Bepaal die waarde(s) van x : \( \frac{{b - 2}}{{b - 1}} - \frac{5}{{{b^2} - 4b + 3}} + 1 = \frac{{10}}{{3 - b}} \)
\( \begin{align} \frac{{b - 2}}{{b - 1}} - \frac{5}{{{b^2} - 4b + 3}} + 1 &= \frac{{10}}{{3 - b}}\\ \frac{{\left( {b - 2} \right)}}{{\left( {b - 1} \right)}} - \frac{5}{{\left( {b - 3} \right)\left( {b - 1} \right)}} + \frac{1}{1} &=
\frac{{ - 10}}{{\left( {b - 3} \right)}}\\ {\rm{KGV}} &= \left( {b - 1} \right)\left( {b - 3} \right) \\ \left( {b - 2} \right)\left( {b - 3} \right) - 5 + 1\left( {b - 1} \right)\left( {b - 3} \right) &= - 10\left( {b - 1} \right)\\
{b^2} - 5b + 6 - 5 + {b^2} - 4b + 3 &= - 10b + 10\\ 2{b^2} - 9b + 4 &= - 10b + 10\\ 2{b^2} - 9b + 10b + 4 - 10 &= 0\\ 2{b^2} + b - 6 &= 0\\ \left( {2b - 3} \right)\left( {b + 2} \right) &= 0\\ 2b - 3 &= 0 &of&
&b + 2 &= 0\\ b &= \frac{3}{2} && &b &= - 2 \end{align} \)
Beperkings:
\( [\begin{align} b - 1 &\ne 0\\ b &\ne 1 \end{align} \)
\( \begin{align} b - 3 &\ne 0\\ b &\ne 3 \end{align} \)