7. Suppose that you open a bank account and after 3 months you have \$200 in the account, and after 7 months you have \$400. Let   x   denote the number of months and   y   denote the amount of money you have in the bank. Find a linear equation which will give you these amounts of money at these times. Use that equation to predict how much money you would have in the bank after 12 months if the balance continued to grow linearly at this rate.

If   x   denotes the number of months and   y   denotes the amount of money at that time, then having \$200 at the end of 3 months corresponds to the ordered pair

(x1, y1) = (3, 200)

and having \$400 after 7 months corresponds to the ordered pair

(x2, y2) = (7, 400)

First we find the slope. We substitute these numbers into the formula for the slope

which gives us

A slope of \$50/mo means that you are putting \$50 per month into the bank. All we need to finish the equation is the y-intercept. For this we can use the point-slope form of the equation.

y - y1 = m(x - x1)

When we substitute our numbers into the equation we get

y - 200 = 50(x -3)

Remove parentheses

y - 200 = 50x -150

y = 50x + 50

We see that the y-intercept is also 50. That means that your bank balance can be explained by a linear model where you opened your account with \$50 and put \$50 in each month thereafter.

We can check that this works. After 3 months our equation would say

y = 50(3) + 50 = 150 + 50 = 200

After 7 months we would get

y = 50(7) + 50 = 350 + 50 = 400

and it checks.

Since there is only one linear equation which will agree with the two pieces of information, this must be the one. We can use this equation to predict how much money you would have in the bank after any number of months. The problem asks us to predict how much money would have after 12 months. Substitute a 12 for the   x   in the equation and we get

y = 50(12) + 50 = 600 + 50 = 650