We have
[tex]\displaystyle\lim_{x\to5^-}f(x)=\lim_{x\to5}b-2x=b-10[/tex]
[tex]\displaystyle\lim_{x\to5^+}f(x)=\lim_{x\to5}-\frac{150}{x-b}=\frac{150}{b-5}[/tex]
with [tex]f(5)[/tex] equal to the limit from the right as [tex]x\to5[/tex].
We need to get
[tex]b-10=\dfrac{150}{b-5}\implies(b-5)(b-10)=150\implies b^2-15b-100=(b-20)(b+5)=0[/tex]
[tex]\implies b=20\text{ or }b=-5[/tex]
The larger solution in absolute value is [tex]b=20[/tex].
