Samantha invests a total of $\$ 21,500$ in two accounts. The first account earned a rate of return of $6 \%$ (after a year). However, the second account suffered a $3 \%$ loss in the same time period. At the end of one year, the total amount of money gained was $\$ 345.00$ . How much was invested into each account?

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Grade 7

Percent of change: word problems

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Q. Samantha invests a total of

\$ 21,500

in two accounts. The first account earned a rate of return of

6 \%

(after a year). However, the second account suffered a

3 \%

loss in the same time period. At the end of one year, the total amount of money gained was

\$ 345.00

. How much was invested into each account?

Denote Investments: Let's denote the amount invested in the first account as $x$ and the amount invested in the second account as $y$ . We know that the total investment is $\$21,500$ , so we can write the first equation as: $\newline$ $x + y = 21,500$
Equations Setup: The first account earned a $6$ \% return, so the gain from the first account is $0.06x$ . The second account had a $3$ \% loss, so the loss from the second account is $0.03y$ . The total gain is $\$345$ , so we can write the second equation as: $\newline$ $0.06x - 0.03y = 345$
Substitution Method: Now we have a system of two equations with two variables: $\newline$ $1$ ) $x + y = 21,500$ $\newline$ $2$ ) $0.06x - 0.03y = 345$ $\newline$ We can solve this system using the substitution or elimination method. Let's use the substitution method. We can express $y$ from the first equation as: $\newline$ $y = 21,500 - x$
Solve for $x$ : Substitute $y$ in the second equation with the expression we found in the previous step: $\newline$ $0.06x - 0.03(21,500 - x) = 345$ $\newline$ Now, let's solve for $x$ .
Calculate x: Distribute the $-0.03$ to both terms in the parentheses: $\newline$ $0.06x - 0.03 \times 21,500 + 0.03x = 345$ $\newline$ Combine like terms: $\newline$ $0.06x + 0.03x = 345 + 0.03 \times 21,500$ $\newline$ $0.09x = 345 + 645$ $\newline$ $0.09x = 990$
Find $y$ : Divide both sides by $0.09$ to solve for $x$ :
$x = \frac{990}{0.09}$
$x = 11,000$
So, $\$11,000$ was invested in the first account.
Find y: Divide both sides by $0.09$ to solve for $x$ : $\newline$ $x = \frac{990}{0.09}$ $\newline$ $x = 11,000$ $\newline$ So, $\$11,000$ was invested in the first account.Now, let's find $y$ by substituting $x$ back into the first equation: $\newline$ $y = 21,500 - x$ $\newline$ $y = 21,500 - 11,000$ $\newline$ $y = 10,500$ $\newline$ So, $\$10,500$ was invested in the second account.