1. **State the problem:** Solve the system of equations by substitution: $$\begin{cases} 5x - 2y + 3z = 6 \\ -4x + 6y - 7z = -3 \\ 3x + 2y - z = 6 \end{cases}$$ 2. **Isolate a variable:** From the third equation, solve for $z$: $$3x + 2y - z = 6 \implies z = 3x + 2y - 6$$ 3. **Substitute $z$ into the first two equations:** First equation: $$5x - 2y + 3(3x + 2y - 6) = 6$$ Simplify: $$5x - 2y + 9x + 6y - 18 = 6$$ $$14x + 4y = 24$$ Divide by 2: $$7x + 2y = 12$$ Second equation: $$-4x + 6y - 7(3x + 2y - 6) = -3$$ Simplify: $$-4x + 6y - 21x - 14y + 42 = -3$$ $$-25x - 8y = -45$$ 4. **Solve the system of two equations:** $$\begin{cases} 7x + 2y = 12 \\ -25x - 8y = -45 \end{cases}$$ Multiply the first equation by 4: $$28x + 8y = 48$$ Add to the second equation: $$(-25x - 8y) + (28x + 8y) = -45 + 48$$ $$3x = 3 \implies x = 1$$ 5. **Find $y$:** Substitute $x=1$ into $7x + 2y = 12$: $$7(1) + 2y = 12 \implies 7 + 2y = 12 \implies 2y = 5 \implies y = \frac{5}{2}$$ 6. **Find $z$:** Substitute $x=1$, $y=\frac{5}{2}$ into $z = 3x + 2y - 6$: $$z = 3(1) + 2\left(\frac{5}{2}\right) - 6 = 3 + 5 - 6 = 2$$ **Final solution:** $$(x, y, z) = \left(1, \frac{5}{2}, 2\right)$$