Question 907917
We don't know the total invested. We know the difference in the two accounts.
We know that the interest for the two accounts is $1830
0.06*x+0.03*y=1830
We know that the account at 6% has $8000 more.
x=8000+y
We substitute for x
0.06*(8000+y)+0.03*y=1830
We multiply out
480+0.06y+0.03*y=1830
We combine like terms.
0.09*y=1350
Isolate y
y=$15000 at 3%
x=8000+y
Calculate x
x=$23000 at 6%
Total invested $23000+$15000=$38000
We check
0.06*23000+0.03*15000=1830
1380+450=1830
1830=1830
Since this statement is TRUE then it is ok