[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

If the lines x - y - 1 = 0, 4x + 3y = k and 2x – 3y +1 = 0 are concurrent, then k is

Find x and y if :- [(x+3y, y),(7-x, 4)]=[(4, -1),(0, 4)] .

Simplify : (i) (5x - 9y) - (-7x + y) (ii) (x^(2) -x) -(1)/(2)(x - 3 + 3x^(2)) (iii) [7 - 2x + 5y - (x -y)]-(5x + 3y -7) (iv) ((1)/(3)y^(2) - (4)/(7)y + 5) - ((2)/(7)y - (2)/(3)y^(2) + 2) - ((1)/(7)y - 3 + 2y^(2))

x-3y=1,3x-2y+4=0

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

Find the greatest value of x^3y^4 if 2x + 3y = 7 and x>=0,y>=0 .

Find the greatest value of x^3y^4 if 2x + 3y = 7 and x>=0,y>=0 .