A student has money in three accounts that pay 5%, 7%, and 8% in annual interest. She has three times as much invested at 8% as she does a 5%. If the total amount she has invested is $1600 and her interest for the year comes to $115, how much money does she have in each account?
a. Set up a linear system for the amount invested in each account.
b. Solve the System.
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Define the variables
x: the amount in the 5% interest account
y: the amount in the 7% interest account
z: the amount in the 8% interest account
She invested a total of $1600 and split it up among the 3 accounts, so you have this equation
x + y + z = 1600 [this says that the sum of money in all 3 accounts totals the $1600 she invested]
To find simple interest use the formula I = PRT where P: Principal ; R: Interest Rate; T: Time
For each account you have
I = PRT = (x)(0.05)(1 year) = .05x [for the 5% account]
I = PRT = (y)(0.07)(1 year) = .07y [for the 7% account]
I = PRT = (z)(0.08)(1 year) = .08z [for the 8% account]
The interest adds up to $115 across the 3 accounts so
.05x + .07y + .08z = 115
We also know that "She has three times as much invested at 8% as she does a 5%" so
z = 3x
Solve the system
x + y + z = 1600
.05x + .07y + .08z = 115
z = 3x
I get x = $300; y = $400; and z = $900
2 approaches to sparkling up: #a million: removal x+2y=0 2x-6y=5 Multiply equation a million by using 3 3x+6y=0 2x-6y=5 -----------upload 5x=5 x=a million replace a million for x into the two of the two unique equations. x+2y=0 a million+2y=0 2y=-a million y= -a million/2 #2: Substitution x+2y=0 2x-6y=5 From Equation No.a million, x=-2y replace x with -2y in Equation No.2 2(-2y)-6y=5 -4y-6y=5 -10y=5 y= -a million/2 As till now, replace y with -a million/2 in the two unique equation. 2x-6y=5 2x-6(-a million/2)=5 2x+3=5 2x=2 x=a million Take your p.c..!