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Question 1132311: a certain amount of money is invested at 9% per year. a second amount is Php50,000 larger than the first and is invested at 12% per year. the interest from the investment at the higher rate exceeds the income from the lower investment by Php13,500. find the investment at each rate. (show table if possible and use only 1 variable) (thanks in advance)
Answer by ikleyn(52776)   (Show Source):
You can put this solution on YOUR website! .
Let X be the amount invested at 9%.
Then amount invested at 12% is (x+50000) Php.
Your equation, from the condition, is
interest at 12% - interest at 9% = 13500, or
0.12*(x+50000) - 0.09*x = 13500,
Simplify and find x =  = 250000 Php.
Answer. The amount of Php 250000 was invested at 9%.
The amount of 25000+50000 Php = 300000 Php was invested at 12%.
Check. 0.12*300000 - 0.09*250000 = 13500. ! Exactly correct !
Solved.
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Making a Table for such simple problem is SPENDING YOUR TIME for nothing.
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It is a standard and typical problem on investments.
If you need more details, or if you want to see other similar problems solved by different methods, 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".
Save the link to this online textbook together with its description
Free of charge online textbook in ALGEBRA-I
https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson
to your archive and use it when it is needed.
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