Question

A unique solution?

Answer 1

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

To have unique equations, $\frac{a_1}{a_2}\ne \frac{b1}{b2}$

where, $a_1$ and $a_2$ are X-coefficients

$b_1$ and $b_2$ are Y-coefficients of the given equations.

Therefore,$\frac{1}{1}\ne a^2-3a+3$

$a^2-3a+2\ne 0$

$a\ne 2$ or $a\ne 1$

The values of $a\in R-\{1,2\}$ for the unique solutions.