To ensure that the function $f(x)$ is continuous at $x = 5$, the left-hand limit and the right-hand limit of $f(x)$ at $x = 5$ must be equal to the function's value at $x = 5$. Here's how we approach this problem step-by-step:

Step 1: Left-hand limit

For $x < 5$, we are given: $f(x) = b - 2x$ The left-hand limit of $f(x)$ as $x$ approaches 5 is: $\lim_{x \to 5^-} f(x) = b - 2(5) = b - 10$

Step 2: Right-hand limit

For $x \geq 5$, we are given: $f(x) = \frac{150}{x - b}$ The right-hand limit of $f(x)$ as $x$ approaches 5 is: $\lim_{x \to 5^+} f(x) = \frac{150}{5 - b}$

Step 3: Continuity at $x = 5$

For $f(x)$ to be continuous at $x = 5$, we need the left-hand limit to equal the right-hand limit: $b - 10 = \frac{150}{5 - b}$

Step 4: Solve for $b$

Multiply both sides of the equation by $5 - b$ to eliminate the denominator: $(b - 10)(5 - b) = 150$ Expand the left-hand side: $(b - 10)(5 - b) = 5b - b^2 - 50 + 10b = -b^2 + 15b - 50$ Thus, the equation becomes: $-b^2 + 15b - 50 = 150$ Move all terms to one side: $-b^2 + 15b - 200 = 0$ Multiply through by -1 to simplify: $b^2 - 15b + 200 = 0$ Now, solve this quadratic equation using the quadratic formula: $b = \frac{-(-15) \pm \sqrt{(-15)^2 - 4(1)(200)}}{2(1)}$ $b = \frac{15 \pm \sqrt{225 - 800}}{2}$ $b = \frac{15 \pm \sqrt{-575}}{2}$ Since the discriminant is negative, this equation has no real solutions, meaning that the approach we have taken to solve the problem does not lead to real solutions for $b$.

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