### Video Transcript

To calculate the remaining balance on loan, you can use the formula π
π΅ equals ππ΅ times one plus π to the π power minus π times one plus π to the π power minus one over π, where π
π΅ is the remaining balance, ππ΅ is the original balance, π is the payment, and π is the rate of payment and π is the number of payments. Which of the following gives a formula for π in terms of π
π΅, ππ΅, π, and π?

To get π in terms of π
π΅, ππ΅, π, and π, weβll need to isolate the π variable. Before we do that, we could notice that a lot is happening in this formula, particularly with these rates. To make our work easier and to help insure we donβt make small sign mistakes or drop numbers, we could let π₯ be equal to one plus π to the π power and let π¦ be equal to one plus π to the π power minus one over π. This allows us to write this equation as π
π΅ equals ππ΅ times π₯ minus π times π¦. After we isolate π, we can plug back in the values for π₯ and π¦. Because π is negative, we can add ππ¦ to both sides. π becomes positive. Now, we have ππ¦ plus π
π΅ equals ππ΅π₯. And so, we subtract π
π΅ from both sides. And weβll have ππ¦ equals ππ΅π₯ minus π
π΅. To get π by itself, we need to divide both sides by π¦. And so, we can say that π equals ππ΅π₯ minus π
π΅ divided by π¦.

At this point, weβll have to plug back in what we know π₯ and π¦ are equal to. Instead of ππ΅π₯, weβll have ππ΅ times one plus π to the π power minus π
π΅ over one plus π to the π power minus one over π. This is pretty messy because we have a fraction in our denominator. If we write it out horizontally, we notice that weβre dividing by a fraction. And to simplify this, we need to multiply by that fractionβs reciprocal. In this case, weβll now be multiplying by π over one plus π to the π power minus one. We multiply π times whatβs already in the numerator. π times ππ΅ one plus π to the π power minus π
π΅. And the denominator stays the same: one plus π to the π power minus one. This is π in terms of π
π΅, ππ΅, π, and π.

If we look at our options, we see that A and D have π as the denominator. So we know that they cannot be correct. Both π΅ and C have the correct denominators. But if we look closely, only one of them is subtracting π
π΅ from ππ΅ times one plus π to the π power. And thatβs option C. Option C has correctly stated π in terms of π
π΅, ππ΅, π, and π.