For the system of equations `x+y+z=4, y+2z=5 and x+y+pz=q` to have no solution (A) `p=1 and q=4` (B) `p=1 and q!=4` (C) `p!=1 and q=4` (D) `p!=1 and q!=4`

For what values of p and q are the system of equations : 2x+5y+pz=q;x+2y+3z=14x+y+z=6 are consistent?

SECTION MULTIPLE CORRECT CHOICE TYPE Example 41 Let a, b, c , p, q be five different non-zero real numbers and x, y, z be three numbers satisfying the system of equations a a-p aq then (a) x + y + z = a + b + c-p-q abc (b) x = pq 0 in (c) y = (a-p)(h-p)(c-p) p(p-q) (a-q)(b-q)(c-q) aq-p) (d) z = b.

The value of p and q(p!=0,q!=0) for which p ,q are the roots of the equation x^2+p x+q=0 are (a) p=1,q=-2 (b) p=-1,q=-2 (c) p=-1,q=2 (d) p=1,q=2

If z=x+iy and z^(1/3) = p+iq, then (x/p+y/q) ÷ ( p^2 - q^2 )=

If 2x+1=15;3y-2=16 if x=p(mod 5),y=q(mod 5) then the value of p, q are (i) p=2,q=3 (ii) p=2,q=1 (iii) p=3,q=2 (iv) p=4,q=1

If H.C.F. of p and q is x and q=x y , then the L.C.M. of p and q is p q (b) q y (c) x y (d) p y

If the lines p_1 x + q_1 y = 1, p_2 x + q_2 y=1 and p_3 x + q_3 y = 1 be concurrent, show that the points (p_1 , q_1), (p_2 , q_2 ) and (p_3 , q_3) are colliner.

If x^(1/p)=y^(1/q)=z^(1/r) and xyz=1, then the value of p+q+r=