Olympiad Question: Solve this System of Equations | Step-by-Step Tutorial

#OlympiadMathematics #OlympiadPreparation #CollegeEntranceExam

[zplusacademy]

Your way of progression in solving ❤️🙏 any problem in the blink of eye in simple and easy pathway is mind-blowing, interesting, enjoying and really tickling sir 🙏❤️🙏🙏🙏🙏🙏so much thanks and love 💕💕😘💕💕💕 from India 🙏❤️

[xyz]

Once solved a+b+c, I did it differently. By adding equation 1 and 3, minus 2, to get 2a(a+b+c) = 6, thus a=+/- 1, likewise to get b and c

[LarzB]

Eq 1-eq3 then divide by eq2 gives b = -2c
Eq 2/eq1 with b gives a= -1/2 c
Fill a and b in eq 2 gives c²=4
Rest is just filling in the -2 and 2 for the 2 solutions

So i didn't calculate a+b+c, but eliminated it all the time by dividing

[sigmamaleslogokijalegi]

Easy but long process, thanks for sharing

[furzaanullah]

Very well done. This video impressed me a lot. I want another video of similar type.

[wackojacko]

Archimedes Thoughtful....
The greatest Mathematician of all time ... Archimedes.
You Sir are an inspiration to us all as well. 🙂

[keithwood]

I used the substitution x = (a+b+c), then added and worked through it. It seemed to help.

[mustafizrahman]

a=1, b=4 and c=-2. It was a great question. It took about 3 minutes to solve.

[HemantPandey]

You can also assume a+b = x, b+c = y etc. Then divide the equations and solve for x, yz, etc.

[opytmx]

First set a+b+c = z, then sum all 3 equations: 2(a+b+c) * z = 18, so z² = 9, and finally z = 3 or z = - 3. The solution z = 3 leads to a new system of equations: a+b = 5, b+c = 2 and a+c = - 1, solution: a = 1, b = 4 and c = - 2. Finally, for z = - 3 you just flip the signs, as the square root of z² = + / - z and - z = (- a) + (- b) + (- c), hence the 2nd solution is: a = - 1, b = - 4 and c = 2.

[dominiquebercot]

Bonjour
Poser à+b+c=k, puis résoudre le système à+b= trouve a, b, c en fonction de k. On reporte dans a +b+c=k, cela donne k^2=9, et on a fini!

[-swayamphanse-f]

There's an easier way to solve this
→
Because the numbers on right of = are all multiple of 3
we can assume that
#1 a+b+c = 3
if a+b+c = 3
then a+b must be equal to 5
#2 a+b = 5
and b+c must be equal to 2
#3 b+c = 2
and c+a must be equal to -1
#4 c+a = -1


#1 - #3
a+b+c = 3
b+c= 2
we get a = 1
from #1
a+b = 5
so b = 4
from #4
c+a = -1
so c = -2

a = 1, b = 4, c = -1

There's also a 2nd Case where
a+b+c = -3
and a+b = -5
we get the same answer just the signs are inverted

[sngmn]

Add them up and we will get (a+b+c)^2 = 9. After that its easy to find a, b, c.

[andylee]

I was so pleased with myself because I did it in my head -- but oops, I forgot to handle the case where a + b + c = -3. Oh well, I can try to do better next time. :)

[harrymattah]

So we have a system of three equations, with three variables... and two solutions ?

[tanujvrbgs]

I used x, y and z as substitutions. The substitutions were:
x=a+b
y=b+c and,
z=c+a

[piyushabhay]

If I divide 1st equation by 2nd equation I get something....similarly on dividing 2nd by 3rd I get on solving I get a=b=c=0....why is it so?

[rmela]

Why doesn't it work if i first divide eq1 by eq2, then eq1by eq3 and e2 by eq3 to cancel out a+b+c? After rearranging new equations they are all set equal to zero which gives me all zeros

[zplusacademy]

Cordial jubilance welcoming your every upload ❤️🙏❤️🙏❤️🙏🙏❤️