No, it is not.
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y=2x-4 y=2x-5 y=1 1=2x-4 -2x = -5 x=2/5 the solution is (x,y) = (2/5,1)
2x - 3y = 142x - y = 10eqn (1) - eqn(2) gives: -2y = 4 so that y = -2Then, by eqn (2): 2x -(-2) = 102x + 2 = 102x = 8 and so x = 4The solution, therefore, is (x, y) = (4, -2)2x - 3y = 142x - y = 10eqn (1) - eqn(2) gives: -2y = 4 so that y = -2Then, by eqn (2): 2x -(-2) = 102x + 2 = 102x = 8 and so x = 4The solution, therefore, is (x, y) = (4, -2)2x - 3y = 142x - y = 10eqn (1) - eqn(2) gives: -2y = 4 so that y = -2Then, by eqn (2): 2x -(-2) = 102x + 2 = 102x = 8 and so x = 4The solution, therefore, is (x, y) = (4, -2)2x - 3y = 142x - y = 10eqn (1) - eqn(2) gives: -2y = 4 so that y = -2Then, by eqn (2): 2x -(-2) = 102x + 2 = 102x = 8 and so x = 4The solution, therefore, is (x, y) = (4, -2)
y = -1/(3x7 + 2x - 9)y = -(3x7 + 2x - 9)-1y' = -(-1)(3x7 + 2x - 9)-2(3x7 + 2x - 9)'y' = (3x7 + 2x - 9)-2(21x6 + 2)y' = (21x6 + 2)/(3x7 + 2x - 9)2
y-2x=3 -y -y -2x=3-y -3 -3 -2x-3=-y /-1 /-1 2x+3=y y=2x and y=2x+3 have the same slope of 2, so they are parallel. Hope this helps! ;D
2x / (y-2) + 2x / (2-y)First get a common denominator: (y-2) (2-y)Then you have:2x(2-y) / (y-2)(2-y) + 2x(y-2) / (y-2)(2-y), summing and simplifying, you get:(4x-4) / (y-2)(2-y)