Question

The ordered pair (p, q) satisfies the simultaneous
equations (a + b)x + (b + c)y+ (c + a) = 0 and (b + c)
x + (c + a)y + (a + b) = 0 such that p and
in the ratio 1 : 2, then which of the following is
correct?
(a) a2 + 2ac + 32 = 262 + 3ab + bc
(b) a² + 12 + 2 = ab + bc + ca
(c) a2 + 3ac + 32 = 362 + 3ab + bc
(d) a3 + b3 + 3 = 3abc

Answer 1

Answer:

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Answer 2

Answer:

Correct option is A)

Given: p+q=1.....(1)

Now (p,q) satisfy 3x+2y=1

so, 3p+2q=1......(2)

Multiply equation (1) by 3

3p+3q = 3....(1)

3p+2q =1....(2)

Subtracting eq(2) from eq(1) we get

q = 2

Substitute q = 2 in equation (1)

p+2 = 1  

p = -1

Now, p =- 1 and q = 2

Putting the value in option(A)

3(-1)+4(2) = -3 + 8 = 5

Hence (p'q) satisfy option (A)

Step-by-step explanation:

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