Questions on Algebra in Finance answered by real tutors!

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Let = amount invested @ 8%
Let = amount invested @ 6%
Let = amount invested @ 4%
It is given that
(1)
(2) annually
(3)
This is 3 equations and 3 unknowns, so it should
be solvable
Multiply both sides of (2) by
(2)
Multiply both sides of (1) by and
subtract from (2)
(1)
(4)
Since
(4)
(4)



and
(1)
(1)


$10,400 was invested @ 8%
$5,200 was invested @ 6%
$9,400 was invested @ 4%
check:
(1)
(1)

(2)
(2)
(2)

(3)
{3)
(3)
OK

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Let the amounts in three different accounts be x,y,z respectively.
Then x + y + z = 25000 ------------(1)
and annual interest of x is 8%
y is 6%
z is 4%
The amount deposited in the 8% account is twice the amount in the 6% account.
That is x=2y --------------------(2)
the three accounts earns total annual interest of $1520
that gives 8%x + 6%y + 4%z = 1520
8x + 6y + 4z = 152000
4x + 3y + 2z = 76000 --------------(3)
Solving the 3 eqns for x,y,z
plug x=2y in eqns 1 and 3
3y + z = 25000 ----------(4)
11y + 2z = 76000 ----------(5)
subtract eqns (5) and 2* eqn (4)
5y = 26000
y = 5200
x = 10400 from eqn (2)
and z = 9400 from eqn (1)
The money deposited in the 3 accounts is 10400, 5200 and 9400