`y=|1+{x}| and 3y=|-2x^2+5x+3|` for `-1/2 < x <= 1.` where {*} denotes fractional part of x.

Complete the following activity to solve the simultaneous equations. 5x-3y=13 and 2x+3y=1 {:(5x -" 3y = 13 ...(1)"),(2x+" 3y = 1 ...(2)"),(bar(square x " = 14")" "therefore x = square):} Substituting x=2 in equation (2) , 2xx2+3y=1 " " therefore 3y = square " " therefore y = square

2x-y=1 ; 6x+3y=5

1 + 5x + 3y = 9 2x 3y = 12 X = (x, y) - 4

x+y+z=1 x-2y+3z=2 5x-3y+z=3

2x + 5y = 1 , 2x + 3y = 3 .

If y=2+|x| -|x| -|x-1|-|x+1| , then y'(-1/2)+y'(3/2) +y'(5/2)=

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))