03:42 PM 23/03/2019
unanswerable
03:43 PM 23/03/2019
I didn’t learn thzt yet
03:43 PM 23/03/2019
1.
03:43 PM 23/03/2019
Given equations are,
2x² - xy - y² + 3x + 2y - 3 = 0. ...(1)
x² - 3x + 2 = 0. ...(2)
From equation (2),
x² - 3x + 2 = 0
or, x² - 2x - x + 2 = 0
or,. ( x - 2) ( x - 1 ) = 0
So, values of x are
x = 1, 2.
On putting the values of x in equation(1), we get
For x = 1, the value of y is
2(1)² - (1)y - y² + 3(1) + 2y - 3 = 0
- y² + y + 2 = 0
or,
y² - y - 2 = 0
y = (1±3)/2
On taking +ve sign,
y = 2
On -ve sign,
y = -1.
And for x = 2, the value of y is
2(4) - 2y - y² + 3(2) + 2y - 3 = 0
8 - y² + 6 - 3 = 0
or, y² = 11
or, y = ±√11
So, the values of x&y are
x = 1. y = -1, 2.
x = 2. y = ±√11
How i did it: searched it on quora
2x² - xy - y² + 3x + 2y - 3 = 0. ...(1)
x² - 3x + 2 = 0. ...(2)
From equation (2),
x² - 3x + 2 = 0
or, x² - 2x - x + 2 = 0
or,. ( x - 2) ( x - 1 ) = 0
So, values of x are
x = 1, 2.
On putting the values of x in equation(1), we get
For x = 1, the value of y is
2(1)² - (1)y - y² + 3(1) + 2y - 3 = 0
- y² + y + 2 = 0
or,
y² - y - 2 = 0
y = (1±3)/2
On taking +ve sign,
y = 2
On -ve sign,
y = -1.
And for x = 2, the value of y is
2(4) - 2y - y² + 3(2) + 2y - 3 = 0
8 - y² + 6 - 3 = 0
or, y² = 11
or, y = ±√11
So, the values of x&y are
x = 1. y = -1, 2.
x = 2. y = ±√11
How i did it: searched it on quora
03:44 PM 23/03/2019
Quote from Voodoo , 03:43 PM 23/03/2019Hint: you're not mean to find the values of x, y or z.
Given equations are,
2x² - xy - y² + 3x + 2y - 3 = 0. ...(1)
x² - 3x + 2 = 0. ...(2)
From equation (2),
x² - 3x + 2 = 0
or, x² - 2x - x + 2 = 0
or,. ( x - 2) ( x - 1 ) = 0
So, values of x are
x = 1, 2.
On putting the values of x in equation(1), we get
For x = 1, the value of y is
2(1)² - (1)y - y² + 3(1) + 2y - 3 = 0
- y² + y + 2 = 0
or,
y² - y - 2 = 0
y = (1±3)/2
On taking +ve sign,
y = 2
On -ve sign,
y = -1.
And for x = 2, the value of y is
2(4) - 2y - y² + 3(2) + 2y - 3 = 0
8 - y² + 6 - 3 = 0
or, y² = 11
or, y = ±√11
So, the values of x&y are
x = 1. y = -1, 2.
x = 2. y = ±√11
How i did it: searched it on quora
03:44 PM 23/03/2019
i did with a human
Given equations are,
2x² - xy - y² + 3x + 2y - 3 = 0. ...(1)
x² - 3x + 2 = 0. ...(2)
From equation (2),
x² - 3x + 2 = 0
or, x² - 2x - x + 2 = 0
or,. ( x - 2) ( x - 1 ) = 0
So, values of x are
x = 1, 2.
On putting the values of x in equation(1), we get
For x = 1, the value of y is
2(1)² - (1)y - y² + 3(1) + 2y - 3 = 0
- y² + y + 2 = 0
or,
y² - y - 2 = 0
y = (1±3)/2
On taking +ve sign,
y = 2
On -ve sign,
y = -1.
And for x = 2, the value of y is
2(4) - 2y - y² + 3(2) + 2y - 3 = 0
8 - y² + 6 - 3 = 0
or, y² = 11
or, y = ±√11
So, the values of x&y are
x = 1. y = -1, 2.
x = 2. y = ±√11
How i did it: copied
Given equations are,
2x² - xy - y² + 3x + 2y - 3 = 0. ...(1)
x² - 3x + 2 = 0. ...(2)
From equation (2),
x² - 3x + 2 = 0
or, x² - 2x - x + 2 = 0
or,. ( x - 2) ( x - 1 ) = 0
So, values of x are
x = 1, 2.
On putting the values of x in equation(1), we get
For x = 1, the value of y is
2(1)² - (1)y - y² + 3(1) + 2y - 3 = 0
- y² + y + 2 = 0
or,
y² - y - 2 = 0
y = (1±3)/2
On taking +ve sign,
y = 2
On -ve sign,
y = -1.
And for x = 2, the value of y is
2(4) - 2y - y² + 3(2) + 2y - 3 = 0
8 - y² + 6 - 3 = 0
or, y² = 11
or, y = ±√11
So, the values of x&y are
x = 1. y = -1, 2.
x = 2. y = ±√11
How i did it: copied
03:44 PM 23/03/2019
2, 3, 1
03:45 PM 23/03/2019
Quote from ivo , 03:44 PM 23/03/2019answer is nothingQuote from Voodoo , 03:43 PM 23/03/2019
Given equations are,
2x² - xy - y² + 3x + 2y - 3 = 0. ...(1)
x² - 3x + 2 = 0. ...(2)
From equation (2),
x² - 3x + 2 = 0
or, x² - 2x - x + 2 = 0
or,. ( x - 2) ( x - 1 ) = 0
So, values of x are
x = 1, 2.
On putting the values of x in equation(1), we get
For x = 1, the value of y is
2(1)² - (1)y - y² + 3(1) + 2y - 3 = 0
- y² + y + 2 = 0
or,
y² - y - 2 = 0
y = (1±3)/2
On taking +ve sign,
y = 2
On -ve sign,
y = -1.
And for x = 2, the value of y is
2(4) - 2y - y² + 3(2) + 2y - 3 = 0
8 - y² + 6 - 3 = 0
or, y² = 11
or, y = ±√11
So, the values of x&y are
x = 1. y = -1, 2.
x = 2. y = ±√11
How i did it: searched it on quora
Hint: you're not mean to find the values of x, y or z.