Strategies to Solve Multi Step Linear Equations with Fractions

I'm Anil Kumar welcome to my series on solving equations and thanks a lot for another request you want to understand how to solve equations with fractions here are eight examples and I hope by the end of the video you will have all your concepts absolutely clear I like you to pause the video copy these questions and then we are going to take them one by one I'm going to use two strategies here to solve all these questions let me call strategy one as cross multiplication and strategy to has get rid of fractions now cross multiplication also helps to get rid of fractions let's say the number is three here we'll just take it on the other side so we'll get rid of three from the left side now sometimes what happens is we have different numbers right for example here we have three terms here how do we get rid of fractions so in this case we'll look for the LCM lowest common multiple will multiply both sides by lowest common multiple so the strategy will be LCM times both sides and then we will solve as a linear equation without fractions so I hope the strategy is absolutely clear now let's apply the strategy to all these questions and solve them one by one now in such case which is kind of a ratio you could do cross multiplication let's understand what we are trying to do think like this we have our equation as X plus 2 over 3 equals to 4 over nothing so think we have one here so cross multiplication means this 3 gets multiplied with the term on the right side and this term the whole term gets multiplied by 1 you get an idea that means it remains kind of same that's what we mean so cross multiplication we're left with X plus 2 on the left side and on the right side we get 4 times 3 or 3 times 4 1 of the same thing so what we have here is X plus 2 equals to 12 and now this is without fractions easy to solve x equals 2 12 minus 2 which is 10 so we get x equals to 10 some of you can check the answer by placing 10 what do we get we get 10 plus 2 over 3 which is 12 over 3 which is indeed 4 so that works perfect so you may or may not check for the time being but it's a good practice so you may check ok so I think the technique is absolutely clear less applied once again so we get 1 we get X minus 6 equals 2 2 times 3 so that is X minus 6 equals to 6 X is equals to 6 plus 6 so X is equals to 12 and you can check your answer 12 minus 6 over 3 is 6 over 3 which is indeed - so that's what you expected correct okay so we have another similar kind of a question now here we'll apply both the techniques so this one I think now you can do easily we can write this as minus 3x equals to 4 times 5 so we have 2 minus 3x equals to 20 minus 3x equals to 20 minus 2 so minus 3x is equal to 18 X is 18 divided by minus 3 wedges minus 6 so we get x equals 2 minus 6 perfect next one now here what we see is that we see three different terms so this technique is not going to work we have to find the lowest common multiple so in this case what is the LCM so we are using the second strategy we are finding lowest common multiple for denominator so this is for denominator so 3 6 & 2 the lowest common denominator is 6 so we are going to multiply each term by 6 right so let's rewrite this so we get 4x over 3 equals 2 the equation is 7 over 6 minus 5x over 2 we are going to multiply each term by 6 on both the sides do you see this now since you do the same operation on both sides you don't change the equation but you could simplify it so so when you do so you could get rid of fractions that's the whole idea right so here 3 goes 2 times 6 one time and that goes 3 time so basically you get linear equation without fractions 4 times 2 is 8 so I grab this as a tax equals to 7 minus 3 times 5 15 X correct so 4 gets multiplied by 2 we get 8x several remains our search + 5 - x 3 LCD also we call lowest common denominator so we will also call this lowest common denominator and you get a linear equation now it is simpler to solve bring it to one side 8x + 15 x equals to 7 and when you add this 8 plus 5 is 13 so get 23 x equals to 7 so X is equal to 7 over 23 so that becomes the solution for the given equation so I hope these steps are absolutely clear so in this page we have seen both the techniques here we did cross multiplication and then in D we found the lowest common denominator which is actually the lowest common multiple of the denominators multiplying the same gets rid of the denominator and then you can solve as shown here so so I hope the strategy is absolutely clear let's move on and take a rest of the questions now I would like you to pause this video apply the right strategy and then solve so these two questions we again have three terms not just two terms right so what do you expect in this case we have to find the lowest common denominator right so let's find the least common denominator so lowest common denominator which is 6 - 4 is what as 12 right so is 12 now for some of you how do we figure this out let's take that now so to find this we have a ladder division method you could write all these numbers 6 2 & 4 / some common factors let's say - in this case we get 3 times 1 & 2 so 2 goes 3 times in 6 1 times in 2 2 times in full so I left with these numbers so the lowest common denominator I should say multiple now because we found that is 2 times 3 times 1 times 2 which is 12 so that is the technique of finding the lowest common denominator perfect so once we find that we have to multiply by this number to all the terms so we have X plus 7 over 6 plus so we are going to multiply this by 12 so that is this 12 now 12 times 1/2 equals to 12 times X minus 2 over 4 as expected all the denominators can now be cancelled with 12 so 6 goes 2 times 2 goes 6 times 4 goes 3 times now this 3 should be multiplied to both remember this part right otherwise you will get wrong answer same here 2 has to be multiplied with both X and 7 so when you open this bracket you get 2x plus 14 plus 6 equals to 3x minus 6 I hope this step is clear we are multiplying this 2 with both the numbers right applying the distributive property now you have to solve bringing the X terms together and the constants together sometimes we know 3x is higher we will bring X to the right side we have 14 plus 6 let me write 14 plus 6 this X coming this side will become it positive 6 3x minus 2 X so we get x equals 2 26 so x equals 2 26 is our answer is that clear now let's do the next one so here what is the lowest common denominator 4 & 2 it is 4 right since 4 is multiple of 2 so we'll multiply everything by 4 so what do we get we get 4 times 1 over 4x plus 1 in brackets equals to 4 times x over 2 plus 4 times 3 so here 4 & 4 cancels in this case we get 2 times here it is 4 times so when you open the bracket you get X plus 1 equals 2 2 X plus 12 again I will take this X to the right side now since I see that 2x is greater bring 12 on this side so we have 1 minus 12 equals 2 2 X minus X that gives me x equals 2 minus 11 so I hope you appreciate this strategy of avoiding the negative sign with X it also helps saves time perfect now let's move on and take the last two questions I hope by now you have learned this strategy apply and solve these two questions then check with my solution here 5 & 2 so what is the lowest common denominator in this case the lowest common denominator is 10 so we'll multiply every term by 10 so what we get here is 10 times 2 over 5 times 3x minus 1 equals to 4 times 10 minus 10 times 1/2 X plus 2 now when you simplify 10 goes to x with 5 and 5 times with 2 so 2 times 2 is 4 so we're at 4 times 3x minus 1 equals 240 minus so when you open this bracket remember to apply minus 5 into both right so minus 5x and minus 10 so take care of minus sign also right so you have to multiply with both the terms bring XS together let's open this bracket also so we get 12 X minus 4 equals 2 let's combine them 40 minus 10 is 30 minus 5x bring 5x to the left side it 12x plus 5x equals 230 plus 4 and that is 17 x equals 234 or X is equals 234 over 17 which is too perfect so that is how you're going to solve it now let's take the last example this is kind of different since you have X in the denominator to see this part X is in the denominator so in this case what is the lowest common denominator it is 6 X you see that not just 6 6 X so we are going to multiply each term by 6 X so we have 6 X times 1 over 3 equals 2 6 x times 2 over X minus 6 x over 6 so that gives you 6 divided by 3 is 2 so we get 2x equals to 6 times 2 is 12 X and X cancel rights we get 12 and here we get minus X so see what happens X and X cancels this goes to x + 6 and 6 cans so we get our equation which is 2x equal to 12 minus X now bringing X to the left we have 2x + x equals to 12 3x equals to 12 X is equal to 12 over 3 which is 4 so I hope the solutions are absolutely clear that is how it should be so go through these examples once again so we have different types of strategies applied and these are sufficient to answer or solve any equation involving fractions feel free to write your comments and share your views if you really like and subscribe to my videos that we create thanks for watching and all the best

Channel: Anil Kumar

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Keywords: Anil kumar, khandelwal, aan learning centre, advance functions, mhf4u, khan academy, CBSC, O-level, grade 12, mcr3u, grade 11, sat, NCERT, high school math, ib math, grade 12 math, university math, ap math, DSSSB Math TGT Exam tricks Preparation, tricks and examples, Pre Calculus, domain, linear equations, multi-step linear equations, equations with fractions, simple solution, global math institute, globalmathinstitute, linear, equations, solve, simple, multistep, rational equations

Id: F-aqjOfs_Cw

Length: 15min 40sec (940 seconds)

Published: Sat Oct 20 2018