Question

Infinitely many solutions?

Answer 1

Given equation

$x+(a^2-3a+3)y=a^2$ and $x+y=1$

To have infinity no. of solutions for equations

$a_{1}x+b_{1}y=c_1$ and $a_{2}x+b_{2}y=c_2$

$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$

So, $\frac{1}{1}=\frac{(a^2-3a+3)}{1}=\frac{a_2}{1}$

$\Rightarrow a^2-3a+3=a^2$

$⟹a=1$

So, the value of $a=1$