Question

For what value of will the following system of equations have no solution? (2p − 1)x +
(p − 1)y = 2p + 1; y + 3x − 1 = 0

Answer 1

Step-by-step explanation:

$2(p - 1)x + (p - 1)y = 2p + 1$

$y + 3x - 1 = 0$

$3x + y - 1 = 0$

$for \: no \: solution \: \frac{a1}{a2} = \frac{b1}{b2} ≠ \frac{c1}{c2}$

$a1 = 2(p - 1), \: a2 = 3$

$b1 = p - 1, \: b2 = 1$

$c1 = - (2p + 1), \: c2 = - 1$

$\frac{a1}{a2} = \frac{2(p - 1)}{3}, \: \frac{b1}{b2} = \frac{p - 1}{1}, \: \frac{c1}{c2} = \frac{ - (2p + 1)}{ - 1}$

$\frac{2(p - 1)}{3} = \frac{p - 1}{1}$

$2p - 2 = 3(p - 1)$

$2p - 2 = 3p - 3$

$2p - 3p = - 3 + 2$

$- p = - 1$

$p = 1$

Therefore the value of p is 1