[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

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) .

Solve for x and y if [(x^(2)),(y^(2))]+2[(2x), (3y)]=3[(7),(-3)]

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

Simplify: (3x+2y)(4x+3y)-(2x-y)(7x-3y)

{:(3x + 2y = 4),(2x - 3y = 7):}

{:(3x + 2y = 4),(2x - 3y = 7):}

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