Y_3 = abX_1 + acX_2
Y_3 = b(aX_1) + c(aX_2)
Y_3 = bY_1 + cY_2++
Excellent! Superposition holds true so we know that the system must be linear! Now, let’s try that for a linear equation $Y = aX + b$. Let’s define the output for two different signals: $Y_1 = aX_1 + b$ and $Y_2 = aX_2 + b$. And let’s also define a third signal as the sum of the first two inputs: $X_3 = mX_1 + nX_2$. (Like before, m and n are arbitrary constants). So for an input of $X_3$ we should expect $Y_3 = aX_3 + b$. Plugging in for X_3 yields:
++Y_3 = a(mX_1 + nX_2) + b++
Now here I’m going to do something a bit different than the first example. I’m going to solve for $X_1$ and $X_2$ and plug them into the previous equation.
++Y_3 = a(m\frac{Y_1-b}{a} + n\frac{Y_2-b}{a}) + b
Y_3 = m(Y_1-b) + n(Y_2-b) + b
Y_3 = mY_1-mb + nY_2-nb + b
Y_3 = mY_1 + nY_2 + b(1-m-n)++
As can be seen: $mY_1 + nY_2 + b(1-m-n) eq mY_1 + nY_2$ So this system is not linear despite the fact that the system is a linear equation! A pretty important concept (and sure to be a test question on an “Intro to Signals”-type test)!