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(B) The system of equations $AX = B$ is given by: $2x - 5y = 20$ $3x + my = m$ Using Cramer's rule or substitution,we find the determinant $|A| = 2m -

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$450$

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(B) The system of equations $AX = B$ is given by: $2x - 5y = 20$ $3x + my = m$ Using Cramer's rule or substitution,we find the determinant $|A| = 2m - (-15) = 2m + 15$. For a unique solution,$|A| 
eq 0$,so $m 
eq -15/2$. Solving for $x$ and $y$: $x = \frac{\begin{vmatrix} 20 & -5 \ m & m \end{vmatrix}}{|A|} = \frac{20m + 5m}{2m + 15} = \frac{25m}{2m + 15}$ $y = \frac{\begin{vmatrix} 2 & 20 \ 3 & m \end{vmatrix}}{|A|} = \frac{2m - 60}{2m + 15}$ For $x For $y The intersection of these intervals is $m \in (-\frac{15}{2}, 0)$,so $a = -15/2$ and $b = 0$. Now,calculate $8 \int_{-15/2}^0 (2m + 15) dm$: $8 [m^2 + 15m]_{-15/2}^0 = 8 [0 - ((\frac{-15}{2})^2 + 15(\frac{-15}{2}))]$ $= 8 [0 - (\frac{225}{4} - \frac{225}{2})] = 8 [0 - (\frac{225 - 450}{4})] = 8 [\frac{225}{4}] = 450$.

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