Question

Question #71843

Answer 1

`"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->eq.1`
   `0.14x+0.10y=890->eq.2`

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)`

3. 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)`

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)`

Answer 2

see a solution process below...

Explanation:

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..