1. **Stating the problem:** We need to find the unknown $x$ from the equation involving determinants: $$\left|\begin{matrix} 1 & 8 & 1 \\ -1 & 5 & 4 \\ 3 & 3x + 8 & -4 \end{matrix}\right| = \left|\begin{matrix} 7 & 3x + 6 \\ 1 & 1 \end{matrix}\right|$$ 2. **Recall the determinant formulas:** - For a $3 \times 3$ matrix $M = \begin{bmatrix}a & b & c \\ d & e & f \\ g & h & i\end{bmatrix}$, the determinant is: $$\det(M) = a(ei - fh) - b(di - fg) + c(dh - eg)$$ - For a $2 \times 2$ matrix $N = \begin{bmatrix}p & q \\ r & s\end{bmatrix}$, the determinant is: $$\det(N) = ps - rq$$ 3. **Calculate the determinant of the left $3 \times 3$ matrix:** Let $$a=1, b=8, c=1, d=-1, e=5, f=4, g=3, h=3x+8, i=-4$$ Then $$\det = 1 \times (5 \times (-4) - 4 \times (3x+8)) - 8 \times (-1 \times (-4) - 4 \times 3) + 1 \times (-1 \times (3x+8) - 5 \times 3)$$ Calculate each term: - $5 \times (-4) = -20$ - $4 \times (3x+8) = 12x + 32$ - $-1 \times (-4) = 4$ - $4 \times 3 = 12$ - $-1 \times (3x+8) = -3x - 8$ - $5 \times 3 = 15$ So, $$\det = 1 \times (-20 - (12x + 32)) - 8 \times (4 - 12) + 1 \times (-3x - 8 - 15)$$ Simplify inside parentheses: $$= 1 \times (-20 - 12x - 32) - 8 \times (-8) + 1 \times (-3x - 23)$$ $$= (-52 - 12x) + 64 - 3x - 23$$ Combine like terms: $$(-52 + 64 - 23) + (-12x - 3x) = (-11) - 15x$$ 4. **Calculate the determinant of the right $2 \times 2$ matrix:** $$\det = 7 \times 1 - (3x + 6) \times 1 = 7 - 3x - 6 = 1 - 3x$$ 5. **Set the determinants equal and solve for $x$:** $$-11 - 15x = 1 - 3x$$ Bring all terms to one side: $$-11 - 15x - 1 + 3x = 0$$ $$-12 - 12x = 0$$ $$-12x = 12$$ $$x = -1$$ **Final answer:** $$\boxed{x = -1}$$