SOLUTION: Last year, Tony had $20,000 to invest. He invested some of it in an account that paid 5% simple interest per year, and he invested the rest in an account that paid 6% simple

Question 1049016: Last year, Tony had $20,000 to invest. He invested some of it in an account that paid 5% simple interest per year, and he invested the rest in an account that paid 6% simple interest per year. After one year, he received a total of $1170 in interest. How much did he invest in each account?
Found 2 solutions by jorel555, advanced_Learner:
Answer by jorel555(1290) (Show Source): You can put this solution on YOUR website!
Let n be the amount invested at 6%. Then:
.06n+.05(20000-n)=1170
.01n+1000=1170
.01n=170
n=17000
$17000 was invested at 6%, $3000 was invested at 5%. ☺☺☺☺
Answer by advanced_Learner(501) (Show Source): You can put this solution on YOUR website!
let stand for amount invested at then paid
+=

+()=
+()=
+()=
+()=
=
=-
=-
=
=
you solve for
at and
which means at

Solved by pluggable solver: EXPLAIN simplification of an expression

Your Result:

YOUR ANSWER

This is an equation! Solutions: a=3000.

Graphical form: Equation was fully solved.
Text form: 5a+120000-6a=117000 simplifies to 0=0
Cartoon (animation) form:
For tutors: simplify_cartoon( 5a+120000-6a=117000 )

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DETAILED EXPLANATION

Look at .
Moved to the right of expression
It becomes .

Look at .
Eliminated similar terms , replacing them with
It becomes .

Look at .
Added fractions or integers together
It becomes .

Look at .
Removed extra sign in front of
It becomes .

Look at .
Remove unneeded parentheses around factor
It becomes .

Look at .
Remove extraneous '1' from product
It becomes .

Look at .
Moved these terms to the left
It becomes .

Look at .
Added fractions or integers together
It becomes .

Look at .
Solved linear equation equivalent to -a+3000 =0
It becomes .
Result:
This is an equation! Solutions: a=3000.

Universal Simplifier and Solver

Done!

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