Solve for B
\left\{\begin{matrix}\\B=0\text{, }&\text{unconditionally}\\B\in \mathrm{R}\text{, }&T=0\end{matrix}\right.
Solve for K
K\in \mathrm{R}
T=0\text{ or }B=0
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1TB-\frac{\mathrm{d}}{\mathrm{d}x}(M)KB=0
Subtract \frac{\mathrm{d}}{\mathrm{d}x}(M)KB from both sides.
BT-BK\frac{\mathrm{d}}{\mathrm{d}x}(M)=0
Reorder the terms.
-BK\frac{\mathrm{d}}{\mathrm{d}x}(M)+BT=0
Reorder the terms.
\left(-K\frac{\mathrm{d}}{\mathrm{d}x}(M)+T\right)B=0
Combine all terms containing B.
TB=0
The equation is in standard form.
B=0
Divide 0 by T.
\frac{\mathrm{d}}{\mathrm{d}x}(M)KB=1TB
Swap sides so that all variable terms are on the left hand side.
BK\frac{\mathrm{d}}{\mathrm{d}x}(M)=BT
Reorder the terms.
0=BT
The equation is in standard form.
K\in
This is false for any K.
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