You invested $7000 between two accounts paying 4% and 7% annual interest, respectively. If the total interest earned for the year was $400, how much was invested at each rate?

Question

Those are all the details

Answer 1

Hi Luis,

so let the two accounts be X and Y. You have two variables so you need two equations.

First we know that .04X + .07Y = 400

The second thing we can deduce is that the *net* interest rate is 400/7000 = 2/35 (~0.0571428),

so your second equation is

2/35X + 2/35Y = 400

So now you have a system of linear equations and can solve as usual. Can you do the math from here? If not I can walk you through it, step by step. (Answer hint: you can divide both accounts by $1000 and have no cents).

Answer 2

Hi Luis

$3000 is invested at 4%

$4000 is invested at 7%

A system equations using the Substitution Method fits this situation well

You are given a total principal and total interest earned for the year

Let

x = amount invested at 4%

y = amount invested at 7%

You have two equations that you can use

Equation 1 for the total principal

x + y = 7000

Equation 2 for the total interest earned, remember to write the rates as decimals

.04x + .07y = 400

A substitution can easily be set up from Equation 1

x + y = 7000

y = 7000 - x

We plug this substitution into Equation 2, putting everything in terms of x

Equation 2

.04 x + .07y = 400

.04x + .07(7000 - x) = 400

Multiply

.04x + 490 - .07x = 400

Combine like terms

490 - .03x = 400

Subtract 490 from both sides of the equation.

-.03x = 400 - 490

-.03x = -90

Divide both sides by -.03 to solve for x

x = -90/-.03x

Remember a negative divided by a negative is a positive

x = 3000

We already know from our substitution set up using Equation 1

y = 7000 -x

y = 7000 - 3000

y = 4000

We can check it all in both Equations

Equation 1

x + y = 7000

3000 + 4000 = 7000

Equation 2

.04x + .07y = 400

.04(3000) + .07(4000) = 400

120 + 280 = 400

I hope you find this useful and if you have any questions send me a message.

You can also try using the Elimination Method on the System Equations as well, give it a try.