Question #71843

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Answer 1 · 2018-02-16T06:54:48

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

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;

$x+y=6900->eq.1$
$0.14x+0.10y=890->eq.2$
Now, solve the equations simultaneously. Multiply $eq.1$ 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.

$x+y=6900color(red)]color(red)(-0.14$
$ul(0.14x+0.10y=890)$
$cancel(-0.14x)-0.14y=-966$
$ul(cancel(0.14x)+0.10y=890)$
$0-0.04y=-76$
$(cancel(-0.04)y)/(cancel(-0.04))=(-76)/(-0.04)$
$color(red)(y=1900)$
Now, find the value of $x$ by using $eq.1$ as reflected in $"step" 1$ ; that is,

$x+y=6900->eq. 1$
$where$
$y=1900$
$x+1900=6900$ ; subtract both sides by 1900 to isolate the variable $x$
$x+1900-1900=6900-1900$ ; simplify
$color(blue)(x=5000)$
Therefore, the amount invested in these accounts are:

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

1st account(x)= $$5,000xx0.14=$700$
2nd account(y)= $($1,900xx0.10=$190)/("Interest earned" ...$890)$

Answer 2 · 2018-02-22T10:44:17

see a solution process below...

Let both accounts represent, $x and y$

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..

$x + y = 6900 - - - eqn1$

$14%x + 10%y = 860$

$0.14x + 0.1y = 890 - - - eqn2$

From $eqn1$

$x + y = 6900$

$x = 6900 – y - - - eqn3$

Substituting $x$ into $eqn2$

$0.14x + 0.1y = 890$

$0.14(6900 – y) + 0.1y = 890$

$966 – 0.14y + 0.1y = 890$

$0.04y = 76$

$y = 76/0.04$

$y = 1900$

Substituting the value of $y$ into $eqn3$

$x = 6900 – y$

$x = 6900 - 1900$

$x = 5000$

Hence the amount that is in both accounts are $5000 and 1900$ respectively..