Peter invested some money at 6% annual interest and martha invested some at 12%. If their investments is $6000 and their combined interest was $540. How much money did martha invest?

3000

Let Peter's investment be x

then Martha's investment is 6000-x

solve:

.06x + .12(6000-x) = 540
6x + 12x(6000-x) = 54000

I am sure you can finish it.

To find out how much money Martha invested, we can start by setting up a system of equations based on the given information.

Let's assume that Peter invested x dollars and Martha invested y dollars.

According to the first piece of information, their combined investment is $6000, so we have the equation:

x + y = 6000

Next, we know that Peter's investment grew at an annual interest rate of 6%, and Martha's investment grew at an annual interest rate of 12%. The combined interest earned by both investments is $540.

The interest earned on Peter's investment is 6% of x dollars, which can be expressed as 0.06x. Similarly, Martha's investment earned interest of 12% of y dollars, which can be expressed as 0.12y.

So, our second equation is:

0.06x + 0.12y = $540

Now, we have a system of two equations:

x + y = 6000
0.06x + 0.12y = $540

To solve this system of equations, we can use either substitution or elimination method. In this case, let's use elimination:

Multiply both sides of the first equation by 0.06 to get:

0.06x + 0.06y = 360

Now subtract this equation from the second equation:

0.06x + 0.12y - (0.06x + 0.06y) = $540 - $360

Simplifying, we get:

0.06y = $180

Divide both sides by 0.06:

y = $3000

Therefore, Martha invested $3000.

4000