can you solve these two problems, please?

Question

explain how u got ur answer, show ur work

Liz invested $3000,part at an annual interest rate of 5% and the rest at an annual interest rate of 12%. How much did she invest at each rate ifher total income on the investment for one year is $220?

Andrew invested a total of $15,000 in two local businesses. He received 6% per year on one investment and 8% per year on the other. If the total income from these two investments for one year was $1090, how much did andrew invest in each business?

THANK YOU SO MUCH

Answer 1

1)

220 = .05x + .12(3000 - x)
220 = .05x + 360 - .12x
220 = 360 - .07x
.07x = 360 - 220
.07x = 140
x = 140/.07
x = 2000

Liz invested $2,000 at 5% and $1,000 at 12%.

2)

1090 = .06x + .08(15000 - x)
1090 = .06x + 1200 - .08x
1090 = 1200 - .02x
.02x = 1200 - 1090
.02x = 110
x = 110/.02
x = 5500

Andrew invested $5,500 at 6% and $9,500 at 8%.

Answer 2

x=amount invested at 5%
3000-x = amount invested at 12%
.05x +.12(3000-x) = 220
.05x +360 -.12x =220
-.07x = -140
x = 2000 invested at 5%
3000-x = 1000 invested at 12%

Let x = amount at 6%
15,000-x = amount at 8 %
.06x+.08(15,000-x) = 1090
.06x + 1200 -.08x = 1090
-.02x = - 110
x =5500 invested at 6%
15000-5500 = 9500 invested at 8%

Answer 3

$150 is 5% of $3000
$220 - $150 = $70
$70/.12 = $583 + $3000 = $5383

Answer 4

I am jazzed! This is my first yahoo answer. I will answer the first problem.

0.05a + 0.12b = 220
a+b = 3000
a = 3000 - b
0.05(3000-b) + 0.12b = 220
150 - 0.05b + 0.12b = 220
150 + 0.07b = 220
0.07b = 70
b = 1000
a = 2000

Answer 5

Look at FRO!!