You invested $5000 between two accounts paying 4% and 9% annual interest, respectively. If the total interest earned for theyear was $350, how much was invested at each rate?$was invested at 4% andwas invested at 9%.​

Accepted Solution

Answer:Part 1) The amount invested at 4% was $2,000Part 2) The amount invested at 9% was $3,000Step-by-step explanation:we know that The simple interest formula is equal to [tex]I=P(rt)[/tex] where I is the Final Interest Value P is the Principal amount of money to be invested r is the rate of interest  t is Number of Time Periods Letx------> the amount invested at 4%5,000-x ----> the amount invested at 9%in this problem we have [tex]t=1\ year\\ P=\$5,000\\I=\$350\\r1=0.04\\r2=0.09[/tex] substitute in the formula above [tex]350=x(0.04*1)+(5,000-x)(0.09*1)[/tex] [tex]350=0.04x+450-0.09x[/tex] [tex]0.05x=450-350[/tex] [tex]0.05x=100[/tex] [tex]x=2,000[/tex] so[tex]5,000-x=5,000-2,000=3,000[/tex] thereforeThe amount invested at 4% was $2,000The amount invested at 9% was $3,000