Question

solve the following word problem.
money is invested at two rates of interest. one rate is 8% and the other is 5%. if there is $1400 more invested at 8% than at 5%, find the amount invested at each rate if the total annual interest received is $840. let ( x = ) amount invested at 8% and ( y = ) amount invested at 5%. then the system that models the problem is ( \begin{cases} x = y + 1400 \\ 0.08x + 0.05y = 840 end{cases} ). solve the system by using the method of addition.
answer
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$ \underline{\quad\quad\quad} $ invested at 8%
$ \underline{\quad\quad\quad} $ invested at 5%

Explanation:

Step1: Rewrite the first equation

Rewrite \( x = y + 1400 \) as \( x - y = 1400 \).

Step2: Multiply the second equation

Multiply the equation \( 0.08x + 0.05y = 840 \) by 20 to eliminate decimals: \( 20\times(0.08x + 0.05y)=20\times840 \), which gives \( 1.6x + y = 16800 \).

Step3: Add the two equations

Add \( x - y = 1400 \) and \( 1.6x + y = 16800 \) together: \( (x - y)+(1.6x + y)=1400 + 16800 \). Simplifying the left side: \( x - y+1.6x + y = 2.6x \), and the right side: \( 18200 \). So \( 2.6x = 18200 \).

Step4: Solve for x

Divide both sides by 2.6: \( x=\frac{18200}{2.6}=7000 \).

Step5: Solve for y

Substitute \( x = 7000 \) into \( x = y + 1400 \): \( 7000=y + 1400 \). Subtract 1400 from both sides: \( y=7000 - 1400=5600 \).

Answer:

\$7000 invested at 8%
\$5600 invested at 5%