Question

Use two equations in two variables to solve the application.

Peter invested some money at 6% annual interest, and Martha invested some at 12%. If their combined investment was $4,000 and their combined interest was $360, how much money (in dollars) did Martha invest?

Answer 1

Answer: Let's set up two equations to represent the given information:

Let x be the amount of money Peter invested at 6% interest.

Let y be the amount of money Martha invested at 12% interest.

The first equation represents the total amount of money they invested:

x + y = 4000

The second equation represents the total interest they earned from their investments:

0.06x + 0.12y = 360

Now, we have a system of equations:

x + y = 4000

0.06x + 0.12y = 360

To solve this system of equations, we can use substitution or elimination. Let's use substitution to find the value of y:

From the first equation, we can express x in terms of y:

x = 4000 - y

Now, substitute this value of x into the second equation:

0.06(4000 - y) + 0.12y = 360

Simplify and solve for y:

240 - 0.06y + 0.12y = 360

Combine like terms:

0.06y = 360 - 240

0.06y = 120

Divide both sides by 0.06:

y = 120 / 0.06

y = 2000

So, Martha invested $2000 at 12% interest.