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kondaur [170]
3 years ago
8

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

Mathematics
1 answer:
kramer3 years ago
5 0
<h2>Steps</h2>

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 <em>increasing,</em> this would translate to 1.04 in decimal form) and y's interest is 6% (aka 1.06 in decimal form since its <em>increasing</em>) 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

<h2>Answer</h2>

In short:

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