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It is then 2y
x = 2y y = x - 2 or x = 2y subtract 2y to both sides x - 2 = y subtract y and add 2 to both sides x - 2y = 0 multiply by -1 to both sides x - y = 2 -x + 2y = 0 x - y = 2 add these two equations together y = 2 substitute 2 for y into the first equation, x = 2y x = 2(2) = 4 Thus the solution of these linear system of equations is the point (4, 2). Check:
x would equal -1 and y would equal 1
3xy-2y=0 3xy=2y y=2y (3x) y/2y=3x 1/2=3x multiply across by 2 1=6x 1/6=x therefore substituting x=1/6 into 3xy-2y; 3(1/6)y-2y=0 1/2y=2y y=2y/0.5 0.5 aka 1/2 y=1
(2y + 1) + (y^2 + 1) = 2y + 1 + y^2 + 1 = y^2 + 2y + 2