Question 71843

Feb 16, 2018

"1st account"=$5,000 "2nd account"=$1,900

Explanation:

1. To answer this, set-up two 2-variable equations to relate the amounts invested in two different interest rates and the amount earned during the first year of its investment; i.e.,

Let:
$x$=the 1st account invested with an interest rate of 14% per year
$y$=the 2nd account invested with an interest rate of 10% per year
$6900=the total amount invested in two accounts $890=the interest earned for the 1st year of operation
so that;

$x + y = 6900 \to e q .1$
$0.14 x + 0.10 y = 890 \to e q .2$

2. Now, solve the equations simultaneously. Multiply $e q .1$ by $- 0.14$ to cancel out the $x$ terms. Add the remaining $y$ terms and the numerical values of both equations. color(red)("{Don't forget to attach the correct sign in the answer"}). Then, isolate $y$ by dividing both sides of the equation by $- 0.04$ as shown below.

x+y=6900color(red)]color(red)(-0.14
$\underline{0.14 x + 0.10 y = 890}$
$\cancel{- 0.14 x} - 0.14 y = - 966$
$\underline{\cancel{0.14 x} + 0.10 y = 890}$
$0 - 0.04 y = - 76$
$\frac{\cancel{- 0.04} y}{\cancel{- 0.04}} = \frac{- 76}{- 0.04}$
$\textcolor{red}{y = 1900}$

3. Now, find the value of $x$ by using $e q .1$ as reflected in $\text{step} 1$; that is,

$x + y = 6900 \to e q . 1$
$w h e r e$
$y = 1900$
$x + 1900 = 6900$; subtract both sides by 1900 to isolate the variable $x$
$x + 1900 - 1900 = 6900 - 1900$; simplify
$\textcolor{b l u e}{x = 5000}$

4. Therefore, the amount invested in these accounts are:

x=$5,000 at 14% interest rate y=$1,900 at 10% interest rate

5. Checking:

1st account(x)=$5,000xx0.14=$700
2nd account(y)=($1,900xx0.10=$190)/("Interest earned" ...$890) Feb 22, 2018 see a solution process below... Explanation: Let both accounts represent, $x \mathmr{and} y$First statement; Total money invested in both accounts; x + y =$6900

Second statement;

End of first year he invested in both accounts; 14%x + 10%y = $860 Solving simultaneously.. $x + y = 6900 - - - e q n 1$14%x + 10%y = 860 $0.14 x + 0.1 y = 890 - - - e q n 2$From $e q n 1$$x + y = 6900$x = 6900 – y - - - eqn3 Substituting $x$into $e q n 2$$0.14 x + 0.1 y = 890$0.14(6900 – y) + 0.1y = 890 966 – 0.14y + 0.1y = 890 $0.04 y = 76$$y = \frac{76}{0.04}$$y = 1900$Substituting the value of $y$into $e q n 3$x = 6900 – y $x = 6900 - 1900$$x = 5000$Hence the amount that is in both accounts are $5000 and 1900\$# respectively..