Question

To buy a car for $42,000, Stacy uses her credit card which charges 20% interest and borrows therest from an online bank that charges 4.5% interest. If she pays $2510 in interest the first year,how much did she put on the credit card?

Answer 1

Let x be the interest on the first bank with 20% interest and y be the amount of the interest on the onilne bank with 4.5% interest.

Thus, the total amount on the banks can be represented as follows:

`x+y=42000`

On the other hand, the total interest can also be represented as follows:

`0.2x+0.045y=2510`

Using the first equation, solve for the value of x.

`x=42000-y`

Substitute the obtained value of x into the second equation.

`0.2(42000-y)+0.045y=2510`

Simplify the left side of the equation. Distribute 0.2 to 42000-y and then combine like terms.

`\begin{gathered} 8400-0.2y+0.045y=2510 \\ 8400-0.155y=2510 \end{gathered}`

Add 0.155y and subtract 2510 on both sides of the equation.

`\begin{gathered} 8400-2510=0.155y \\ 0.155y=8400-2510 \\ 0.155y=5890 \end{gathered}`

Divide both sides of the equation by 0.155.

`\begin{gathered} (0.155y)/(0.155)=(5890)/(0.155) \\ y=38000 \end{gathered}`

To solve for x, substitute the obtained value of x into the first obtained equation.

`\begin{gathered} x+y=42000 \\ x+38000=42000 \end{gathered}`

Subtract 38000 from both sides of the equation.

`\begin{gathered} x=42000-38000 \\ x=4000 \end{gathered}`

Thus, she put $4000 on the credit card which charges 20% interest and put $38000 on the credit card which charges 4.5%.