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A college student borrowed $50,000 to pay for tuition, room, and board. Some of the money was borrowed at 4%, some at 6%, and some at 8%. How much (in dollars) was borrowed at each simple interest rate, given that the annual interest was $3050 and the amount at 8% was three times the amount at 6%?

4% $
6% $
8% $

[tex]\boxed{\begin{minipage}{7 cm}\underline{Simple Interest Formula}\\\\$ I = Prt$\\\\where:\\\\ \phantom{ww}$\bullet$ $I =$ interest accrued \\ \phantom{ww}$\bullet$ $P =$ principal amount \\ \phantom{ww}$\bullet$ $r =$ interest rate (in decimal form) \\ \phantom{ww}$\bullet$ $t =$ time (in years) \\ \end{minipage}}[/tex]

Let:

P₁ be the amount borrowed at 4%
P₂ be the amount borrowed at 6%
P₃ be the amount borrowed at 8%

Given:

P₁ + P₂ + P₃ = $50,000
P₃ = 3 P₂

Therefore:

⇒ P₁ + P₂ + P₃ = 50000

⇒ P₁ + P₂ + 3 P₂ = 50000

⇒ P₁ + 4 P₂ = 50000

⇒ P₁ = 50000 - 4 P₂

Let:

r₁ = 4% = 0.04
r₂ = 6% = 0.06
r₃ = 8% = 0.08

Using the simple interest formula:

⇒ I₁ = 0.04 P₁

⇒ I₂ = 0.06 P₂

⇒ I₃ = 0.08 P₃

Given:

I₁ + I₂ + I₃ = $3,050
P₃ = 3 P₂

⇒ I₁ + I₂ + I₃ = 3050

⇒ 0.04 P₁ + 0.06 P₂ + 0.08 P₃ = 3050

⇒ 0.04 P₁ + 0.06 P₂ + 0.08 · 3 P₂ = 3050

⇒ 0.04 P₁ + 0.3 P₂ = 3050

Now we have created a system of equations:

[tex]\begin{cases}\sf P_1= 50000 - 4\: P_2\\ \sf 0.04\: P_1 + 0.3\: P_2 = 3050\end{cases}[/tex]

Substitute the first equation into the second equation and solve for P₂:

⇒ 0.04 (50000 - 4 P₂) + 0.3 P₂ = 3050

⇒ 2000 - 0.16 P₂ + 0.3 P₂ = 3050

⇒ 0.14 P₂ = 1050

⇒ P₂ = 7500

Substitute the found value of P₂ into the first equation and solve for P₁:

⇒ P₁ = 50000 - 4 (7500)

⇒ P₁ = 50000 - 30000

⇒ P₁ = 20000

Subtract the found values of P₁ and P₂ from $50,000 to find the value of P₃:

⇒ P₃ = 50000 - P₁ - P₂

⇒ P₃ = 50000 - 20000 - 7500

⇒ P₃ = 22500