A system of equations with a parameter t. A challenge in Algebra...

This video is about a system of equations in two variables! (Wait, did I say two?)
Radical expressions, algebra, algebraic equations, SyberMath, algebraic manipulations, equations, substitution, Challenging Math Problems, Non-routine Math Problems, radical equations, ...

[leech]

I think there was an easier way - using a+b = 1- t -> t = 1-a-b. Replace t in the second equation and you'll wind up with 2ab = 8 - 4a - 4b + 2ab + a^2 + b^2. The ab terms cancel, and you can rewrite the equation as 0 = (a-2)^2 + (b-2)^2, which implies a=2 and b=2, which will yield x=2 and y=3

[Drk]

I tried by myself first. I realized the substitutions, I'm learning your tricks xD. My face was "wtf" when I saw the discriminant "-(t+3)^2" and I forgot the trivial case =0. Very important the difference between ">0" AND ">=0". Thanks for you great video!

[stephomn]

Love your videos man and your style of teaching ❤

[damiennortier]

I think that when you have a + b = 1 - t and 2ab = 5 + t^2 + 2t, you just have to add the 2 equation to get (a+b)^2 = 6 + t + t^2 = (1 - t)^2 = 1 - 2t + t^2
So 6 + t + t^2 = 1 - 2t + t^2
=> 5 + 3t = 0 (I don't know why I got an different solution)
T = - 5/3
...
Otherwise, subtract those
a + b = 1 - t and 2ab = 5 + t^2 + 2t,
(a - b)^2 = - 4 - t^2 - 3t
So a - b = I sqrt[(t + 2)^2 - t] (if you want real solutions, set (t + 2)^2 - t = 0 and then, t + 2 = -+ sqrt t...)
And since a + b = 1 - t, you'll get a nice equation...

[GlasiaVD]

You can solve all systems of the form {xy = q AND x+y = -p} with the theoreme of Vieta.

[GirishManjunathMusic]

Before watching the video:

a + b = 1 - t
2ab = 5 + t² + 2t
adding,
(a + b)² = t² + t + 6
t² + 1 - 2t = t² + t + 6
3t + 5 = 0
t = -5/3

substituting
a + b = 8/3
2ab = 5 + 25/9 - 5/3 = 55/9
2b(8/3 - b) = 55/9
16b/3 - b² = 55/9
48b - 9b² - 55 = 0
9b² - 48b + 55 = 0
b² = (48 ± sqrt(48² - 36×55))/18
= 5/3 or 11/3
correspondingly, a = 1, or a = -1
As sqrt is a positive only function, a = -1 can be ignored.

b = 5/3, a = 1

Setting sqrt(x + 2) = 5/3 we get x = 7/9.
Setting sqrt(y + 1) = 1 we get y = 0.

Seeing as neither of my answers depended on the parameter, I think I just f'ed up. Or you can take b = -t or b = t + 16/3 and get parameter-dependant answers I guess.
b = -t still gives a = 1:
x = t² - 2
y = 0

b = t + 16/3 gives a = -(13/3 + 2t):
x = t² + 32t/3 + 238/9
y = 160/9 + 52t/3 + 4t².

[paultoutounji]

I enjoyed the video but had difficulty to understand how does Z worked ….How does Z can substitute 😀