(4y+1)/(3)+2y-1quad 3y-(7)/(5)=(47)/(10)

(4y+1)/(3)+(2y-1)/(2)-(3y-7)/(5)=(47)/(10)

Find (x+y)+(x-y), if x=(5)/(4),y=(-1)/(3) (ii) x=(2)/(7),y=(4)/(3)( iii) x=(1)/(4),y=(3)/(2)

Find the S.D. between the lines : (i) (x)/(2) = (y)/(-3) = (z)/(1) and (x -2)/(3) = (y - 1)/(-5) = (z + 4)/(2) (ii) (x -1)/(2) = (y - 2)/(3) = (z - 3)/(2) and (x + 1)/(3) = (y - 1)/(2) = (z - 1)/(5) (iii) (x + 1)/(7) = (y + 1)/(-6) = (z + 1)/(1) and (x -3)/(1) = (y -5)/(-2) = (z - 7)/(1) (iv) (x - 3)/(3) = (y - 8)/(-1) = (z-3)/(1) and (x + 3)/(-3) = (y +7)/(2) = (z -6)/(4) .

The distance of point of intersection of lines (x-4)/(1)=(x+3)/(-4)=(z-1)/(7) and (x-1)/(2)=(y+1)/(-3)=(z+10)/(8) from (1,-4,7) is

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))

Simplify: ((1)/(3)y^(2)-(4)/(7)y+11)-((1)/(7)y-3+2y^(2))-((2)/(7)y-(2)/(3)y^(2)+2)

(1)/(2(3x+4y))=(1)/(5(2x-3y))=(1)/(4),

(1) / (6) (4y + 5) - (2) / (3) (2y + 7) = (3) / (2)

(1) / (6) (4y + 5) - (2) / (3) (2y + 7) = (3) / (2)