# Video: US-SAT04S3-Q10-976106731342

To calculate the remaining balance on loan, you can use the formula ππ΅ = ππ΅(1 + π)^(π) β π(((1 + π)^(π) β 1)/π), where ππ΅ is the remaining balance, ππ΅ is the original balance, π is the payment, 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΅, ππ΅, π and π? [A] π = (((1 + π)^(π) β 1) ππ΅ β ππ΅(1 + π)^(π))/π [B] π = (π(ππ΅ β ππ΅(1 + π)^(π))/((1 + π)^(π) β 1) [C] π = (π(ππ΅(1 + π)^(π) β ππ΅)/((1 + π)^(π) β 1) [D] π = (((1 + π)^(π) β 1) (ππ΅(1 + π)^(π) β ππ΅))/π

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### Video Transcript

To calculate the remaining balance on loan, you can use the formula ππ΅ equals ππ΅ 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, ππ΅ 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΅, ππ΅, π, and π?

To get π in terms of ππ΅, ππ΅, π, and π, weβll 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 donβt 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 ππ΅ 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 ππ¦ to both sides. π becomes positive. Now, we have ππ¦ plus ππ΅ equals ππ΅π₯. And so, we subtract ππ΅ from both sides. And weβll have ππ¦ equals ππ΅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 ππ΅π₯ minus ππ΅ divided by π¦.

At this point, weβll have to plug back in what we know π₯ and π¦ are equal to. Instead of ππ΅π₯, weβll have ππ΅ 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 weβre dividing by a fraction. And to simplify this, we need to multiply by that fractionβs reciprocal. In this case, weβll now be multiplying by π over one plus π to the π power minus one. We multiply π times whatβs already in the numerator. π times ππ΅ 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΅, ππ΅, 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 ππ΅ times one plus π to the π power. And thatβs option C. Option C has correctly stated π in terms of ππ΅, ππ΅, π, and π.