Question

Pauline invested $10,000 at two different

Answer 1

Answer:

So for this, I will be doing a system of equations. Let x = money invested at 4% and y = money invested at 6%. We know that x and y will total up to 10k since that's how much he initially invested, so that'll be one of our equations.

Our other equation will represent the total balance he has after investing. We know that x's interest is 4% (since the money is increasing, this would translate to 1.04 in decimal form) and y's interest is 6% (aka 1.06 in decimal form since its increasing) and that the total balance is 10470 (initial balance + interest income).

Step-by-step explanation:

Using all this info, these are our two equations:

X+Y=10000

1.04x+1.06=10470

Next, substitute x in the second equation for (10000 - y) and solve for y as such:

1.04(10000-y) +1.06y =10470

10400 - 1.04y + 1.60y = 10470

10400 + 0.02y = 10470

0.02y =70

y = 3500

Now that we have the value of y, substitute it into either equation to solve for x:

x + 3500 = 10000

x= 6500

1.04x + 1.06(3500)= 10470

1.04x = 6500

Answer

In short:

$6,500 was invested in the 4% rate

$3,500 was invested in the 6% rate