Problem 27 - Intermediate Algebra

Question

For Problems \(27-40\), solve each problem by setting up and solving an appropriate system of equations. (Objective 2 ) Doris invested some money at \(7 \%\) and some money at \(8 \%\). She invested \(\$ 6000\) more at \(8 \%\) than she did at \(7 \%\). Her total yearly interest from the two investments was \(\$ 780\). How much did Doris invest at each rate?

Step-by-Step Solution

Answer

Doris invested \( \$2000 \) at \(7\%\) and \( \$8000 \) at \(8\%\).

Step 1: Define Variables

Let \( x \) be the amount Doris invested at \(7\%\) and \( y \) be the amount Doris invested at \(8\%\).

Step 2: Set Up Equations

Using the problem information, we set up the system of equations. From the statement "She invested \( \\(6000 \) more at \( 8\% \) than she did at \( 7\% \)", we get the equation \( y = x + 6000 \). From "Her total yearly interest was \( \\)780 \)", calculated by \( 0.07x + 0.08y = 780 \).

Step 3: Substitute and Simplify

Substitute \( y = x + 6000 \) into the interest equation: \[ 0.07x + 0.08(x + 6000) = 780 \]Simplify this to:\[ 0.07x + 0.08x + 480 = 780 \]\[ 0.15x + 480 = 780 \].

Step 4: Solve for x

Subtract \( 480 \) from both sides to find:\[ 0.15x = 300 \]Divide both sides by \( 0.15 \):\[ x = \frac{300}{0.15} = 2000 \].

Step 5: Solve for y

Now that we know \( x = 2000 \), substitute back into \( y = x + 6000 \):\[ y = 2000 + 6000 = 8000 \].

Key Concepts

Investment problems

Investment problems often involve determining how much money is allocated to different financial vehicles to achieve a goal, such as maximizing returns.
In problems like the one Doris faces, different sums are invested at varying interest rates.
One common objective is to figure out the specific amounts invested in each fund that lead to a desired outcome, such as a particular total interest or return.

• These problems typically involve several variables representing different investment schemes.
• A clear plan to separate the investment amount into different parts is crucial.
• Understanding how these parts contribute to overall interest or return is important.

Conceptually, approaching investment problems requires breaking them down systematically.
We often use systems of equations to model the situation and then proceed by solving these equations.

Interest calculation

Interest calculation is key in understanding investment problems.
The interest for any given investment is determined by the formula: interest = principal × rate × time.
In Doris's case, since we are dealing with yearly interest, we simplify it to: interest = principal × rate.

• The principal is the amount invested.
• The rate is the percentage of interest per annum.

Adding up the interests from all investments gives us the total interest.
For Doris, the interest from both investments adds up to $780.
It is broken down into $0.07x from the 7% investment and $0.08y from the 8% investment, where these represent the interest contributions from each respective investment.

Algebraic problem solving

Solving investment problems involves arithmetic and algebraic skills, particularly with systems of equations.
The goal is to find unknown variables by setting up and solving equations derived from problem conditions.

• Create equations based on the problem description, like total investment and goals.
• Substitution or elimination methods often help solve these equations.

In Doris's problem, we set up two equations: one for the difference in investment amounts and another for the total interest.
By substituting one variable in terms of the other, we arrive at a single variable equation that we can then solve.

Variable substitution

Variable substitution is a strategic method used to solve systems of equations.
It involves solving one equation for one variable and then substituting it into another equation.
This reduces the system down to one equation with a single variable.

• Choose an equation that is simple and straightforward to solve for one variable.
• Substitute this variable's expression into the other equation.

This technique simplifies the process.
For Doris's situation, after setting up the equations, we expressed y in terms of x.
This expression was then substituted into the total interest equation, allowing us to solve for x.
Subsequently, we found y using the relationship between x and y, simplifying the entire solving process.