Bob invested $12,350, part at 7% interest and part at 6% interest. How much money was invested at the 7% rate if each account earned the same amount of interest money?

$5,700 was invested in the 7% account

Step-by-step explanation:

Let the amount invested in 7% interest account be x while the other account is y

Total amount invested is &12,350

Thus:

x + y = 12,350 ••••(i)

interest on 7% = 7% of x = 7/100 * x = 7x/100

Interest on 6% = 6% of y = 6/100 * y = 6y/100

Since both have same interest;

7x/100 = 6y/100

7x = 6y

y = 7x/6

Put this into equation i

x + 7x/6 = 12,350

Multiply through by 6

6x + 7x = 74,100

13x = 74,100

x = 74,100/13 = $5,700

1e+27

step-by-step explanation:

a) y = -3x +2

b) y = 1/2x -6

these 2 can be combined into one equation.

-3x +2 = .5x -6

8 = 3.5x

x = 8 / 3.5 = 2.2857142857

y = -3 * (8 / 3.5) +2

y = (-24 / 3.5) + 2

y = -4.8571428571

step-by-step explanation: