# Question #71843

##### 2 Answers

#### Explanation:

- 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)#

see a solution process below...

#### Explanation:

Let both accounts represent,

First statement;

Total money invested in both accounts;

Second statement;

End of first year he invested in both accounts;

Solving simultaneously..

From

Substituting

Substituting the value of

Hence the amount that is in both accounts are