Answer:
The first year Sam invested $2,000. The third year sally invested $1,900.
Step-by-step explanation:
Let $x be the amount of money Sam invested the first year. The second year, he invested $2,000 less than 5/2 times the amount he invested the first year, then the second year he invested
[tex]\$\left(\dfrac{5}{2}x-2,000\right).[/tex]
The third year, he invested $1,000 more than 1/5 of the amount he invested the first year, then the third year he invested
[tex]\$\left(\dfrac{1}{5}x+1,000\right).[/tex]
During three years Sam invested
[tex]\$\left(x+\dfrac{5}{2}x-2,000+\dfrac{1}{5}x+1,000\right)=\$\left(\dfrac{37}{10}x-1,000\right).[/tex]
The first year, Sally invested $1,000 less than 3/2 times the amount Sam invested the first year, then the first year she invested
[tex]\$\left(\dfrac{3}{2}x-1,000\right).[/tex]
The second year, she invested $1,500 less than 2 times the amount Sam invested the first year, then the second year she invested
[tex]\$(2x-1,500).[/tex]
The third year, she invested $1,400 more than 1/4 of the amount Sam invested the first year, then the third year she invested
[tex]\$\left(\dfrac{1}{4}x+1,400\right).[/tex]
During three years Sally invested
[tex]\$\left(\dfrac{3}{2}x-1,000+2x-1,500+\dfrac{1}{4}x+1,400\right)=\$\left(\dfrac{15}{4}x-1,100\right).[/tex]
If Sam and Sally invested the same total amount at the end of three years, then
[tex]\dfrac{37}{10}x-1,000=\dfrac{15}{4}x-1,100,\\ \\\dfrac{37}{10}x-\dfrac{15}{4}x=1,000-1,100,\\ \\-\dfrac{1}{20}x=-100,\\ \\x=\$2,000.[/tex]
Thus,
[tex]\$\left(\dfrac{1}{4}\cdot 2,000+1,400\right)=\$1,900.[/tex]
