Suggested: (x+2y)^3 - 7x-15y=2 x+2y=3 by substitution method - 7x-15y=2 x+2y=3 vilopan vidhi - show that the straight lines x-2y+3=0 and 6x+3y+8=0 are perpendicular - 7x-15y=2 x+2y=3 by elimination method - dy/dx=x+2y-3/2x+y-3 - 7x-15y=2 x+2y=3 pratisthapan vidhi - 2x+3y=8 x-2y+3=0 solve graphically - dy/dx(x^2y^3+xy)=1 - without finding the cubes factorise (x-2y)^3+(2y-3z)^3+(3z-x)^3 - dy/dx=2x-y+1/x+2y-3 - (x+2y+3z)2 - x+2y=3 4x+3y=2 - (x+2y^3)dy/dx=y - (x+2y)^3 Browse related:
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