Question 1111648
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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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<pre>
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 = {{{100/0.02}}} = {{{10000/2}}} = 5000.


<U>Answer</U>.  Php 5000 was invested at 6% and  Php 10000 was invested at 8%.


<U>Check</U>.   0.08*10000 - 0.06*5000 = 500 Php.   ! Correct !
</pre>


Solved.


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Any other answer is &nbsp;&nbsp;<U>I N C O R R E C T</U>.



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To see many other similar solved problems on investment, &nbsp;look into the lesson

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/coordinate/lessons/Using-systems-of-equations-to-solve-problems-on-investment.lesson>Using systems of equations to solve problems on investment</A>

in this site.


You will find there different approaches &nbsp;(using one equation or a system of two equations in two unknowns), &nbsp;as well as 
different methods of solution to the equations &nbsp;(Substitution, &nbsp;Elimination).


Also, &nbsp;you have this free of charge online textbook in ALGEBRA-I in this site

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson>ALGEBRA-I - YOUR ONLINE TEXTBOOK</A>.


The referred lesson is the part of this online textbook under the topic &nbsp;"<U>Systems of two linear equations in two unknowns</U>".



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.