2x - 3y = 14
2x - y = 10
eqn (1) - eqn(2) gives: -2y = 4 so that y = -2
Then, by eqn (2): 2x -(-2) = 10
2x + 2 = 10
2x = 8 and so x = 4
The solution, therefore, is (x, y) = (4, -2)
2x - 3y = 14
2x - y = 10
eqn (1) - eqn(2) gives: -2y = 4 so that y = -2
Then, by eqn (2): 2x -(-2) = 10
2x + 2 = 10
2x = 8 and so x = 4
The solution, therefore, is (x, y) = (4, -2)
2x - 3y = 14
2x - y = 10
eqn (1) - eqn(2) gives: -2y = 4 so that y = -2
Then, by eqn (2): 2x -(-2) = 10
2x + 2 = 10
2x = 8 and so x = 4
The solution, therefore, is (x, y) = (4, -2)
2x - 3y = 14
2x - y = 10
eqn (1) - eqn(2) gives: -2y = 4 so that y = -2
Then, by eqn (2): 2x -(-2) = 10
2x + 2 = 10
2x = 8 and so x = 4
The solution, therefore, is (x, y) = (4, -2)
2x - 3y = 14
2x - y = 10
eqn (1) - eqn(2) gives: -2y = 4 so that y = -2
Then, by eqn (2): 2x -(-2) = 10
2x + 2 = 10
2x = 8 and so x = 4
The solution, therefore, is (x, y) = (4, -2)
2xy substitute 3 for y and 5 for x 2(5)(3) = 30
(2xy)x=2x2y
L = 3x + 2y
That equals 2xy + 3y which factors to y(2x + 3)
8-x2y 8-2xy 2xy-8
It's an hyperbola equation.
2xy=2*30*2=120
-2x - 2y = -122x + 2y = 122y = 12 - 2xy = 6 - x
2xy substitute 3 for y and 5 for x 2(5)(3) = 30
15xy-2xy-7x+x=13xy-6x
4
No, it equals -2xy. lrn2math
Because 2x+y is addition not multiplication
Y=15/(2c+2x)
x2 + y2 = x2 - 2xy + y2 + 2xy = (x - y)2 + 2xy = 72 + 2*8 = 49 + 16 = 65 You could, instead, solve the two equations for x and y and substitute, but the above method is simpler.
2xy
2xy*9 if x=3 and y = 2 2(3)(2)*9 12*9 108 You substitute x and y for their equivalent (3 and 2) then simply multiply 2*x*y, then multiply the answer of that by 9.