1. **State the problem:** We are given the system of linear equations: $$2x + 3y = 5$$ $$3y = -2x + 8$$ We need to use substitution to determine how many solutions this system has and find the solution if there is exactly one. 2. **Rewrite the second equation:** The second equation is already solved for $3y$: $$3y = -2x + 8$$ Divide both sides by 3 to solve for $y$: $$y = \frac{-2x + 8}{3}$$ 3. **Substitute $y$ into the first equation:** Replace $y$ in the first equation with the expression from step 2: $$2x + 3\left(\frac{-2x + 8}{3}\right) = 5$$ 4. **Simplify the substitution:** $$2x + \cancel{3} \times \frac{-2x + 8}{\cancel{3}} = 5$$ This simplifies to: $$2x + (-2x + 8) = 5$$ 5. **Combine like terms:** $$2x - 2x + 8 = 5$$ $$0 + 8 = 5$$ $$8 = 5$$ 6. **Analyze the result:** The statement $8 = 5$ is false, which means the system has no solution. **Conclusion:** The system of equations has **no solution** because the lines are parallel and never intersect.