A man invests a total of $9,493in two savings accounts. One account yields 9% simple interest and the other 10% simple interest. He earned a total of $930.80interest for the year. How much was invested in the 9% account?

Answer:$1850Step-by-step explanation:GivenThe person invests a total of $9493Amount invested in 9% Simple interest account =xAmount invested in 10% Simple interest account= 9493-xEarnings in Interest=$930.8The simple interest formula[tex]S.I=\frac{P\times T\times R}{100}[/tex]whereS.I is simple interestP is principle investedT is time periodR is rate of interestThere fore total interest[tex]=930.8= \frac{x\times 1\times 9}{100} +\frac{(9493-x)\times 1\times 10}{100}\\930.8=\frac{9x+94930-10x}{100}\\93080=94930-x\\x=94930-93080\\x=1850[/tex]Therefore x=1850⇒Amount invested in 9% account= $1850