[" 5."7(y+3)-2(x+2),=14],[4(y-2)+3(x-3),=2]

Solve the following system of equations: 7(y+3)-2(x+2)=14,quad 4(y-2)+3(x-3)=2

Solve the following system of equations: 7(y+3)-2(x+2)=14 ,\ \ \ \ \ 4(y-2)+3(x-3)=2

Find x and y if :- ((x+2y, 14),(-3, y-2))=((4, 14),(-3, 7+3x)) .

3x+2y=14; x+4y=7

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

Find the following products: (x+2)(x+3) (ii) (x+7)(x-2)y-4)(y-3)( iv )(y-7)(y+3)(2x-3)(2x+5)( vi )(3x+4)(3x-5)

Factorise (2x+3y)^(2)+14(2x+3y)(3x-y)-32(3x-y)^(2) .

Factorise (2x+3y)^(2)+14(2x+3y)(3x-y)-32(3x-y)^(2) .

The equation of the transvers and conjugate axes of a hyperbola are, respectively, x+2y-3=0 and 2x-y+4=0 , and their respective lengths are sqrt(2) and 2sqrt(3)dot The equation of the hyperbola is 2/5(x+2y-3)^2-3/5(2x-y+4)^2=1 2/5(x-y-4)^2-3/5(x+2y-3)^2=1 2(2x-y+4)^2-3(x+2y-3)^2=1 2(x+2y-3)^2-3(2x-y+4)^2=1