Video: US-SAT04S3-Q10-976106731342

To calculate the remaining balance on loan, you can use the formula 𝑅𝐡 = 𝑂𝐡(1 + π‘Ÿ)^(𝑛) βˆ’ 𝑃(((1 + π‘Ÿ)^(𝑛) βˆ’ 1)/π‘Ÿ), where 𝑅𝐡 is the remaining balance, 𝑂𝐡 is the original balance, 𝑃 is the payment, π‘Ÿ is the rate of payment, and 𝑛 is the number of payments. Which of the following gives a formula for 𝑃 in terms of 𝑅𝐡, 𝑂𝐡, π‘Ÿ and 𝑛? [A] 𝑃 = (((1 + π‘Ÿ)^(𝑛) βˆ’ 1) 𝑅𝐡 βˆ’ 𝑂𝐡(1 + π‘Ÿ)^(𝑛))/π‘Ÿ [B] 𝑃 = (π‘Ÿ(𝑅𝐡 βˆ’ 𝑂𝐡(1 + π‘Ÿ)^(𝑛))/((1 + π‘Ÿ)^(𝑛) βˆ’ 1) [C] 𝑃 = (π‘Ÿ(𝑂𝐡(1 + π‘Ÿ)^(𝑛) βˆ’ 𝑅𝐡)/((1 + π‘Ÿ)^(𝑛) βˆ’ 1) [D] 𝑃 = (((1 + π‘Ÿ)^(𝑛) βˆ’ 1) (𝑂𝐡(1 + π‘Ÿ)^(𝑛) βˆ’ 𝑅𝐡))/π‘Ÿ

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Video Transcript

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

To get 𝑃 in terms of 𝑅𝐡, 𝑂𝐡, π‘Ÿ, and 𝑛, we’ll need to isolate the 𝑃 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 don’t make small sign mistakes or drop numbers, we could let π‘₯ be equal to one plus π‘Ÿ to the 𝑛 power and let 𝑦 be equal to one plus π‘Ÿ to the 𝑛 power minus one over π‘Ÿ. This allows us to write this equation as 𝑅𝐡 equals 𝑂𝐡 times π‘₯ minus 𝑃 times 𝑦. After we isolate 𝑃, we can plug back in the values for π‘₯ and 𝑦. Because 𝑃 is negative, we can add 𝑃𝑦 to both sides. 𝑃 becomes positive. Now, we have 𝑃𝑦 plus 𝑅𝐡 equals 𝑂𝐡π‘₯. And so, we subtract 𝑅𝐡 from both sides. And we’ll have 𝑃𝑦 equals 𝑂𝐡π‘₯ minus 𝑅𝐡. To get 𝑃 by itself, we need to divide both sides by 𝑦. And so, we can say that 𝑃 equals 𝑂𝐡π‘₯ minus 𝑅𝐡 divided by 𝑦.

At this point, we’ll have to plug back in what we know π‘₯ and 𝑦 are equal to. Instead of 𝑂𝐡π‘₯, we’ll have 𝑂𝐡 times one plus π‘Ÿ to the 𝑛 power minus 𝑅𝐡 over one plus π‘Ÿ to the 𝑛 power minus one over π‘Ÿ. This is pretty messy because we have a fraction in our denominator. If we write it out horizontally, we notice that we’re dividing by a fraction. And to simplify this, we need to multiply by that fraction’s reciprocal. In this case, we’ll now be multiplying by π‘Ÿ over one plus π‘Ÿ to the 𝑛 power minus one. We multiply π‘Ÿ times what’s already in the numerator. π‘Ÿ times 𝑂𝐡 one plus π‘Ÿ to the 𝑛 power minus 𝑅𝐡. And the denominator stays the same: one plus π‘Ÿ to the 𝑛 power minus one. This is 𝑃 in terms of 𝑅𝐡, 𝑂𝐡, π‘Ÿ, and 𝑛.

If we look at our options, we see that A and D have π‘Ÿ as the denominator. So we know that they cannot be correct. Both 𝐡 and C have the correct denominators. But if we look closely, only one of them is subtracting 𝑅𝐡 from 𝑂𝐡 times one plus π‘Ÿ to the 𝑛 power. And that’s option C. Option C has correctly stated 𝑃 in terms of 𝑅𝐡, 𝑂𝐡, π‘Ÿ, and 𝑛.

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