Phyllis invested 41000 dollars,

Question

Phyllis invested 41000 dollars, a portion earning a simple interest rate of 4 percent per year and the rest earning a rate of 7 percent per year. After one year the total interest earned on these investments was 1940 dollars. How much money did she invest at each rate?

At rate 4 percent :
At rate 7 percent :

Answer 1

This a problem of two equations in two variables.  The first step in solving the problem is defining the variables.  I will define them as follows:

x = Amount of money invested at 4 percent

y = Amount of money invested at 7 percent

Now we need to write two equations using the two variables.

The amount invested at 4 percent (x) plus (+) the amount invested at 7 percent (y) equals (=) 41,000.

x + y = 41,000             Equation 1

The amount of interest from the money invested at 4 percent (.04x) plus (+) the amount of interest from the money invested at 7 percent (.07y) equals (=) 1940.

.04x + .07y = 1940        Equation 2

You can use either substitution or elimination method to solve.  I am going to use substitution.

x + y = 41,000                              Equation 1

x + y - y = 41,000 - y                    Subtract y from each side

x = 41,000 - y                                Simplify

.04x + .07y = 1940                         Equation 2

.04(41,000 - y) + .07y = 1940          Substitution (value of x from equation 1)

1640 - .04y + .07y = 1940               Distribute the .04

1640 + .03y = 1940                         Simplify

1640 + .03y - 1640 = 1940 - 1640     Subtract 1640 from each side

.03y = 300                                       Simplify

.03y/.03 = 300/.03                            Divide each side by .03

y = 10,000                                        Simplify

x = 41,000 - y                                   Equation 1 solved for x

x = 41,000 - 10,000                          Substitution

x = 31,000                                        Simplify

Phyllis invested $31,000 at 4% and $10,000 at 7%.