Answer/Step-by-step explanation:
A man invests $5,200, part at 4% and the balance at 3%. If his total income for the two investments is $194, how much money did he invest at each rate?
Let x = amount ($) invested at 4%
then
5200-x = amount ($) invested at 3%
.04x + .03(5200-x) = 194
.04x + 156-.03x = 194
.01x + 156 = 194
.01x = 38
x = $3800 (amount invested at 4%)
amount invested at 3%:
5200-3800 = $1400
Or
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:
x+y=5,200
0.03x+0.04y=194
From the 1st equation express x and substitute it into the 2nd equation:
0.03(5,200-y)+0.04y=194,
156-0.03y+0.04y=194,
0.01y=38
y=3,800
Then x=5,200-3,800=1400
Answer: $1,400 at 3% and $3,800 at 4%.
