A Polynomial System
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Fun problem, also the graph looks really pretty :0
If you multiply the second equation by 9 and sum you get x^2+9y^2-6xy=81
Or
(x-3y)^2 =81
And you can combine it with (x+y)^2=49
And finding the solution is now pretty straightforward
I'm sorry, but I just glanced at the thumbnail and thought, "oh, that's (x+y)^2 = 49". Finding the actual solution sets did take some scratch paper, though.
There is a sign error in your second solution pair at 4:54. Of the three methods you discuss the second method is obviously the best choice and this method is easily spotted, since, like you said, this is a contrived problem. Your quadratic in y at 8:36 is easily factored by grouping and I was a bit surprised that you did not use that here because you did do factoring by grouping of your quadratic in t at 2:32.
Here's a fourth method. The left hand sides or your original equations are homogeneous, and since it is evident that neither x nor y can be zero we can let y/x = k, i.e. we can substitute
y = kx
which gives
x²(1 + 3k) = 45
x²(k² − k) = 4
Multiplying both sides of the first equation by 4 and both sides of the second equation by 45 and noting that x² ≠ 0 this gives
45(k² − k) = 4(1 + 3k)
45k² − 57k − 4 = 0
45k² − 60k + 3k − 4 = 0
15k(3k − 4) + (3k − 4) = 0
(15k + 1)(3k − 4) = 0
k = −¹⁄₁₅ ⋁ k = ⁴⁄₃
Substituting k = −¹⁄₁₅ in x²(1 + 3k) = 45 we get
⁴⁄₅x² = 45
x² = ²²⁵⁄₄
x = ¹⁵⁄₂ ⋁ x = −¹⁵⁄₂
and since y = kx = −¹⁄₁₅x this gives the solution pairs
(x, y) = (¹⁵⁄₂, −¹⁄₂) ⋁ (x, y) = (−¹⁵⁄₂, ¹⁄₂)
Substituting k = ⁴⁄₃ in x²(1 + 3k) = 45 we get
5x² = 45
x² = 9
x = 3 ⋁ x = −3
and since y = kx = ⁴⁄₃x this gives the solution pairs
(x, y) = (3, 4) ⋁ (x, y) = (−3, −4)
A fifth method which is related to the previous method is to multiply both sides of the first equation by 4 and both sides of the second equation by 45. Then, when we subtract the second equation from the first we have
4x² + 12xy − 45y² + 45xy = 0
4x² + 57xy − 45y² = 0
4x² + 60xy − 3xy − 45y² = 0
4x(x + 15y) − 3y(x + 15y) = 0
(x + 15y)(4x − 3y) = 0
x + 15y = 0 ⋁ 4x − 3y = 0
y = −¹⁄₁₅x ⋁ y = ⁴⁄₃x
The solution can now be completed by substituting either expression for y in the first equation x² + 3xy = 45 which gives x² = ²²⁵⁄₄ with y = −¹⁄₁₅x and x² = 9 with y = ⁴⁄₃x exactly as with the previous method.
Here's a sixth method. Rewrite the system of equations as
x² = 45 − 3xy
y² = 4 + xy
Multiply both equations to get
x²y² = (45 − 3xy)(4 + xy)
x²y² = 180 + 33xy − 3x²y²
4x²y² − 33xy x²y² − 180 = 0
Now let
xy = p
then we have
4p² − 33p − 180 = 0
Let's try to factor this quadratic over the integers using factoring by grouping. Factoring 4·(−180) = −720 = −2⁴·3²·5 and noting that for an integer factorization we need to distribute these prime factors over two integers with sum −33 it follows that one of these integers needs to be positive and the other negative since their product is negative. Also, one of these integers needs to be even and the other odd since their sum is odd, so all factors 2 must be contained in one of the integers which is the even integer. Knowing this it is not hard to see that the two numbers with product −720 and sum −33 we are looking for are −2⁴·3 = −48 and 3·5 = 15. So, we get
4p² − 48p + 15p − 180 = 0
4p(p − 12) + 15(p − 12) = 0
(4p + 15)(p − 12) = 0
p = −¹⁵⁄₄ ⋁ p = 12
Substituting xy = p = −¹⁵⁄₄ in x² = 45 − 3xy gives
x² = ²²⁵⁄₄
x = ¹⁵⁄₂ ⋁ x = −¹⁵⁄₂
and since xy = −¹⁵⁄₄ this gives the solution pairs
(x, y) = (¹⁵⁄₂, −¹⁄₂) ⋁ (x, y) = (−¹⁵⁄₂, ¹⁄₂)
Substituting xy = p = 12 in x² = 45 − 3xy gives
x² = 9
x = 3 ⋁ x = −3
and since xy = 12 this gives the solution pairs
(x, y) = (3, 4) ⋁ (x, y) = (−3, −4)
Seventh method. Adding both equations gives
(x + y)² = 49
and multiplying both sides of the second equation by 9 and _then_ adding both equations gives
(x − 3y)² = 81
So we have
(x + y)² = 7² ⋀ (x − 3y)² = 9²
(x + y = 7 ⋁ x + y = −7) ⋀ (x − 3y = 9 ⋁ x − 3y = −9)
(x + y = 7 ⋀ x − 3y = 9) ⋁ (x + y = −7 ⋀ x − 3y = 9) ⋁ (x + y = 7 ⋀ x − 3y = −9) ⋁ (x + y = −7 ⋀ x − 3y = −9)
and then we only need to solve these four linear systems in x and y to get
(x, y) = (¹⁵⁄₂, −¹⁄₂) ⋁ (x, y) = (−¹⁵⁄₂, ¹⁄₂) ⋁ (x, y) = (3, 4) ⋁ (x, y) = (−3, −4)
(x + y)² = 49
x + y = ± 7 => y = ± 7 - x
y = 7 - x
x² + 3x(7 - x) = 45
x² - 3x² + 21x - 45 = 0
2x² - 21x + 45 = 0
x = (21 ± 9)/4
*x = 15/2* => *y = -1/2*
*x = 3* => *y = 4*
y = -7 - x
x² + 3x(-7 - x) = 45
x² - 3x² - 21x - 45 = 0
2x² + 21x + 45 = 0
x = (-21 ± 9)/4
*x = -3* => *y = -4*
*x = -15/2* => *y = 1/2*
Why dont u take lectures on arts of problem solving books?
x = 3, y = 4
There are 4 solutions ... of course you can inverse signs ... if (3,4) is solution then (-3,-4) is obviously one too. But there is another pair ...
Iyi iş biraderim
I used the second method, and my solutions are (15/2,-1/2), (-15/2,29/2), (3,4) and (-3,10). Do they look correct?
I haven't seen the video but i just checked my solutions on desmos and they're correct, looks like you mixed up a ± somewhere because half of your solutions are fine.
you should have gotten
(-15/2, 1/2) (15/2, - 1/2) (-3,-4) (3,4)
I did mine by making a perfect square and setting y = -x ± 7, then i plugged it in to the second equation, you get a quadratic equation with two ± signs then just clean it up ect.
Unfortunately not. For x = -15/2, you should have y = 1/2, not 29/2, and for x = -3, you should have y = -4, not y = 10. the other two solutions are correct, though.
Wouw ... you're complicated ... it's a very simple biquadratic ... there are obviously 4 solutions
(ii) xy = y² - 4 and x = y - 4/y (y=0 is impossible)
x² = y² -8 + 16/y²
(i) y² - 8 + 16/y² + 3y² - 12 - 45 = 0
4y² - 65 + 16/y² = 0
4y4 - 65y² + 16 = 0 delta = 65²-256 = 3969 = 63²
y² = (65 +/- 63)/8
y² = 1/4 or 16
y = +/- 1/2 , +/- 4
4 solutions :
1) y = 1/2 => x = 1/2 - 8 = -15/2
2) y = -1/2 => x = 15/2
3) y = 4 => x = 4 - 1 = 3
4) y = -4 => x = -3