The sum of $3,000 is deposited into an account paying 10% annually. if $1,206 is withdrawn at the end of years 1 and 2, how much then remains in the account? "

What we have so far: 
INITIAL CASH AMOUNT IN THE BANK: USD3,000
ANNUAL INCREASE OF THE CASH AMOUNT IN THE BANK: 10%
YEARS THE CASH STAYED IN THE BANK: 2 years.
AMOUNT WITHDRAWN AT THE END OF YEAR 1: USD1,206
AMOUNT WITHDRAWN AT THE END OF YEAR 2: USD1,206

First, we need to solve for YEAR 1:
FOR YEAR 1: 
Initial Deposit * Annual Increase Rate = Annual Increase
3,000 * 0.10 = Year 1's Annual Increase
Year 1's  Annual Increase = USD300
∴The YEAR 1'S ANNUAL INCREASE IS USD300.
∴The NEW AMOUNT is now USD3,300.

BUT NOT SO FAST! After the year, you took out USD1,206.
New Amount - USD1,206 = Year 1 Amount
3,300 - 1,206 = Year 1 Amount
Year 1 Amount = USD2094
∴The YEAR 1 AMOUNT which will carry over to YEAR 2 is USD2094.

Now, let us solve for the REMAINING BALANCE.
FOR YEAR 2's Annual Increase: 
YEAR 1 AMOUNT * Annual Increase = Year 2's Annual Increase
2094*0.10 = Year 2's Annual Increase
Year 2's Annual Increase = USD209.4
∴The YEAR 1'S ANNUAL INCREASE IS USD209.4.
∴The NEW AMOUNT is now USD2,303.4.

But you took out USD1,206
USD2,303.4 - USD1,206 = Remaining Balance
Remaining Balance = USD1097.4

∴The Answer is: USD1097.4

0=6x+2y+10 is equivalent to the simplified form: y= -3x - 5• parallel lines always have the same slope, different y intercept knowing this, we can say the slope, or m of the line parallel to y= -3x - 5 is -3.m= -3 hope this could !