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Let x be the amount invested at 8% (in dollars).
Then the amount invested at 12% is (10000-x) dollars.
The amount x produces the interest equal 0.08*x.
The amount (10000-x) produces the interest equal 0.12*(10000-x).
Together, these investments produce the total interest of 0.08x + 0.12*(10000-x).
We are given that the total interest is 944 dollars.
It gives you an equation
0.08x + 0.12*(10000-x) = 944.
Simplify and solve it for x:
0.08x + 1200 - 0.12x = 944
-0.04x = 944 - 1200 = -256
x = = 6400.
Answer. $6400 was invested at 8%. The rest (10000-6400) = 3600 dollars was invested at 12%.
Check. 0.08*6400 + 0.12*3600 = 944. ! Correct !
Solved.
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It is a typical and standard problem on investment.
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".
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.