Question

Leroy invested 12,000 into two different bank accounts. One account (x) gives Leroy 1.5% interest yearly and the other account (y) give Leroy 3% interest yearly. At the end of one year Leroy had earned $300 in interest. Which system of equations could be used to solve for how much money Leroy invested in the two accounts?

Answer 1

0.015x+0.03(12000-x)=300
Solve for x
X=4000 at 1.5%

12000-4000=8000 at 3%

Answer 2

Answer: The system of equations is,

$0.15x+0.03y=300$

$x+y=12000$

Step-by-step explanation:

For first account,

Investment = x,

Annual rate of interest = 1.5 %

Time = 1 year

Thus, the total interest by one account,

$I_1=\frac{x\times 1.5\times 1}{100}=0.015x$

Now, for second account,

Investment = y

Annual rate of interest = 1.5%

Time = 1,

Hence, the interest by the second account,

$I_2=\frac{y\times 3\times 1}{100}=0.03y$

Thus, the total interest,

$I=I_1+I_2=0.015 x + 0.03y$

According to the question,

$I=300$

$\implies 0.15x+0.03y = 300$

Now,  total investment = 12000,

⇒ x + y = 12000

Hence, the required system of equations, to solve for how much money Leroy invested in the two accounts is,

$0.015x+0.03y= 300$,

$x+y=12000$