Question 1012655: A checking account is set up with an initial balance of $9,400 and $800 is removed from the account at eh end of each month for rent. Write an inequality whose solutions are the months, m, in which the account balance is greater than $3,000. Write the solution to your equation by identifying all of the solutions.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! i believe the equqation will be 9400 - 800m > 3000
subtract 9400 from both sides of this equation to get -800m > 3000 - 9400
simplify to get -800m > -6400
divide both sides of the equation by -800 to get m < 8.
when you multiply or divide both sides of an inequality by a negative number, the inequality reverses.
you had -800m > -6400 and you divided by -800 which reversed the inequality and got you m < 8.
the balance in the account will be > 3000 when m is < 8.
you can confirm using the original equation.
the original equation is 9400 - 800m > 3000
when m = 8, you get 9400 - 800*8 > 3000 which results in 9400 - 6400 > 3000 which results in 3000 > 3000 which is NOT true because 3000 is equal to 3000.
so m can't be 8, which is ok because the solution says that m had to be less than 8.
you can also look at the remaining balance at the end of each month as shown below:
month balance
0 9400
1 8600
2 7800
3 7000
4 6200
5 5400
6 4600
7 3800
8 3000 *****
month 7 is the last month when the remaining balance is > 3000.
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