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Two-step equation word problem: computers

Learn how to construct and solve a basic linear equation to solve a word problem. Created by Sal Khan.

Video transcript

Marcia has just opened her new computer store. She makes $27 on every computer she sells and her monthly expenses are $10,000. What is the minimum number of computer she needs to sell in a month to make a profit? So I'll let you think about that for a second. Well, let's think about what we have to figure out. We have to figure out the minimum number of computers she needs to sell. So let's set that to a variable or set a variable to represent that. So let's let x equal the number of computers she sells. Number of computers sold. Now, let's think about how much net profit she will make in a month. And that's what we're thinking about. How many computers, the minimum number she needs to sell in order to make a net profit? So I'll write her profit is going to be how much money she brings in from selling the computers. And she makes $27 on every computer she sells. So her profit is going to be $27 times the number of computer she sells. She gets $27 per computer times the number she sells. But we're not done yet. She still has expenses of $10,000 per month. So we're going to have to subtract out the $10,000. What we care about is making a profit. We want this number right over here to be greater than 0. So let's just think about what number of computers would get us to 0. And then, maybe she needs to sell a little bit more than that. So let's see what gets her to break even. So break even-- that's 0 profit. Neither positive or negative-- is equal to 27 times-- and I'll do it all in one color now. 27x minus 10,000. Well, we've seen equations like this before. We can add 10,000 to both sides. Add 10,000 to both sides, so it's no longer on the right-hand side. And we are left with 10,000. 10,000 is equal to 27x. And then to solve for x, we just have to divide both sides by 27. Let's do that. Divide both sides by 27. On our right-hand side, we have x. So let me just write this down. So we have x on our right-hand side is going to be equal to 10,000/27. I switched the right and the left-hand sides here. Now, what is this going to be? Well, we can do a little bit of long division to handle that. So 27 goes into 10,000. So 27 doesn't go into 1. Doesn't go into 10. It goes into 100 three times. 3 times 27 is what? 81. 100 minus 81 is 19. Then we can bring down a 0. 27 goes into 190? It looks like it will go into it about six times. Let's see if that's right. 6 times 7 is 42. 6 times 2 is 12. Plus 4. 16. Let's see, 90 minus 62 is actually 28. Let me turn this back. So it goes 7 times. 7 times 7 is 49. 7 times 2 is 14. Plus 4 is 18. There you go. Look at that. So 190 minus 189. We get 1. Let's bring down another 0. We bring down another 0. We have a 0 right there. 27 goes into 10 how many times? Well, it doesn't go into 10 at all. So we'll put a 0 right there. 0 times 27 is 0. Then we subtract. And then, we get 10 again. And now we're in the decimal range. Or we're going to start getting decimal values. We can bring down another 0. We could get 27 goes into 100 three times. So our x-value is going to be approximately 370.3. And then we're going to keep going on and on and on and on. But this is enough information for us to answer our question. What is the minimum number of computers she needs to sell in a month to make a profit? Well, she can't sell a decimal number of computers, a third of a computer. She could either sell 370 computers. If she sells 370 computers, then she's not going to get to break even because that's less than the quantity she needs for break even. So she needs to sell 371. She needs to sell 371 computers in a month to make a profit.