derive limit to make a function continuous Here's the problem:
Choose the value of k that makes the following 
function continuous at $x = 1$:
$f(x)=\begin{cases}
\frac{-8x^2 + 48x - 40}{x - 1} & x < 1\\
-2x + k &x \geq 1
\end{cases}$
My steps:
$\lim_{x\uparrow 1} f(x) = \lim_{x\downarrow 1} f(x)$
$\frac{-8x^2 + 48x - 40}{x - 1} = -2x + k$
$\frac{-8(x^2 - 6x + 5)}{x - 1} = -2x + k$
$\frac{-8(x - 5)(x - 1)}{x - 1} = -2x + k$
$-8(x - 5) = -2x + k$
$-8x - (-8)\cdot5 = -2x + k$
$-8x + 40 = -2x + k$
$-8x + 2x + 40 = k$ (See note at bottom)
$-6x + 40 = k$
$-6\cdot1 + 40 = k$
$-6 + 40 = k$
$36 = k$
The solution steps:
For f to be continuous at $x = 1$ we need $\lim_{x\uparrow 1} f(x) = \lim_{x\downarrow1} f(x) = f(1)$
First lets evaluate lim x→1− f(x). Since x < 1 as x approaches 1 from the left, 
f(x) = (−8x^2 + 48x − 40)/(x − 1) as x→1−
So lim x→1− f(x) = lim x→1−  (−8x^2 + 48x − 40)/(x − 1)
Start by factoring the numerator. 
In this case we find: 
lim x→1− (−8x^2 + 48x − 40)/(−3x^2 − 15x + 18) = lim x→1− (−8x + 40)(x − 1)/(x − 1)
Aside: can anyone explain (−3x^2 − 15x + 18) in the denominator?
Since we are taking the limit as x→1−, we may assume that x ≠ 1. 
Canceling factors we see: 
\lim x→1− (−8x + 40)(x − 1)/(−3x − 18)(x − 1) = \lim x→1− (−8x + 40)
We can evaluate this linear function at x = 1. 
Hence, 
lim x→1− −8x^2 + 48x − 40)/(−3x^2 − 15x + 18) = −8(1) + 40. 
Thus, lim x→1− f(x) = 32. 
Thus lim x→1− f(x) = 32 
lim x→1+ f(x) = lim x→ (−2x + k), because x > 1 as x approaches 1 from the right. 
lim x→ (−2x + k) = −2⋅1 + k = −2 + k = f(1) because linear functions are continuous. 
For f to be continuous, these limits have to be equal. Thus 32 = −2 + k 
When k = 34, f is continuous.

NOTE: This is where I diverged from the solution by moving -2x to the left. My question is, assuming the given solution is correct, why is this nominally valid algebraic step not permitted?

You solution is correct, but it's an overkill. After you have correctly found limit from the above, set limit from the below $f(1) = -2 \cdot 1 +k = 34$ and solve for $k$. Continuity is ensured by the fact that $1$ is included in the definition of the limit from below. 
