For $x, y$ are real numbers
$x^2 - xy + y^2 = 1$
$x^2 + xy + 2y^2 = 4$
#Could someone help me solve this system of equations?
blissful bridge
formal yachtBOT
velvet berryBOT
1 divided by 0 equals Infinity
blissful bridge
Could someone help me solve this system of equations?
my teacher teaches this method that for every reflecting expressions of x and y, then you can turn it into something only has numbers, x + y and xy
idk if i am saying the right term
i can convert the first expression, but the second expression cannot be converted
north basin
2xy + y^2 = 3
solve for y
blissful bridge
y(2x + y) = 3
2x + y = 3/y
2x = 3/y - y
x = (3/y - y)/2
you want me to sub that into one of the equations
north basin
you want me to sub that into one of the equations
quadratic formula
north basin
2xy + y² = 3
y² + 2xy - 3 = 0
blissful bridge
but i knew it already
y^2 + 2xy - 3 = 0
y^2 + 2xy + 1 - 4 = 0
(y + 1)^2 = 4
oh thanks
north basin
y = (-2x (+-) sqrt(4x² - 12)) / 2
blissful bridge
💀
can you prove the formula?
because i haven't learned it yet
OH WAIT
THAT DOES NOT WORK 😭
steep hornet
For $x, y$ are real numbers
$x^2 - xy + y^2 = 1$
$x^2 + xy + 2y^2 = 4$
Okay try this. Multiply the first equation by 4
steep hornet
4x^2 - 4xy + 4y^2 = 4
Now equate them together
blissful bridge
4x^2 - 4xy + 4y^2 = x^2 + xy + 2y^2
steep hornet
4x^2 - 4xy + 4y^2 = x^2 + xy + 2y^2
Ok, move things over to the LHS
steep hornet
3x^2 - 5xy + 2y^2 = 0
Now try factor this
blissful bridge
3x^2 - 3xy - 2xy + 2y^2 = 0
3x(x - y) - 2y(x - y) = 0
(x - y)(3x - 2y) = 0
uneven prairie
I would just use quadratic formula to get x in terms of y then equate them
steep hornet
or 3x - 2y = 0
Yes both are correct!
blissful bridge
okay
x = y then x^2 - x^2 + y^2 = 1
y^2 = 1 => x = y = 1 or -1
3x - 2y = 0
x = 2y/3
thanks so much!
steep hornet
thanks so much!
No problem!
blissful bridge
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Solved