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 than3/2 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 $1,400 more than 1/4of the amount Sam invested the first year. If Sam and Sally invested the same total amount at the end of three years, the amount Sam invested the first year is ___$ and the amount Sally invested the last year is ___$.

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]