Helene invested a total of $1,750 in two simple-interest bank accounts. One account paid 4% annual interest; the other paid 6% annual interest. The total amount of interest she earned after 1 year was $82. Find the amount invested in each account.

The amount invested in the account paid 4% is $1150

The amount invested in the account paid 6% is $600

Step-by-step explanation:

The rule of the simple interest is I = Prt, where

∵ Helene invested a total of $1,750 in two simple-interest bank accounts

∴ P1 + P2 = 1750 ⇒ (1)

∵ One account paid 4% annual interest

∴ r1 = 4% = 4/100 = 0.04

∵ The other paid 6% annual interest

∴ r2 = 6% = 6/100 = 0.06

∵ The total amount of interest she earned after 1 year was $82

∴ I1 + I2 = 82 ⇒ (2)

∵ t = 1

∵ I1 = P1 × (0.04) × 1

∴ I1 = 0.04 P1

∵ I2 = P2 × (0.06) × 1

∴ I2 = 0.06 P2

→ Substitute them in equation (2)

∴ 0.04 P1 + 0.06 P2 = 82 ⇒ (3)

Now we have a system of equations to solve it

→ Multiply equation (1) by -0.06 to make the coefficients of y equal in

   values and different in signs to eliminate it

∵ -0.06(P1) + -0.06(P2) = -0.06(1750)

∴ -0.06 P1 + -0.06 P2 = -105 ⇒ (4)

→ Add equations (3) and (4)

∵ (0.04 P1 + -0.06 P2) + (0.06 P2 + -0.06 P2) = (82 + -105)

∴ -0.02 P1 + 0 P2 = -23

∴ -0.02 P1 = -23

→ Divide both sides by -0.02

∴ P1 = 1150

→ Substitute the value of P1 in equation (1) to find P2

∵ 1150 + P2 = 1750

→ Subtract both sides by 1150 to find P2

∴ P2 = 600

∴ The amount invested in the account paid 4% is $1150

∴ The amount invested in the account paid 6% is $600

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