Question

Benjamin has $6000 invested in two accounts. One earns 8% interest per year, and the other pays 7.5% interest per year. If his total interest for the year is $472.50, how much is invested at 8%?

Answer 1

0.08x+0.075 (6000-x)=472.5
Solve for x
0.08x+450-0.075x=472.5
0.08x-0.075x=472.5-450
0.005x=22.5
X=22.5/0.005
X=4500 invested at 8%

Answer 2

Answer:

Benjamin invested $4,500 at 8%.

Step-by-step explanation:

We know that Benjamin has invested $6,000 in two different accounts.

Let account 1 earn 8% interest per year be defined as $x$ and account 2 earn 7.5% interest per year as $y$.

The sum of both accounts must be $6,000:

$x+y=6,000$

We also know that at the end of the year Benjamin earned $472.50, which is the sum of the interest he earned of the amount he invested in account 1 and in account 2, with their different interest rates (in this case, it is useful for us to transform the expression of the interest rate by dividing it by 100, so that it can be simplified):

$(0.08x)+(0.075y)=472.5$

We need to express the $y$ in terms of $x$.

From our first expression, we know that:

$y=6,000-x$

So we substitute this value in our second equation and solve it:

$(0.08x)+(0.075(6,000-x))=472.5$

$(0.08x)+((0.075*6,000)-0.075x)=472.5$

$(0.08x)+((0.075*6,000)-0.075x)=472.5$

$0.08x+450-0.075x=472.5$

$0.08x-0.075x=472.5-450$

$0.005x=22.5$

$x=\frac{22.5}{0.005}$

$x=4,500$

This way, we know that the amount that Benjamin invested in the account that earns 8% interest per year is $4,500.