Set up an algebraic equation and then solve. Marty put last...

Set up an algebraic equation and then solve. Marty put last year's bonus of $\$ 2,400$ into two accounts. He invested part in a CD with $2.5 \%$ annual interest and the rest in a money market fund with $1.3 \%$ annual interest. His total interest for the year was $\$ 42.00 .$ How much did he invest in each account?

Set up an algebraic equation and then solve.
Marty put last year's bonus of $\$ 2,400$ into two accounts. He invested part in a CD with $2.5 \%$ annual interest and the rest in a money market fund with $1.3 \%$ annual interest. His total interest for the year was $\$ 42.00 .$ How much did he invest in each account?

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Key Concepts

Linear Equations

Linear equations are mathematical expressions set equal to some value that involve only the first power of the unknown. In the context of word problems, formulating and solving a linear equation allows one to determine the value of unknown variables. This involves standard techniques such as combining like terms, isolating the variable, and applying inverse operations.

Interest Calculation

Understanding interest calculation is key in problems involving investments or loans. This includes grasping how annual interest rates are applied to principal amounts, converting percentage rates into decimals, and setting up expressions that describe how the individual interests contribute to a total interest amount.

Word Problem Translation

This concept involves interpreting a real-world scenario and converting the information provided into an algebraic equation. It is the process of identifying the unknown quantities, assigning variables to them, and translating situational statements into mathematical expressions, which is a crucial step in solving word problems in algebra.

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