Suggested: 3x/2-5y/3=-2 x/3+y/2=13/6 by substitution method - 2/x+3/y=13 5/x-4/y=-2 - d2+d+1)y=x 3 - d^2+d+1)y=x^3 - x(x-1)dy/dx-(x-2)y=x^3(2x-1) - x^2d^2y/dx^2-2x(1+x)dy/dx+2(1+x)y=x^3 - draw the graph of the equation x-y+1=0 and 3x+2y-12=0 - verify that x3+y3+z3-3xyz=1/2(x+y+z) (x-y)2+(y-z)2+(z-x)2 - x/2+2y/3=-1 and x-y/3=3 - if u=tan^-1(x^3+y^3/x-y) - x-y+1=0 3x+2y-12=0 - discuss the maxima and minima of f(x y)=x^3y^2(1-x-y) - if u=xlog xy where x^3+y^3+3xy=1 find du/dx - y=(x-1)(x-2)(x-3) - 1)y=x^3 Browse related:
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