Question

For what value of p, is the system of equations :
$p^3x + (p + 1)^3 y = (p + 2)^3$
$px + (p + 1) y = p + 2$
$x + y = 1$
consistent ?

• A. p = 0
• B. p = 1
• C. p = - 1
• D. For all p > 1

Answer 1

Answer: (C)

The given system of equations are :
$p^{3}x + \left(p + 1\right)^{3} y = \left(p + 2\right)^{3}\quad...\left(1\right)$
$px + \left(p + 1\right) y = p + 2\quad ...\left(2\right)$
$x + y = 1\quad ...\left(3\right)$
This system is consistent, if values of x and y from first two equation satisfy the third equation.
which $\Rightarrow \begin{vmatrix}p^{3}&\left(p+1\right)^{3}&\left(p+2\right)^{3}\\ p&\left(p+1\right)&\left(p+2\right)\\ 1&1&1\end{vmatrix} = 0$
$\Rightarrow \begin{vmatrix}p^{3}&\left(p+1\right)^{3}-p^{3}&\left(p+2\right)^{3} -p^{3}\\ p&1&2\\ 1&0&0\end{vmatrix} = 0$
$\Rightarrow 2 \left(p + 1\right)^{3} - 2p^{3} - \left(p + 2\right)^{3 }+ p^{3} = 0$
$\Rightarrow 2 \left(p^{3} +1 + 3p^{2} + 3p\right) - 2p^{3} - \left(p^{3} + 8 + 12p + 6p^{2}\right) + p^{3} = 0$
$\Rightarrow 2p^{3} + 2 + 6p^{2} + 6p - 2p^{3}- p^{3} - 8 - 12p - 6p^{2 }+ p^{3} = 0$
$\Rightarrow \quad- 6 - 6p = 0$
$\Rightarrow \quad p = - 1$