Loving your videos man. I'm an old man trying to learn a bit of math to keep the old brain going. Also to help me make sense of the amazing things modern science is showing us. Thanks for your efforts, much appreciated.
Subtract the x over in the top equation and multiply both sides by x so you can replace the left side of the bottom equation then you get 3x - x^2 = 2 where x =2 and x =1 and plugging in for y you get (2,-1) and (1,-1/2).
x / (x + y) = 2, from here we could find that x = -2y. Then it is substitution of the first equation of x with -2y. This gives quadtaric 2y^2 + 3y +1 =0, which gives y = [-1, -0.5}. so x = {2, 1}
This problem can actually be solved by substitution really easily, without any special tricks at all. Solve the second equation by hand, plug it into the first, and you get a trinomial that is simple to factor just using inspection
Take second equation first. From it we get --> y= -x/2 Then put the value of y in first equation. Then we will get a quadratic equation ie --> x^2 - 3x +2 =0------> we will get x= 1 or 2. Then we will put the value of x in y= -x/2. Hence we will get (x,y) == (1, -1/2) & (2, -1)
You are a genius with far more difficult problems, but this one is really simple even without factor assumptions. From eqn 2, X=2(x+y) => x=-2y. Use it n eqn 1, it becomes 1/(-y) +(-2y)=3 => 2y.y +3y+1=0 (2y+1)(y+1)=0
This is extremely simple. Multiply the top eq by x. Sub in eq 2. Solve the quadratic for x, solve for y. Simple
The video you posted prior to this video makes the solution plain to see.
It is what we call learning
Loving your videos man. I'm an old man trying to learn a bit of math to keep the old brain going. Also to help me make sense of the amazing things modern science is showing us.
Thanks for your efforts, much appreciated.
Just multiply the first equation by X and subtract the second equation and you get X^2-3X+2=0
Hence X=2,1
Same, well by -x and added, but that was more intuitive to me
x and 1/(x+y) are root of the t²-3t +2 = 0 or (t-1)(t-2) means
x = 1 or x = 2, substitute to find y
Subtract the x over in the top equation and multiply both sides by x so you can replace the left side of the bottom equation then you get 3x - x^2 = 2 where x =2 and x =1 and plugging in for y you get (2,-1) and (1,-1/2).
x / (x + y) = 2, from here we could find that x = -2y.
Then it is substitution of the first equation of x with -2y. This gives quadtaric 2y^2 + 3y +1 =0, which gives y = [-1, -0.5}. so x = {2, 1}
That bracket took me to the continuity.
Nice! And the condition x+y 0 is ok in both solutions.
This problem can actually be solved by substitution really easily, without any special tricks at all. Solve the second equation by hand, plug it into the first, and you get a trinomial that is simple to factor just using inspection
From the second equation y = -0.5*x, x != 0.
Then substitute y in the first equation:
2/x + x = 3
x = 1 or x = 2
Then y = -0.5 or y = -1
(x,y) = (1,-1/2) orally
cool!
Take second equation first. From it we get --> y= -x/2
Then put the value of y in first equation. Then we will get a quadratic equation ie --> x^2 - 3x +2 =0------> we will get x= 1 or 2. Then we will put the value of x in y= -x/2. Hence we will get (x,y) == (1, -1/2) & (2, -1)
Best intro !!! Love from India. can you share the name of the music or youtube link? by the way awesome explanation.
Thank you. The music is my original music.
You are a genius with far more difficult problems, but this one is really simple even without factor assumptions. From eqn 2, X=2(x+y) => x=-2y.
Use it n eqn 1, it becomes
1/(-y) +(-2y)=3 => 2y.y +3y+1=0
(2y+1)(y+1)=0
Thanks Sir 👍
x/x+y + x² = 3x
x²-3x+2 = 0
(x-2)(x-1) = 0
x = 2 doesn't satisfy
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(1/(x + y)) + x = 3
1/(x + y) = a
a + x = 3
ax = 2
x = 3 - a
(3 - a)a = 2
3a - a² = 2
-a² + 3a - 2 = 0
3±1/2
a₁ = 1
a₂ = 2
a₁x = 2
x
x₁= 2
a₂x = 2
2x = 2
x₂ = 1
1/(x₁ + y₁) = 1
1/(2 + y₁) = 1
2 + y₁ = 1
y₁ = -1
1/( x₂ + y₂) = 2
1/(1 + y₂) = 2
2 + 2y₂ = 1
2y₂ = -1
y₂ = -½
check for x₁ and y₁
1/(x + y) + x = 3
x/(x + y) = 2
1/(2 - 1) + 2
1/1 + 2
1 + 2 = 3
2/(2 - 1)
2/1 = 2
check for x₂ and y₂
1/(x + y) + x = 2
x/(x + y) = 3
1/(1 - ½) + 1
1/(½) + 1
2 + 1 = 3
x/(x + y)
1/(1 - ½)
1/½ = 2
answers:
x₁ = 2, y₁ = -1
x₂ = 1, y₂ = -½