## Olympiad Mathematics | Learn How to Solve the System of Equations Fast | Math Olympiad Preparation

Thanks! Share it with your friends! Learn how to solve the Olympiad Question quickly with these tips and tricks for x, y, and z variables. The system of equation involves x, y, and z variables.

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Subscribe Now as the ultimate shots of Math doses are on their way to fill your minds with the knowledge and wisdom once again. 이호석 says:

Thank you so much sir I am very grateful to learn from you. You teach so well:) Bheri Raju says:

xy=35 and only 7 and 5 are prime factors
X=5
Y=7
Z=8
Ans Raoul Watchueng says:

Good morning. Please how do you do to have viewers ?
I have à channel since ont year. Il have more than 100 vidéo. But I dont have viewers and siscribers.
Please anser me. I wan to descorage my self. Sorry for my english. Il am francophone Asokraj Ts says:

Multiply all the 3 equations.
Then Square of XYZ = 78400.
So XYZ = Sq root of 78400 = 280
Divide XYZ = 280 by each of 3 Equations and we will get value as under
X = 5, Y = 7 and Z = 8 h. says:

sir i have a question please help 2x+3y+4z=1 x,y,z real positive numbers what is the minimum value of the
(1/x)+(1/y)+(1/z) Wilfried Bergs says:

5,7,8 within 5 secs Keith Jurena says:

Solved by observation in 6 seconds Viqar Minai says:

Divide second equation by first, giving z/x = 8/5 => z = (8/5)x. Substituting this in the third equation
yields (8/5)x^2 = 40 => x^2 = 5 => x = +/-5. Working backwards, y = 35/x = +/-7, and z = (8/5x) = +/-8. Minh Nhật says:

Nhân cả 3 lại với nhau sau đó căn bậc 2
Thì xyz = 280
Và xz = 40
=> y = 7 XY Z says:

Can just multiply equations 2 and 3, (xy)z^2 = 56 * 40 Christ K says:

Tres Bien !!! Mallaiah Banala says:

Sir, Thanks for clear explanations ans interesting problems solving.

But here, I feel it can be done still simpler form like multiplying equ 1 and 3, then, we can get square of x is 25. So x is +/-5. Samsung J7 says:

Un desarrollo simple pero eficaz. Bart te Molder says:

xy = 35 => y = 35/x. (a)
yz = 56 => apply (a): (35/x)z = 56 => z = 56x/35 => z = 8x/5 (b)
zx = 40 => apply (b): (8x/5)x = 40 => 8x² = 200 => x² = 25 => x = +/- √25 => x = +/- 5 (c)
(c) in (a) => y = 35/(+/- 5) => y = +/- 7; (c) in (b) => z = 8(+/-5)/5 => z = +/-8. Simple substitution is all that is needed. Pervaz Rzayev says:

Abob would be able to find it. Muttley says:

I solved this one using prime decomposition though admittedly I had to assume the solutions would all be integers.

Up to a ± sign there is a unique factorisation of xy = 35 = ±5 x ±7.

Prime decomposition of yz = 56 has ±7 x ±2^3. The number ±7 is then a common factor of xy = 35 and yz = 56. That suggests y = ±7 are possible solutions. That would also mean that we need x = ±5 and z = ±8 to also work out.

To check this could work, we also xz = 40. We need that to equal ±5 x ±8 which it does.

So, x=±5, y=±7, z=±8 solves the simultaneous equations where solutions are either all positive or all negative. William Wingo says:

y = 35/x; z = 40/x; yz = 56; substitute the expressions for y and z into the third equation:
yz = 56 = (40/x)(35/x) = 1400/x^2; solve for x:
x^2 = 1400/56 = 25; x = +/– 5; substitute back into the original equations:
y = 35/(+/– 5) = +/– 7; and
z = 40/(+/– 5) = +/– 8. and x, y, and z must be either all positive or all negative.
Thank you, ladies and gentlemen, I'm here all week.
No applause, please: save it for Miss Whaley and Mr. Clements from 1963. Satrajit Ghosh says:

A bit of work gives yz/(zx) = 56/40
or y/x = 7/5
again xy = 35 Hereby x^2 = 5^2
case I : x = 5, y = 7 , z= 8
case II : x = -5, y = -7 , z= -8 Mendoza Jovy says:

A nice problem to practice 3 equations and 3 unknowns. Math is practice. I am fast in solving these problems bec i practice it (without cheating). Good video sir.