Question

Maggy started a savings account in March of 2003. On March 2007, she had $4,200. On March 2015, she had $10,400. If Maggy's saving is modeled by a linear function, what was her initial deposit?

A) $1,100
B) $1,300
C) $1,500
D) $700

Answer 1

A) $1,100

Explanation:

Let S be savings and t he number of years

S = mt + c

In 2007, t = 4 and S = 4200

4200 = 4m + c (1)

In 2015, t = 12 and S = 10400

10400 = 12m + c (2)

Equation (2) - Equation (1)

6200 = 8m

m = 6200/8

m = 775

(1) 4200 = 4m + c

4200 = 4(775) + c

4200 = 3100 + c

c = 4200 - 3100

c = 1100

S = 775t + 1100

Initial deposit is when t = 0,

S = $1100

Answer 2

1,100

Explanation:

First find the slope of the linear equation...

slope=m=(y2-y1)/(x2-x1)

m=(10400-4200)/(2015-2007)=6200/8=775 so we so far have a line of:

s=775(y-2003)+b, using either original point we can solve for b, the y-intercept which is the initial value as well...I'll use point (2007, 4200) and we get:

4200=775(2007-2003)+b

4200=3100+b

b=$1100

s(y)=775(y-2003)+1100

The initial deposit was $1100.00.