The Hills’ purchased a new Model M car for $39,575. The tax on the vehicle was 3.25% and title and license fees were $210. They were able to get a trade-in of$4,500 on Jackson’s old car. If they financed the remainder at 7.25% for 5 years, what was the monthly payment on the car loan?

[tex]\boxed{\begin{array}{l}\underline{\textsf{Monthly Payment Formula}}\\\\M=P\cdot \dfrac{r\left(1+r\right)^{n}}{\left(1+r\right)^{n}-1}\\\\\textsf{where:}\\\phantom{ww}\bullet\;\;\textsf{$M$ is the monthly payment.}\\\phantom{ww}\bullet\;\;\textsf{$P$ is the principal amount (loan amount).}\\\phantom{ww}\bullet\;\;\textsf{$r$ is the interest rate per month (in decimal form).}\\\phantom{ww}\bullet\;\;\textsf{$n$ is the term of the loan (in months).}\\\end{array}}[/tex]

First, calculate the loan amount (P) by subtracting the trade-in amount from the sum of the purchase price, vehicle tax and fees:

[tex]P=\textsf{Purchase price + vehicle tax + fees $-$ trade-in}[/tex]

[tex]P=39575+39575\cdot 0.0325 + 210-4500[/tex]

[tex]P=39575+1286.1875 + 210-4500[/tex]

[tex]P=36571.1875[/tex]

Given that the Hills financed at 7.25% for 5 years, then:

[tex]P = 36571.1875[/tex]

[tex]r = \dfrac{7.25\%}{12} = \dfrac{0.0725}{12}[/tex]

[tex]n = 5 \;\textsf{years} = 60\;\textsf{months}[/tex]

Substitute these values into the formula and solve for M:

[tex]M=36571.1875\cdot \dfrac{\frac{0.0725}{12}\left(1+\frac{0.0725}{12}\right)^{60}}{\left(1+\frac{0.0725}{12}\right)^{60}-1}[/tex]

[tex]M=36571.1875\cdot \dfrac{0.008671911598...}{0.43535088524...}[/tex]

[tex]M=36571.1875(0.01991936135...)[/tex]

[tex]M=728.4746989...[/tex]

[tex]M=\$728.47[/tex]

Therefore, the monthly payment on the car loan was $728.47.