Answer:
The amount Sam invested the first year = $2000
The amount Sally invested the last year = $1900
Complete question related to this was found at brainly (ID 4527784):
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 5/2 times the amount he invested the first year. The third year, he invested $1,000 more than 1/5 of the amount he invested the first year.
During the same three years, Sally also invested some money at the start of every year. The first year, she invested $1,000 less than 3/2 times the amount Sam invested the first year. The second year, she invested $1,500 less than 2 times the amount Sam invested the first year. The third year, she invested $1,400 more than 1/4 of the amount Sam invested the first year.
If Sam and Sally invested the same total amount at the end of three years, the amount Sam invested the first year is $ and the amount Sally invested the last year is $ .
Step-by-step explanation:
First we would represent the information given with mathematical expressions.
Sam investment for 3 consecutive years:
Year 1 = x dollars
Year 2 = $2,000 less than 5/2 times the amount he invested the first year
Year 2 = (5/2)(x) - 2000
Year 3 = $1,000 more than 1/5 of the amount he invested the first year
Year 3 = (1/5)(x) + 1000
Sally investment for 3 consecutive years:
Year 1 = $1,000 less than 3/2 times the amount Sam invested the first year
Year 1 = (3/2)(x) - 1000
Year 2 = $1,500 less than 2 times the amount Sam invested the first year
Year 2 = 2x - 1500
Year 3 = $1,400 more than 1/4 of the amount Sam invested the first year.
Year 3 = (1/4)(x) + 1400
Since Sam and Sally invested the same total amount at the end of three years, we would equate their sum:
Sum of Sam investment for the 3years = x + (5/2)(x) - 2000 + (1/5)(x) + 1000
= x + 5x/2 -2000 + x/5 + 1000
= (10x+25x+2x)/10 - 1000
= 37x/10 - 1000
Sum of Sally investment for the 3years = (3/2)(x) - 1000 + 2x - 1500 + (1/4)(x) + 1400
= 3x/2 - 1000 + 2x -1500 + x/4 + 1400
= (6x+8x+x)/4 - 1100
= 15x/4 - 1100
37x/10 - 1000 = 15x/4 - 1100
37x/10 - 15x/4 = -100
(148x - 150x)/40 = -100
-2x = -4000
x = 2000
Therefore the amount Sam invested the first year = x = $2000
The amount Sally invested the last year (3rd year) = (1/4)(x) + 1400
(1/4)(2000) + 1400 = 500+1400 = 1900
The amount Sally invested the last year = $1900
