`3-2x=7`
`2y+1=10-2(1)/(2)y`

x+7y=10 3x-2y=7

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

The equation of the circle whose diameter is the common chord of the circles; x^(2)+y^(2)+3x+2y+1=0&x^(2)+y^(2)+3x+4y+2=0 is x^(2)+y^(2)+8x+10y+2=0x^(2)+y^(2)-5x+10y+7=0x^(2)+y^(2)-5x+4y+7=02x^(2)+2y^(2)+6x+2y+1=0 None of these

Find the following products and verify the result for x=-1,y=-2:(3x-5y)(x+y)(2)(x^(2)y-1)(3-2x^(2)y)((1)/(3)x-(y^(2))/(5))((1)/(3)x+(y^(2))/(5))

Factorise: 12(y^(2)+7y)^(2)-8(y^(2)+7y)(2y-1)-15(2y-1)^(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))

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

Subtract: (-2y^(2)+(1)/(2)y-3) om 7y^(2)-2y+10