Question

Pauline invested $10000 at two different rates,4% and6%.If his interest income was $470,how much did he invest in each rate?

Answer 1

Steps

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).

Using all this info, these are our two equations:

$x+y=10000\\1.04x+1.06y=10470$

For this, I will be using the substitution method. For the first equation, subtract y on both sides of the equation:

$x=10000-y\\1.04x+1.06y=10470$

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

$1.04(10000-y)+1.06y=10470\\10400-1.04y+1.06y=10470\\10400+0.02y=10470\\0.02y=70\\y=3500$

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

$x+3500=10000\\x=6500\\\\1.04x+1.06(3500)=10470\\1.04x+3710=10470\\1.04x=6760\\x=6500$

Answer

In short:

• $6,500 was invested in the 4% rate
• $3,500 was invested in the 6% rate