Math problem help please.?

Question

sondra invests $6000. some of the money is invested in stocks paying 6% per year, and some is invested in bonds paying 11% per year. if her total earnings on these investments are $580, how much money was invested in stocks, and how much money was invested in bonds? please show step by step with variables. thanks

Answer 1

Step by Step:

x+y=6000

.11x+.06y=580

x+y=6000

11x+6y=58000

y=-x+6000

11x+6(-x+6000)= 11x-6x+36000=58000

5x=58000-36000

5x=22000

x=22000/5

x= 4400

4400+y=6000

.11(4400)+.06y=580

X= 4400

Y=1600

Therefore, $4400 was invested in 11% interest and 1600 was invested in 6% interest.

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Answer 2

sondra invests $6000.

Money in stocks = X

Money in bonds = The rest, which means 6000-X

X*1,06  +  (6000-X)*1,11 = 6580

1,06X + 6660-1,11X  = 6580

6660 + 1,06X = 6580 + 1,11x

6660 = 6580 + 1,11X - 1,06X

6660 = 6580 + 0,05X

80 = 0.05X

1600 = X

1600 in Stocks, the rest in bonds

Answer 3

* = Mutiplication sign.

:

x+y = 6000$

and after investment and if both investments have the identical time, namely one year, then

x * (6/100) + y * (11/100) = 580 $ >>>

6x + 11y = 58000$ = 6x + 6y + 5y

6 (x+y) + 5y = 58000 $

6 * 6000 + 5y = 58000 >>> 5y = 58000 - 36000

5y = 22000 >>> y = 4400$, and hence we get

x = 6000 - 4400 = 1600$

Answer 4

Amount invested in stocks: x

bonds: 6000 - x

0.06x + 0.11(6000 - x) = 580

-0.05x = -80

x = 1600

6000 - 1600 = 4400

So she invested 1600 in stocks and 4400 in bonds.