Suppose that the functions \( s \) and \( t \) are defined for all real numbers \( x \) as follon \[ \begin{array}{l}s(x)=2 x-4 \\ t(x)=x-3 \\ \text { Write the expressions for }(s+t)(x) \text { and }(s-t)(x) \text { and evaluate }(s \cdot t)(-2) \text {. } \\ (s+t)(x)=\square \\ (s \cdot t)(-2)=\square\end{array} \]

To find the expressions for \( (s+t)(x) \) and \( (s-t)(x) \), we start by performing the operations on the functions \( s(x) \) and \( t(x) \). 1. **Calculating \( (s+t)(x) \)**: \[ (s+t)(x) = s(x) + t(x) = (2x - 4) + (x - 3) = 3x - 7 \] 2. **Calculating \( (s-t)(x) \)**: \[ (s-t)(x) = s(x) - t(x) = (2x - 4) - (x - 3) = 2x - 4 - x + 3 = x - 1 \] Next, we evaluate \( (s \cdot t)(-2) \): \[ s(-2) = 2(-2) - 4 = -4 - 4 = -8 \] \[ t(-2) = -2 - 3 = -5 \] \[ (s \cdot t)(-2) = s(-2) \cdot t(-2) = (-8) \cdot (-5) = 40 \] Putting it all together, we have: \[ (s+t)(x) = 3x - 7 \] \[ (s \cdot t)(-2) = 40 \]