Type the correct answer in each box. For three consecutive years, Sam invested some money at the start of the year. The first year, he invested x dollars. The second year, he invested $2,000 less than times the amount he invested the first year. The third year, he invested $1,000 more than of the amount he invested the first year.

Step-by-step explanation:

The question is incomplete. Find the complete question in the comment section

For Sam:

Initial investment = x

Second year = 5/2x - 2000

Third year = 1/5x + 1000

Total investment = x + 5/2x-2000+1/5x+1000

Total investment = x+5/2x +1/5x-1000

Total investment = (10x+25x+2x)/10 - 1000

Total investment = 37x/10 - 1000

For Sally;

First year = 3/2x - 1000

Second year = 2x-1500

Third year = 1/4 x +1400

Total investment by sally = (3/2x + 2x + 1/4 x) +(-1000-1500+1400)

Total investment by sally = (6x+8x+x)/4 -1100

Total investment = 15x/4 - 1100

If Sam and Sally invested the same total amount at the end of three years then:

37x/10 - 1000 = 15x/4 - 1100

Calculate x;

37x/10-15x/4 = -1100+1000

(148x-150x)/40 = -100

-2x = -4000

x = 4000/2

x = 2000

Since Sam initial investment is x hence Sam investment in his first year is $2000

Sally investment in his last year = 1/4 x + 1400

= 1/4(2000)+1400

= 500+1400

= 1900

Hence sally investment in her last year is $1900

the answer is 13 cm