Let $x be the amount of money that a man invested in 3% account and $y be the amount of money a man invested at 4% account. The problem can be modelled by the system of two equations.
1. The income for the 1st investment is $0.03x and the income for the 2nd investment is $0.04y.
If his total income for the two investments is $194, then
0.03x+0.04y=194.
2. If a man invests $5,200, part at 4% and the balance at 3%, then
x+y=5,200.
3. You get a system of equations:
[tex]\left\{\begin{array}{l}x+y=5,200\\0.03x+0.04y=194\end{array}\right.[/tex]
From the 1st equation express x and substitute it into the 2nd equation:
[tex]0.03(5,200-y)+0.04y=194,\\ \\156-0.03y+0.04y=194,\\ \\0.01y=38,\\ \\y=\$3,800.[/tex]
Then
[tex]x=\$5,200-\$3,800=\$1,400.[/tex]
Answer: $1,400 at 3% and $3,800 at 4%.
