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A certain amount of money is invested at 6% per year. A second amount is Php5000 larger than the first is invested at 8% per year.
The interest from the investment at the higher rate exceeds the income from the lower investment by Php500. Find the investment at each rate.
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Interest - interest = 500.
0.08*(x+5000) - 0.06x = 500 (<<<---=== x = the smaller amount at 6%)
0.08x + 400 - 0.06x = 500
0.02x = 500 - 400 = 100 + x = = = 5000.
Answer. Php 5000 was invested at 6% and Php 10000 was invested at 8%.
Check. 0.08*10000 - 0.06*5000 = 500 Php. ! Correct !
Solved.
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Any other answer is I N C O R R E C T.
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To see many other similar solved problems on investment, look into the lesson
- Using systems of equations to solve problems on investment
in this site.
You will find there different approaches (using one equation or a system of two equations in two unknowns), as well as
different methods of solution to the equations (Substitution, Elimination).
Also, you have this free of charge online textbook in ALGEBRA-I in this site
- ALGEBRA-I - YOUR ONLINE TEXTBOOK.
The referred lesson is the part of this online textbook under the topic "Systems of two linear equations in two unknowns".
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Free of charge online textbook in ALGEBRA-I
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