1. Problem statement: A borrower takes a mortgage with principal $P=500000$ at an annual interest rate of 8% for 20 years and pays monthly; find the monthly payment and the remaining balance after 10 years as an example. 2. Formula and important rules: We use the standard annuity formula for the monthly payment. $$M = P \cdot \dfrac{r(1+r)^n}{(1+r)^n - 1}$$ Here $r$ is the monthly interest rate and $n$ is the total number of monthly payments. Important rules: use the decimal monthly rate $r=\text{annual rate}/12$ and convert years to months via $n=\text{years}\cdot 12$. 3. Intermediate values and monthly-payment computation: Monthly rate $r = 0.08 / 12 = 0.006666666666666667$. Number of payments $n = 20 \cdot 12 = 240$. Compute $(1+r)^n \approx 4.926897$. Substitute into the formula and show the arithmetic steps. $$M = 500000 \cdot \frac{0.006666666666666667 \cdot 4.926897}{4.926897 - 1} \approx 4183.89$$ This gives the monthly payment $M \approx 4183.89$. 4. Remaining balance formula and intermediate work: The outstanding balance after $k$ payments is given by $$B_k = P(1+r)^k - M \cdot \frac{(1+r)^k - 1}{r}$$ For a 10-year example take $k = 10 \cdot 12 = 120$. Compute $(1+r)^{120} \approx 2.219241$. Compute $P(1+r)^{120} = 500000 \cdot 2.219241 \approx 1109620.50$. Compute $\dfrac{(1+r)^{120} - 1}{r} = \dfrac{2.219241 - 1}{0.006666666666666667} \approx 182.88615$. Compute $M \cdot 182.88615 \approx 4183.89 \cdot 182.88615 \approx 764523.60$. Subtract to obtain the remaining balance after 10 years. $$B_{120} \approx 1109620.50 - 764523.60 \approx 345096.90$$ 5. Explanation in learner-friendly terms: The monthly payment formula spreads the loan principal plus interest into equal monthly payments using the monthly rate $r$ and total months $n$. The remaining-balance formula computes how much principal is left after $k$ payments by taking the future value of the original principal and subtracting the future value of the payments made. 6. Final answers: Monthly payment $M \approx 4183.89$. Remaining balance after 10 years $B_{120} \approx 345096.90$. If you want the remaining balance at a different time $k$, replace $k$ in the formula for $B_k$ above and recompute.