Algebra Basics: Solving Basic Equations Part 2 - Math Antics - By Mathantics
Transcript
00:03 | Uh huh . Hi , I'm rob . Welcome to | |
00:07 | Math antics . In our last video we learned how | |
00:10 | to solve basic algebraic equations . That only had one | |
00:13 | addition or one subtraction operation in this video will focus | |
00:17 | on equations that have only one multiplication or one division | |
00:20 | operation . Now before we see some examples , do | |
00:24 | you remember the key strategy for solving an equation with | |
00:27 | an unknown value in it ? Yep . We have | |
00:30 | to use arithmetic to rearrange the equation so that the | |
00:33 | unknown is all by itself on one side of the | |
00:36 | equal sign . And the most important thing to keep | |
00:39 | in mind will rearranging equations is that whenever we do | |
00:42 | something to one side of an equation we have to | |
00:45 | do the same thing to the other side or else | |
00:47 | the other side might get jealous . Hey how come | |
00:52 | he got a cookie And I didn't actually . It's | |
00:58 | to keep the equation in balance . Now remember from | |
01:01 | the last video in equations where a number was being | |
01:04 | added to an unknown , we had to subtract that | |
01:07 | number from both sides . But when a number was | |
01:10 | being subtracted from the unknown , we had to add | |
01:12 | that number to both sides . And that makes sense | |
01:15 | because as we learned in the video called what is | |
01:18 | arithmetic ? Addition and subtraction are inverse operations . They | |
01:23 | undo each other . Well , guess what ? Multiplication | |
01:26 | and division are also inverse operations so we can use | |
01:30 | them to undo each other to If an unknown is | |
01:33 | being multiplied by a number to undo that we need | |
01:37 | to divide both sides by that number . But if | |
01:40 | an unknown is being divided by a number to undo | |
01:43 | that we need to multiply both sides by that number | |
01:46 | . Now don't worry if that sounds a little confusing | |
01:49 | right now it will make it a lot more sense | |
01:51 | . After you've seen a few examples , let's start | |
01:53 | with this 13 X equals 15 . Excuse me . | |
01:57 | Thank you forgot something . I mean didn't you say | |
02:00 | that these equations we're going to have multiplication or division | |
02:03 | in them ? But I don't see any arithmetic operator | |
02:06 | at all in this equation actually . I think you | |
02:09 | forgot something that we learned in the video . What | |
02:11 | is algebra ? You did watch that ? Right ? | |
02:14 | Oh sure sure . Of course . But I you | |
02:18 | know I I just remembered I have something I gotta | |
02:20 | do . I'll be right back . Well I'm sure | |
02:23 | you remember that multiplication is the default operation in algebra | |
02:28 | . So when you see a number and assemble right | |
02:30 | next to each other like this with no operation between | |
02:33 | them . It means they're being multiplied . So three | |
02:37 | X . Is the same as three times X . | |
02:39 | O . And just a side note since in multiplication | |
02:43 | the order of the numbers doesn't matter . You could | |
02:46 | switch the order and right X . Three . But | |
02:49 | it's customary to always list the known number first and | |
02:52 | the unknown number . Second . All right . But | |
02:55 | we need to solve this equation . Right ? That | |
02:57 | means we need to get the unknown X all by | |
03:00 | itself on one side of the equal sign . Right | |
03:03 | now , the X is not by itself because it's | |
03:05 | being multiplied by three . So to undo that operation | |
03:10 | , we need to divide that side by three . | |
03:12 | In algebra , we almost always right , division infraction | |
03:16 | form . So to divide this side by three , | |
03:19 | we just write a fraction line under it and we | |
03:21 | put a three below the line There . This means | |
03:25 | three times x divided by three . Ah But don't | |
03:29 | forget our rule for rearranging equations . We have to | |
03:32 | do the exact same thing to the other side to | |
03:35 | keep the equation balanced . That's better now both sides | |
03:39 | are being divided by three . The next step is | |
03:42 | to simplify the three on the top and the three | |
03:44 | on the bottom on this side cancel because three divided | |
03:47 | by three would just be one . This is just | |
03:50 | like canceling common factors when you're simplifying a fraction . | |
03:54 | That leaves us with just X . On this side | |
03:57 | . And on the other side we have 15 divided | |
04:00 | by three which simplifies to five . There we've solved | |
04:04 | our equation by changing it into the simplified form x | |
04:07 | equals five . Let's try another one just like that | |
04:10 | . 12 x equals 96 . In this problem the | |
04:14 | unknown is being multiplied by 12 . So to get | |
04:17 | the X . All by itself , we're gonna need | |
04:19 | to divide both sides of the equation by 12 On | |
04:22 | the first side . The 12 on top and the | |
04:25 | 12 on bottom cancel out , leaving just X . | |
04:27 | On that side . And on the other side we | |
04:30 | need to divide 96 x 12 . You might be | |
04:33 | able to do that by memory , but if not | |
04:35 | you can use a calculator to divide 96 , divided | |
04:39 | by 12 is eight . So in this problem x | |
04:42 | equals eight . That's pretty easy , isn't it ? | |
04:45 | Are you ready to try a division problem ? Now | |
04:47 | , here we have X divided by two equals three | |
04:51 | . Now when you see division written like this from | |
04:54 | left to right , with the traditional division symbol , | |
04:56 | I want you to rewrite it using the fraction form | |
04:59 | for division . And that's because it's much easier to | |
05:03 | cancel common factors and simplify your equation . When you | |
05:06 | use the fraction for now that we haven't rewritten , | |
05:10 | let's solve it . We can see that the unknown | |
05:12 | is not by itself because it's being divided by two | |
05:16 | . How can we get rid of or undo that | |
05:19 | division , yep , we can undo division with multiplication | |
05:23 | . So we need to multiply both sides of the | |
05:25 | equation by two . Instead of writing the multiplication sign | |
05:29 | . I'm using the parentheses notation that we learned about | |
05:32 | in the video called What is algebra ? Remember the | |
05:35 | multiplication is just implied now to simplify it On the | |
05:39 | first side , the two on top cancels out the | |
05:42 | two on the bottom . Since two divided by two | |
05:45 | is just one . And I know what some of | |
05:47 | you are thinking . How is there a two on | |
05:49 | top ? The two looks like it's really in the | |
05:51 | middle . Kind of like how it makes number looks | |
05:54 | That's true . But don't confuse this with a mixed | |
05:57 | number . Mixed numbers involve addition . But the parentheses | |
06:01 | let you know that the two and the X . | |
06:03 | Over two are being multiplied since multiplication is the default | |
06:07 | operation . Okay , so it's not a mixed number | |
06:10 | , but how is the two on top ? Well | |
06:13 | do you remember how you can turn any number into | |
06:15 | a fraction ? Just by making one the bottom number | |
06:19 | ? That means that too is the same as to | |
06:21 | over one . Ah Now you can see that the | |
06:24 | two really is on top . It's just that we | |
06:26 | don't usually show the one on the bottom . All | |
06:29 | right then . So the two's cancel leaving the X | |
06:32 | all by itself on this side . And on the | |
06:34 | other side we have three times two , which is | |
06:37 | just six . So in this problem x equals six | |
06:41 | . That's not too hard either . Let's try another | |
06:43 | one . X over 10 equals 15 In this problem | |
06:48 | . Since the X is being divided by 10 To | |
06:51 | get it by itself , we're going to need to | |
06:52 | multiply both sides of the equation by 10 . On | |
06:56 | the first side , the tense cancel leaving X all | |
07:00 | by itself . And on the other side we have | |
07:02 | 15 times 10 , which is 150 . So our | |
07:06 | answer is x equals 150 . Great . That's how | |
07:11 | you solve simple equations where an unknown is being multiplied | |
07:15 | by a number or divided by a number . But | |
07:18 | just like with subtraction in the last video with division | |
07:22 | , there's a tricky variation that I need to tell | |
07:24 | you about . What if you have an equation where | |
07:27 | a number is being divided by an unknown since division | |
07:31 | does not have the community of property X over four | |
07:35 | is not the same thing as four over X . | |
07:39 | So what do we do if the unknown is on | |
07:41 | the bottom ? Like in this problem for over X | |
07:44 | equals two . Well , your first thought might be | |
07:48 | to multiply both sides by four , but that won't | |
07:51 | help us here because both of the fours would be | |
07:53 | on top so they wouldn't cancel each other out instead | |
07:57 | what we need to do is multiply both sides by | |
08:00 | ex watch what happens then the X . Is on | |
08:04 | this side of the equation will cancel yep you can | |
08:07 | cancel unknowns and variables exactly like you can regular numbers | |
08:12 | . That will leave us with just four on this | |
08:14 | side of the equation . And on the other side | |
08:17 | we have two times X . Or two X . | |
08:21 | True . That didn't solve the equation but it did | |
08:24 | get rid of the tricky X on the bottom and | |
08:26 | it changed our equation into a problem that we already | |
08:29 | know how to solve . Now to get the X | |
08:31 | . All by itself we just need to divide both | |
08:34 | sides of the equation by two . On the first | |
08:37 | side we have four divided by two which is two | |
08:40 | . And on the other side the 2/2 cancels and | |
08:43 | we're left with just X . So now we know | |
08:46 | that x equals two . Okay , So now that | |
08:50 | you've watched these first three mathematics algebra videos , you | |
08:53 | should be able to solve any simple one step equation | |
08:56 | involving addition , subtraction , multiplication or division . Right | |
09:00 | . Well not unless you practice to really learn how | |
09:03 | to solve equations , you have to try a lot | |
09:06 | of problems on your own to make sure that you | |
09:08 | really understand how to do it . And if you're | |
09:10 | still confused , try re watching these videos a few | |
09:13 | times since they cover so much information as always . | |
09:17 | Thanks for watching Math Antics and I'll see you next | |
09:19 | time . Learn more at Math Antics dot com . | |
09:24 | Hey , I watched that video . You mentioned . | |
09:26 | Hello ? Hello ? Where am I ? |
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