitely many solutions.

[@susennath6035]

@@bentels5340 it has also xyz=240. It should be concider

[@susennath6035]

@@bentels5340 feasibility it must has

[@theophonchana5025]

y^(2) - 46y + 240 = 0 y = (46 + sqrt (1156)) ÷ 2 y = (46 + 34) ÷ 2 y = 80 ÷ 2 y = 40

[@capjus]

I cant believe my first guess within some seconds and try out was correct 4, 6, 10 and i solved in a few seconds by checking with my mind without calculus etc yeheeeyyy ^^^^ Didnt even watch yet to end :d

[@PreMath]

Great job! You are awesome. Keep it up😀

[@theophonchana5025]

#QuadraticEquation #factoring #factor #trinomial #polynomial

[@angelrenemurielpizarro2226]

tambien se puede resolver por el metodo cartesianno

[@ranveeryadav176]

x=4, y=6, z=10 & x=60, y=40, z=1/10 by factor method 👍👍👍

[@PreMath]

Dear Ranveer, z is an integer! => z=1/10 is invalid

[@ranveeryadav176]

Ohk sir 😊 Thank you for giving me this precious information 🙏🙏

[@theophonchana5025]

x^(2) - 64x + 240 = 0 (x - 4) × (x - 60) = 0

[@aminerane3717]

Plz solve: x+y=8=xy

[@kalyanbasak6494]

Answer sharing x=60,y=40,z=0.1 X=4,y=6,z=10 Thanks

[@PreMath]

Thanks Kalyan You are awesome. Keep it up😀

[@broytingaravsol]

one more r contained in the word "integer" in the content

[@PreMath]

Thanks dear. That was an honest typo. You are awesome. Keep it up😀

[@broytingaravsol]

@@PreMath i've come for every episode from u

[@PreMath]

@@broytingaravsol Thanks dear. You are the best 😀

[@_josuke6034]

Can anyone give me tips on how to be better at math. Im able to do math in my school with ease but when it comes to abstract problems like the ones discussed on this channel, im absolutely dumbfounded. Pls help

[@PreMath]

Hello dear, don't get stressed out too much. We are all lifelong learners. Practice, practice, practice is the name of the game in Mathematics! Just keep persevering and you'll see the difference. I wish you all the best dear. You are awesome😀

[@_josuke6034]

@@PreMath thank u so much, I needed that

[@theophonchana5025]

b^(2) - 4ac = 3136

[@mattslupek7988]

Before watching the video, I did the problem by hypothesis, or a conclusion based on facts and evidence. Fact: Neither of the sums of the top 2 equations is a multiple of 5 or 10 Fact: The bottom equation is a multiple of 5 or 10 Fact: The sums of the 2 top equations are in reverse order of each other Fact: z is only found in the multiplication part of the equations Fact: The digits in the sums are 4 and 6 The product of the bottom equation is 240 Fact: 240 is a multiple of 10 Fact: 4x6=24 Fact: The added variable in the first equation is y, and x in the second equation Fact: The added variable in the equation is the last digit of the sum, and the variable multiplied by z is the first digit of the sum x=4; y=6; z=10 Equation 1: xz+y=46 4(10) + 6 = 46 40 + 6 = 46 Equation 2: x + yz =64 4 + 6(10) = 64 4 + 60 = 64 Equation 3: xyz = 240 4(6)(10) = 240

[@mintusaren895]

If x y z must be real number for all equation. Algebra.

[@darellpiper7227]

I got it right

[@PreMath]

Excellent Darell You are awesome. Keep it up😀

[@theophonchana5025]

#binomial #linearequation

[@mathmagicvn]

Hay quá

[@PreMath]

Siêu! Cảm ơn bạn Học! Chúc mừng! Bạn thật tuyệt vời. Giữ nó lên😀

[@johnbrennan3372]

Lines thro script not meant.

[@PreMath]

No worries! Keep it up😀

[@theophonchana5025]

xyz = 240 4 × 6z = 240 24z = 240 z = 10

[@Mathematician6124]

Easy

[@theophonchana5025]

a = 1 b = -64 c = 240

[@venkatesank3841]

Your answer is wrong. Correct answer is X:. 4 Y. 30 Z. 2

[@theophonchana5025]

#distributiveproperty

[@theophonchana5025]

x - 4 = 0 x = 4

[@theophonchana5025]

x - 60 = 0 x = 60

[@giuseppemalaguti435]

A me risulta un unica terna 4,6,10.....magari mi sono sbagliato

[@PreMath]

Hai assolutamente ragione! Lavoro eccellente Grazie! Saluti!😀

[@theophonchana5025]

b^(2) - 4ac = 1156

[@theophonchana5025]

y - 6 = 0 y = 6

[@theophonchana5025]

y - 40 = 0 y = 40