Set up a system to solve $15,000 is invested, part at 5% the rest at 6.2%. If $8

Question

Set up a system to solve

$15,000 is invested, part at 5% the rest at 6.2%. If $855is the interest earned at the end of one year, how much wasinvested at each rate?

I had already set up the system to solve. I just need helpin solving it please. Below is the step I took in setting up thesystem.

Variables to watch for:
x = Amount Invested at 5%
(15000-x) = Amount Invested at 6.2%   <----Weinvest the rest of the 15000
The idea of interest is "rate * investment = interest". Sincewe have two rates and two investments, our equation is modified to:"rate1 * investment1 + rate2 * investment2 = interest"  Our set up then becomes:
0.05 * x + 0.062 * (15000 - x) = 855.

Explanation / Answer

Ok, so just to tell you, normally you have to take into account thetime and interest per time in these types of things, but since itis one year and the interest is in units of per year, it 'cancels'out. so: 0.05 * x + 0.062 * (15000 - x) = 855. distribute the 0.062 * (15000 - x) to get: (.062*15000) - (.062 * x) =930 + .062X so now you have: .05x +930-.062X=855 subtract 930 from both sides to get .05x-.062x=-75 -.012x=-75 .012x=75 x=75/.012 x=6250

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.