Katie has been withdrawing the same amount of money from her savings account every month. The table below shows the amount of money in her account after different amounts of time. Time (months) 8 10 12 14
Money (dollars) 470 390 310 230
How much money (in dollars) was in her account before Katie started withdrawing money?

Group of answer choices

$510

$790

$550

$710

Explanation:

x = number of months

y = account balance

The first column of the table has (x,y) = (8,470) to mean that on month 8, her balance was $470

Then we have (10, 390) meaning x = 10 and y = 39. As x increased by 2, y dropped by 470-390 = 80 dollars. She withdrew $80 over the course of two months. Therefore, she is withdrawing 80/2 = 40 dollars a month. This is the slope value. It represents the rate of change. We could use the slope formula to find the same result.

Let m = -40 be the slope. It's negative to indicate the value of y goes down as x goes up, or vice versa.

With m = -40, x = 8 and y = 470, we can find the y intercept b like so

y = mx+b

470 = -40*8+b

470 = -320+b

470+320 = b

790 = b

b = 790 is the y intercept, and it represents the initial account balance before Katie started withdrawing money.

In other words, it represents the starting amount of money she has.

As a slight alternative, we could use the point slope form to find the y intercept

y - y1 = m(x - x1)

y - 470 = -40(x - 8)

y = -40(x-8) + 470

y = -40x + 320 + 470

y = -40x + 790

You don't have to use (x,y) = (8,470). You can use any column you want from the table of values. The value of m will be m = -40 the whole time however.

As a way to check to see if we have the correct equation, try plugging in various values of x to see if you get the proper y values as outputs as shown in the table.

For instance, if we tried x = 10, then,

y = -40x + 790

y = -40*10 + 790

y = -400+790

y = 390

That confirms column 2. I'll let you verify the other columns.

answer: 48

just remember to use pemdas:

parentheses

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