[" 12."2x+3y+1=0],[(7-4x)/(3)=y]

Solve:(1)/(2(2x+3y))+(12)/(7(3x-2y))=(1)/(2)(7)/(2x+3y)+(4)/(3x-2y)=2 where 2x+3y!=0 and 3x-2y!=0

The equation of the bisector of obtuse angle between the lines 12x - 5y + 7 = 0, 3y-4x-1=0 is

To remove the first degree terms in the following equations origin should be shifted to the another point then calculate the new origins from list - II {:(" List - I "," List - II "),("(A) "x^(2)-y^(2)+2x+4y=0,"(1) (5,-7) "),("(B) "4x^(2) +9y^(2)-8x+36y + 4 = 0,"(2) (1,-2) "),("(C) "x^(2) + 3y^(2) + 2x + 12y + 1 = 0,"(3) (-1,2) "),("(D) "2(x-5)^(2)+3(y+7)^(2)=10,"(4) (-1,-2) "),(,"(5) (-5,7) "):} The correct matching is

Identify the type of conic section for each of the equations 1. 2x^(2) -y^(2) = 7 2. 3x^(2) +3 y^(2) -4x + 3y + 10 =0 3. 3x^(2) + 2y^(2) = 14 4. x^(2) + y^(2) + x-y=0 5. 11x^(2) -25y^(2) -44x + 50y - 256 =0 6. y^(2) + 4x + 3y + 4=0

Solve the following equations by Elimination method : (2x-3y)/3 = 3+(3y-4x)/4 , 1/3(6y+7x)=1/5(7x+12y)+4 .

If [[(x+3y),y],[7-x,4]]=[(4,-1),(0,1)], find the values of x and y.

If |[(x+3y),y],[7-x,4]|=[(4,-1),(0,1)], find the values of x and y.

Solve by cross-multiplication method. (i) 8x-3y=12,5x=2y +7 (ii) 6x+ 7y -11 =0, 5x+ 2y =13 (iii) (2)/(x)+(3)/(y) =5, (3)/(x) -(1)/(y) +9=0

Solve for x and y, using substitution method : 2x + y = 7, 4x - 3y + 1 =0