MA, Stanford University
Teaching in the San Francisco Bay Area
Alissa is currently a teacher in the San Francisco Bay Area and Brightstorm users love her clear, concise explanations of tough concepts
MA, Stanford University
Teaching in the San Francisco Bay Area
This is a word problem that looks like it’s going to be pretty easy but it’s actually pretty challenging. We’re going to try to solve it using a system of equations and substitution. But first let’s make sure were clear on what the problem’s asking for. The sum of the digits of a two digit number is 12.that means like, before we move on I’m, going to show you an example, something like 66 because the sum of those digits is 12. Or I could use 93 or 39, whatever. These are numbers that when I add them together I get a sum of twelve. Add the digits together that is. When the digits are reversed the new number is 54 more than the original number. What is the original number?
A lot of students are going to try to approach this problem using a guess and check chart and that’s okay. Let me show you what it looks like. You have the original number; you’d have the flipped number and you’re going to have to check if one is 54 more than the other. For example, let’s say I picked my original number was 48; my flipped number would be 84 because I’ve flipped the digits and then I would check to see if the new number is 54 more than the original. I would do 84 take away 48 and see if that equals 54 or not. I’ll tell you it doesn’t. So I would this is not correct. And then I’d have to try another digit. Guess and check charts work but they take forever. I don’t know about you guys but I want a shortcut method so I know I’m going to get it right without having to guess.
That’s what I’m going to show you here using a system of equations. I’m going to write my number something like this. And this is kind of cruel of me because it is not totally true. I’m going to start with my original number as xy. But it’s weird, I don’t mean x times y, I mean x for the tens digit and y for the ones digit. So I’m just going to remember that x is going to be my first digit, x equals first digit and y is going to be my second, or my ones-second.
Okay, so when I do that I know that from the first sentence the sum of the digits of the two digit number is 12. So here is the sum, is equal to 12. That’s going to be my first equation in my system of equations.
Now we’ve got to do the second part and it looks like it’s going to be easy but it’s actually pretty tricky. Here we go. When the digits are reversed the new number is 54 more than the original number. Okay. Well, we’ll talk about the value of xy meaning my two digit number. My x was in the tens place so I’m going to call that 10 times my x digit and I’m adding one, one times my y digit. That’s my original number. 10y times my first number and then I have just 1 for my ones places for my second number. In my new, my reversed version I’m going to have now 10 times y. Y is going to be in the tens place. X is going to be in the digits place. So that’s how I’m going to start my equation. This is my original, this is my new and there is one more piece I need to think about. The new number is 54 more than the original.
So I’m going to have to take this guy. Oops wait I wrote 10y plus 1y, I meant to say 10y plus 1x. Excuse me, I’ve flipped the x and y. How come you guys didn’t tell me that viewers? Just teasing. Okay, so I’ve flipped my x and y and I’m dealing with that 54. The new guy is 54 more than the original. That means I’ve to take my original number and add 54 to it like that in order to get what this value would be.
Okay, now we’re ready to go through and solve this system of equations. I have two equations with two unknowns and I’m going to choose to use substitution because looking at this equation; I think I can get y by itself pretty easily. I’m going to say y is equal to 12 take away x so that I can substitute it into my second equations wherever I see a y. Like let’s see, 12 minus x is going to go right there because there is a y. it’s also going to go right there and then I’ll be able to solve. So let’s do the substitution step.
54 plus 10x plus 1 times my new y value is equal to 10 times my y value. Make sure you distribute plus 1x. Now this is just a straight forward Algebra problem that you guys can do and you could have done in your first few weeks of your Algebra class.
This is the most difficult part and guys I’m not playing with you. I know this is hard. When you get to these problems they’re very challenging. It’s hard to remember like what’s the tens, what the ones, whatever. Think it through step by step. The sum of my digits is 12. When I write it out using the value times the digit I get 10x plus 1y. When I flip it, it’s going to be 10 and the y plus 1 times x and I’ve got to add 54 to it.
Okay let’s go through and solve this. I’m going to do it pretty quickly because I know you guys know how to do this on your own. Just so we’ll be able to find the answer. First I’m going to simplify each side of the equation 10x take away x is going to be 9xs, 54 plus 12 is 66. Over here I’m distributing so I’ll have 120 take away 10x plus 1 more x. I’m going to go ahead and combine that into -9xs. Add 9 x to both sides sot hat this will be gone and I’ll have 18x plus 66 equals 120. Subtract 66 from both sides. 18x equals, where is my calculator, sorry, talk amongst yourselves, okay 54.
So my final answer is going to be when I divide both sides by 18, 54 divided by 18 is 3. This tells me my x digit is 3. I know half of the number, it’s going to look like thirty something or other.
In order to know what that something or other is, in my ones place, I’m going to go back to my original equation and remember that 3 plus my second number equals 12. So my second number must be 9 because 3 plus 9 is 12. That’s what I think is the original number. I’m just going to go and make sure and check that that’s correct. If I take 39 and switch the digits to 93, is true that 93 is 54 more than the original number? Well check on your calculator 39 plus 54 is indeed 93 so that’s my original.
Guys again I’m going to say this over and over, these are challenging problems if you don’t get them yet don’t be too hard on yourself, don’t give up on Math. Don’t start ripping out the pages and throwing them in the trash you can totally do this you guys. You just need to slow down, read carefully and keep in mind what you know about digits that you learnt in like second grade. The tens place and the ones place then you flip them, the tens place and the ones place. You guys can do it, just work slowly, set up a good system and keep track of what you’re doing.