Question #71843

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)#

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..