Question

Find the value of ${}^{\prime }{P}^{\prime }$ for which the given system of equation has only one solution (i.e unique solution)
$Px-y=2$ and $6x-2y=3$

• A. $P$ can have all real values except $-4$
• B. $P$ can have all real values except $5$
• C. $P$ can have all real values except $3$
• D. $P$ can have all real values except $-7$

Answer 1

Answer: (C)

$P$ can have all real values except $3$

The system of linear equations $a_{1}x+b_{1}y+c_1$ $=$ 0 and $a_{2}x+b_{2}y+c_2$ $=$ 0 will have a unique solution, if the two lines represented by the equations intersect at a point.
Hence, the lines should not be parallel. i.e., the two lines should have different slopes.
Condition for unique solution is $\frac{a1}{a2}\ne \frac{b1}{b2}$
Hence, $a_1=P,b_1=-1,c_1=-2$
$a_2=6,b_2=-2,c_2=-3$
Condition for unique solution is $\frac{a_1}{a_2}\ne \frac{b_1}{b_2}$
$\Rightarrow$ $\frac{P}{6}\ne \frac{+1}{+2}$
$\Rightarrow$ $P\ne \frac{6}{2}$
$\Rightarrow$ $P\ne 3$
$\therefore$ $P$ can have all real values except $3$.