Addition and Subtraction Operations - Online Tutor, Practice Problems & Exam Prep

When adding or subtracting values in scientific notation, ensure the exponents are the same. The coefficients can be added or subtracted while keeping the exponent constant, resulting in expressions like (a ± b) × 10^x. If the exponents differ, adjust the smaller value to match. The final answer should maintain the least decimal places. This method is essential for accurate calculations without a calculator, emphasizing precision in scientific communication.

Addition and Subtraction

We're told here when you add or subtract values in scientific notation, they must have the same exponents. Now, we're going to say the coefficients add or subtract but the coefficients remain constant. If we're taking a look at these two examples here, we have a×10x−b×10x. The exponent portion must remain constant so our answer at the end will still be times 10 to the x and here we'd subtract our 2 coefficients. It would be a−b.

When we're adding them, it would be a+b times 10 to the x. Now realize here that you could easily put these into your calculator and get an answer at the end. But if your professor is asking you to do that without the use of a calculator, this is the method that we have to use to get the right answer. Now we're going to say if the exponents are not the same, then we transform the smaller value so that they do. We're going to say, remember when adding or subtracting values, that the final answer must have the least decimal places.

Okay. So we'll have the least decimal places when we're doing either a−b or a+b. Using this logic, we'll attempt to do the examples below. Once we've gotten a handle on it, you'll attempt to do some on your own. Okay.

So come back and look at the next video and see how I approach the example problem listed right below us.

Addition and Subtraction

Here it says, using the method discussed above, determine the answer to the following question. Here we have 8.17×108 plus 1.25×109. Remember, the smaller value we have to manipulate its exponent so that it matches up with the larger value. Here, 109 is larger, so I need to change this 108 also to 109. We need this to increase by 1.

In order for that to increase by 1, that means that our coefficient, this portion, has to decrease by 1. That means I'm going to take the decimal point here and move it over by 1. Decreasing that by 1 increases my exponent by 1. Remember, there's an inverse relationship between my coefficient and my exponent. Plugging what we have, we're gonna have 0.817×109 and that's gonna be plus 1.25×109.

Because they're both 109 now, we can just bring that down where it's constant and then we're gonna add this value to this value here. When we do that, that's gonna give me 2.067. But remember, when you're adding or subtracting, it's the least number of decimal points that we have to have for our coefficient. Here, this has 3 decimal points or decimal places, 0.817. This one has 2 decimal places, 1.25.

My answer at the end has to have 2 decimal places. We're gonna keep this portion here. Because this is a 7 here, we'll have to round this one up. At the end, that's gonna give me 2.07×109 as my final answer. So just remember, this is the method that we have to use in order to get the correct answer.

Now example 2. Attempt to do example 2 on your own. If you get stuck, don't worry. Just come back and see how I approach that same exact question. So go ahead.

Try it out. Come back and see what I do.

Addition and Subtraction

Remember, the largest exponent we're going to hold constant and we have to convert the smaller ones to match it. Here, the largest exponent we have is 10-11. We're going to have to convert the other 2 to match it. This one's going to stay still. We're going to keep it as is.

This one, we need to bring it down by 1 so it matches 10-11. It has to increase by going from negative 12 to negative 11 by 1. If it's increasing by 1, that means that the coefficient has to decrease by 1. We're going to move the decimal point one over. So that's going to be minus 0.117 times 10-11.

This one, it's negative 13. So we're going to have to increase it by 2 so that it goes and becomes a negative 11. That means I'm going to have to decrease this one by 2. So this is going to be 0.035 times 10-11.

So you got 10-11 for all of them. That part's going to remain constant. Here, we're going to subtract all of these from one another. So by subtracting from each one another, we have 8.9295 times 10-11. Here, this one has 2 decimal places.

This one here has 3 decimal places and this one here has 4 decimal places. We want 2 decimal places at the end. And because there's a 9 here, we're going to have to round up to 8.93 times 10-11. So that'll be our final answer there. Now we all know in terms of rounding, this is the approach we'd have to take.

But just for future references, let's say we got 8.925 flat. Here, because this is 5, we're traditionally taught to increase it by 1. But when it comes to analytical chemistry, if it ends with just 5 flat with nothing afterwards or like just 50 with a bunch of zeros, we're going to keep this number the same. It'd be 8.92. For it to be rounded up, it'd have to be like 51 or 52 or 53, something larger than 50 here in order for us to round up to the next highest value for this 2.

But again, if it's just 5 or 5 trailed by a bunch of zeros, we wouldn’t round up. We just keep it the same exact value. So it would stay as 8.92. Okay. We haven't seen that example, as of yet but just remember if it did pop up, this is the approach that we have to take in terms of rounding up or holding the number the same.