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please help me solve this word problem.
Monico invests a total of $22,500 in two accounts paying 8% and 6% annual interest, respectively.
How much was invested in each account if, after one year, the total interest was $1,420.00
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Let x be the amount invested at 8%, in dollars.
Then the amount invested at 6% is (22500-x) dollars.
The annual interest produced by the 8% account is 0.08x dollars.
The annual interest produced by the 6% account is 0.06*(22500-x) dollars.
The total annual interest is the sum 0.08x + 0.06*(22500-x) dollars.
Write an equation for the total annual interest
0.08x + 0.06*(22500-x) = 1420 dollars.
Simplify this equation and find x
0.08x + 0.06*22500 - 0.06x = 1420
0.08x - 0.06x = 1420 - 0.06*22500
0.02x = 70
x = 70/0.02
x = 3500.
ANSWER. $3500 was invested at 8% and the rest 22500-3500 = 19000 dollars was invested at 6%.
CHECK. 0.08*3500 + 0.06*19000 = 1420 dollars, total annual interest. ! correct !
Solved, explained and checked.
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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.