Solve the problem.

-An investment is worth $3654 in 1993. By 1997 it has grown to $5394. Let y be the value of the investment in the year x, where x = 0 represents 1993. Write a linear equation that models the
Value of the investment (y) in the year x.
A) $y = \frac { 1 } { 435 } x + 3654$
B) $y = - 435 x + 7134$
C) $y = - 435 x + 3654$
D) $y = 435 x + 3654$

Question

Solve the problem.

-An investment is worth $3654 in 1993. By 1997 it has grown to $5394. Let y be the value of the investment in the year x, where x = 0 represents 1993. Write a linear equation that models the
Value of the investment (y) in the year x. 
A) $y = \frac { 1 } { 435 } x + 3654$
B) $y = - 435 x + 7134$
C) $y = - 435 x + 3654$
D) $y = 435 x + 3654$

Quizplus · Accepted Answer

The investment grows from $3654 in 1993 (x=0) to $5394 in 1997 (x=4). The increase is $5394 - $3654 = $1740 over 4 years. The rate of increase per year is $1740 / 4 = $435. Thus, the equation is $y = 435x + 3654$, where $3654 is the initial value in 1993.

Solve the Problem y=1435x+3654y = \frac { 1 } { 435 } x + 3654y=4351​x+3654

Solve the Problem $y = \frac { 1 } { 435 } x + 3654$