Determine an interval of values for the parameter λ ensuring that the following equation is valid: $−2200=−(45+λ)x_1−(80+λ)x_2$
My attempt : Note that for different values of $\lambda$, the equation represents an equation of a straight line.
Putting $x_1=0\ $ and $x_2=0\ $ we see that for any value of $\lambda\ $ the L.H.S doesn't match with the R.H.S.
Hence for any value of $\lambda\ $ these equations represent straight lines not passing through origin.
So we can write $$−2200=−(45+λ)x_1−(80+λ)x_2 \\ \implies \frac{x_1}{\frac{2200}{(45+\lambda)}}+\frac{x_2}{\frac{2200}{(80+\lambda)}} =1$$
So from this we can see that $\frac{2200}{(45+\lambda)}$ and $\frac{2200}{(80+\lambda)}$ must not be undefined i.e.$\ (45+\lambda)$ and $(80+\lambda)$ must be non-zero.
Hence $\lambda \ne -45,-85$. Thus the admissible values are all real numbers except $-45$ and $-85$.
Is my solution correct?
You have divided by $45+\lambda$ and $80+\lambda$. If you plug $\lambda = -45$ into the original equation you get $-2200=-35x_2$, which is a fine equation with the solution $x_2=\frac {2200}{35}$ and $x_1$ can be anything. If by a valid equation you mean one that does not have any division by $0$ or one that can be satisfied, any $\lambda$ yields a valid equation.