 How are you friends? I'm sure you're all well. You must be watching our videos. We started making other subjects for you. We must be making videos on the chemistry bio portion. I included a chapter of mathematics in this video. In the third chapter of the NCRT Cerebous. You can see the linear equation in two videos. examples, in which we discuss linear equation in two variables, different parts of t and the conditions related to consistent pair तो इस्वो लहिग ख़ने के बाद, यह जाएर चाल फौगी कचापदा है। यह सब दूच सब बाद कोंचे लेग है। अपको ये मी लेगा कि आप कोचाजटों से लेके आप तरक़्साप लगा गोगी। अगर सब से रेलेकेट जोबी सवाल है, वो आप आसानी से करपाएंगे अप दिखेंगे कि अगर अपको कन्टिशन स्याद है, अगर कन्टिशन्ट, इन कन्टिशन्ट सिस्टन्स के, जो देफनीशन में आपको बताने जारागा, अगर उआप को यागा तो इस तरीके के कोई भी सवाल आप से कभी चुटेगा डेग, और बहुती एजी होतिए थोरी सी भारी की से तोगी से दिकना है, तोड़ा सद्यान से को समझना है, और समझने के बाद आपको ये महसुच वा, उसोगा कि यस तरके के कोई बिखवेस्चन आप यह से बूख से उतागे देख लिए यह कही से पुंट करें तो बहुत असानी से आप कर लिए जाएंगे और आप पक एक वि कुई क्छ्टेनाई मेश्भॉस नहीं लोगा जो, मुजे ये पुरी मीद है कि आप यहां से यस विडिय। को आगर देकेंगे जोवे में कन्डिशन्स दिस्पस करूँगा, एक और कन्डिशन्स को आप ज्ँहन से देकेंगे, समजेंगे और विस्तमाल आप यस तरीके के वेस्वें से आगे जागे करेंगे तो एक बार देपनिशन से सुरू करते हैं, यहा पहला कुशन अगर आप देखें इस में यह लिखाओगा एक यह प्यर अप लिन्यर एकवेशन से तु वेर्येवल् जो हम यहा प्यर रहे रहे हैं, तो एकस प्तमत भीवाइ प्लस चीज उग्ल्टो जीरो से हम उस तो अप दिखाते हैं. एक प्यर मगला कुशन दो इकवेशन की स्थडी एक साथ में करे लिएं, उसे हम लिखते हैं, A1 X plus B1 Y plus C1 is equal to 0. और आी 2 X plus B2 Y plus C2 is equal to 0. सु ये मेरा आप आप लिन्यर उक्वेशन्स लोगया। आप इस में कंषिस्तिझन्त और ईनकंसिस्तिं्त कया अवाब? सिस््तामüllण। क� genial कू behavioral पहड राबदा वासह। वभाँ मुञ्श rápido मुऊव आप वाँ work सैंती वाँ work tyre सैंत ब् microwave कfficiency स्शम करूओ और 세상 या जत जुज करूए ईस क vender अगर आपकोप अप निएखोछी तो ड्याठीगागी नहींत के अचाद से तूग, आप ञोब स्तरीए प्पिईट़ागी अपृसोंगी, आप भी एक दीखागी अप पासोंगी तुच आप पोगृसोंगी आप आप पृसोंगी थापिब आप प्घिएखा, आप उची आ� बहुद प्लोट करोगे, तो वो एक लैन आता है, थाज़ा कर खैटिन पोजथीव है, तो अगर उसका गैटिनगटिप है, तो लैन कुच आसे बनके आए, अब आब आप दो लैन आप प्लोट करते हो प्रटीवट आश्सेच्से, तो तीन कन्टीशन बनके आप, पहला कन्ट तो एकवेशन्स आसे हैं कि बे एक तूछरे को इंप्रसेक्त करते हैं अगर वो तोनो एक तूछरे को इंप्रसेक्त कर रहे हैं किसी एक point पे x1-9 पे तो हमें एक सुलूषन लेगा उसो हम यूनिक सूलूषन बोगते हैं आए उसके कन्टीशन क्या है में आगिया आप से दिसकस कर रहें तुसरा कन्तीशन हैं कि आप ने तो लाईंगs drop you, and तोनो parallel हैं अगर तोनो effecton हैं, इस का किया सुभ उऽ? इस का मखब यह होगा, कि यह लाईंगs if parallel हैं, so it will intersect with each other यहा थे है लेज़ा है की ओपनएत से आपकोंग़ कोईई़ई. और तीस्रा केस है जहाते लिंस ये दूसरे पे थरेंगी नहीं कोहलीत करते है गो. तो यह दूसरे ज्योंगे सतेसते रहा है। उआप दूसरी में उब ढ़म कोहलीत कोहले च्रते है। these are the coinciding lines if these are the coinciding lines if this line is L1, red and black line is L2 if I ask you a question in between L1 and L2 how many points are common between L1 and L2 I know your answer because it is coinciding and overlapping it means that infinite points are common between both that is why in this condition whatever condition I am going to discuss in this condition you will get infinitely many points common and in this case the solution of these two linear equations you will get infinitely many solutions so the formula we discussed our definition in that we said that if there is at least one solution then those pair of linear equations will be consistent and if there is no solution then it will be inconsistent so you will know that this first case in which we get a unique solution and this third case where we get infinitely many solutions these are both consistent and this single case where we get no solution this is inconsistent so in this way we define consistent and inconsistent and I have made a graph and tried to explain you by plotting lines how in consistent system you will get a unique solution when there will be one point of intersection when they intersect in one place if they overlap then infinitely many solutions will come even if they are consistent and if they are parallel then you will not get any solution because there will be no common point between both the lines and you will get no solution so if you read this question carefully pair of linear equations in two variables which has a common point and it has only one solution so it will be consistent pair why? because if you get one solution or infinitely many then it will be inconsistent pair if there is no solution then it will be inconsistent pair that in this other question it is asked if a pair of linear equations a1x plus v1y plus c1 is equal to 0 and a2x plus v2y plus ct is equal to 0 represents coincident line coincident line means the line which is overlapping so you will see in that case as I discussed with you that if there is one solution or infinitely many solutions then in the case of coincident line you will get infinitely many solutions and in that case we have solutions one or infinitely many our answer will always be consistent pair until and unless if we get no solution case now we will move on to the third question so in the third question we understand that what are the conditions when we will get a unique solution unique solution means one solution when we will get infinitely many solutions and when we will get no solution so if I get a pair of linear equations a1x plus v1y plus c1 is equal to 0 and a2x plus v2y plus c2 is equal to 0 so in this case the first condition is if a1 by a2 is not equal to b1 by b2 then we will get a unique solution what we will get? unique solution the second is if a1 by a2 is equal to b1 by b2 is equal to c1 by c2 so we will get infinitely many solutions this is the condition of the coincident line this is the condition of one intersection first one is the condition of one intersection here you see infinitely many solutions this is the condition of the coincident lines and the third one in which we get no solution that is a1 by a2 is b1 by b2 but that is not equal to c1 by c2 so we write that as no solution so these are the three cases and the questions we will do in the future are dependent on the three conditions so you can note them and attempt the rest of the questions because you will see that I will use them directly in the rest of the questions so one thing to be aware of if you see these equations then a1x plus b1y is equal to c1 and a2x plus b2y is equal to c2 and if you see here then c1c2 is equal to sin on the left hand side so generally while solving these equations because if I take this on the right hand side then it will be minus c1 so when we write this equation either you take it here then you can write it as k1x minus y minus 2 is equal to 0 and if not then you have to use this formula directly and if this is written in the format of a1x plus b1y plus c then you have to consider c1c2 as negative so this is a very important point when we look at the equations and try to find out the solutions then if this is written in the format of equal to sin then we write this as negative and then we write it as negative so if I solve this question and if there is no unique solution then if I compare it a1x plus b1y is equal to c1 and a2x plus b2y is equal to c2 and I know that my condition is unique solution of a1, a2 is not equal to b1, b2 so the value of a1 is k and the value of a2 is 6 that should not be equal to b1 you will see that it is negative 1 we always write it along with sin and b2 is negative 2 generally I have seen students making mistakes that they forget to write sin here and there are many mistakes so even if you can see that it is getting cancelled while doing it, if you write it along with sin then your calculation mistakes will be much better and I have written it here so I have written it as k by 6 not equal to 1 by 2 and from here k not equal to 6 by 2 which gives you k not equal to 3 so if you will use those 3 conditions like here I used the first condition so your answers will come very soon there is one thing which I have noted earlier that you should write along with sin and then do negative cancels your calculation mistakes will be very less so as we have seen here second option is also coming so we will click on k not equal to 3 option so option b I have selected here for third question and I have changed the first condition now let's go towards fourth question fourth question is infinite number of solutions and it has given two equations so these two equations ax plus by is equal to c and it is given in k form so I write the condition of infinitely many solutions a1 by a2 is equal to b1 by b2 and that is equal to c1 by c and I know that a by b and a1 a2 is equal to 4 b1 b2 is equal to 3k c1 c2 is equal to 510 so that is 5 by 2 is equal to 3 by k 5 by 10 is equal to 1 by k now this is 1 by 2 1 by 2 to get 1 by 2 what should be the value of k so from anywhere you can take if you do cross multiply then you will get k comes out to be 2 into 3 and then we will write k as 6 so from here k value comes to 6 and option c is correct answer so you will see that in this class what I am going to do I will use those 3 conditions again and again and then we will move on now we will move on to question 5 so you will see that the value of k for which the system of equations 2x plus y minus 3 is equal to 0 and 5x plus ky plus 7 is equal to 0 has no solution no solution has any condition so I have told you that you have to write a1 by a2 is equal to b1 by b2 that is not equal to c1 by c2 so we will only get the value of k from here why I wrote this why I wrote this because once our value of k comes then we will have to see if the condition of c1 by c2 is not equal to generally it will be complete you will see in this question that it will be complete but the value of k will be only from this condition so the value of a1 is nothing in front of x so a1 is 1 a2 here is the coefficient of x in the second equation so it is 5 then b1 is the coefficient of y so it is 2 y coefficient in the second equation that is k so from here I will get the value of k you will do cross multiplication you will get k2 into 5 and this is 10 so option a is correct but one thing if you see so this is 1 by 5 this 2 by 10 also becomes 1 by 5 and if you keep c1 c2 then you will see that it is not equal this is minus 3 by 7 so this means there will be no solution these lines are parallel they will not intersect so which option is correct option a is correct and the condition used we used a1 by a2 which is equal to b1 by b2 and this is not equal to c1 by c2 and for the question we used a1 by a2 which is equal to b1 by b2 which is equal to k2 so option a is correct so option a is correct now if we come to question 6 so this is also a very good question you see here it is written that the value of k for which the system of equations 3x plus 5 y is equal to 0 and kx plus 10 y is equal to 0 has a non-zero solution non-zero solution means that its solution is non-zero so for that what we do is from here I want to get the value of k which coefficient is k kx coefficient so what I do is from here I consider the value of y or we can also write a1 by a2 should be equal to b1 by b2 and if we are writing this then you can write 3 by k is equal to 5 by 10 for non-zero solution so this is 2 and if you multiply then k is equal to 2 by 3 so for k then 6 is equal to 6 or from here 3x is equal to minus 5 by 5x and from here you can replace it with y and get the value of k because we need non-zero solution so if you want from here kx and minus 5y is equal to 3x so this is 2 into 5y so 2 into minus 3x is equal to 0 this is the second method of solving so from here kx minus 6x is equal to 0 and x k minus 6 is equal to 0 so from here if x is non-zero so k minus 6 has to be 0 so k is equal to 6 and if we solve this so in such questions you can use one or two ways to confirm your answers I have told you both the ways and whatever method you like you can follow it so I have solved this question option C is correct and after doing this we will move on let's see question 7 so question 7 says if the system of equations 2x plus 3y is equal to 7 and a plus bx plus 2a minus by is equal to 21 has infinitely many solutions okay it has said that I need infinitely many solutions so we will write our condition quickly what was our condition of infinitely many solutions when I am writing this I hope you will say it and then we will write it quickly so a1 by a2 is equal to b1 by b2 is equal to c1 by c I hope all of you must be remembering this and as soon as I have told you infinitely many solutions then this condition should come in your mind so you will write it quickly and then what we will do then we will replace the values so our a1 value is x in the first equation a2 value is x in the second equation a plus b then our b1 value is y in the first equation b2 value is y in the second equation 2a minus b and c1 is 7 and c2 is the second constant so what I know from here I will write 2 by a plus b 3 by 2a minus b and 7 by 21 1 by 3 I do it one by one so first I first and third I will write it together and make an equation 2 by a plus b is equal to 1 by 3 we will multiply the cross it is very simple a plus b is equal to 6 and then I will write 3 by 2a minus b and that is equal to 1 by 3 then I will multiply the cross 2a minus b is equal to 9 this is my equation 1 this is my equation 2 I have 2 equations and I will solve it with elimination method so I have written the first equation under the second equation and I will add it when I add it 2a plus a3a will be 5 plus b is equal to 0 it will be cancelled and 9 plus 6 is equal to 15 so what is a? 5 by 3 and if a5 is so I keep it in the first equation 5 plus b is equal to 6 so b is equal to 6 minus 5 is equal to 1 so what is your answer a5 and b is equal to 1 this is your option b so what we saw here you are solving linear equation you will see when we are studying linear equation in our discussion in elimination method in substitution method you have checked it and after doing this your answer option b is equal to 5 and b is equal to 1 so we have seen if we know the conditions if we hold the conditions if we remember the conditions if we remember and see it carefully if we have done 2-4 questions then we will know how easy these questions are these questions meaning if multiple choice question comes it should not be missed because there are only 3 conditions there is only one of them and the usage of those 3 conditions I am making you here so your this should be an effort if you remember those 3 conditions do these questions and do some more questions so that you can make it your own if the question comes in this way then it should not be missed so with this we will move on to our 8th question you are saying again that this system is inconsistent now you have to go to definition till now we no solution one solution, unique solution infinitely many solution you are using such words here it is not saying no solution here it is to check your deeper concept in the place of no solution it is inconsistent but we do not have any problem we are sitting there what is inconsistent it is inconsistent where you get no solution lines are parallel you do not get intersection it is inconsistent no solution what we will do is a1 by a2 is equal to b1 by b2 which is not equal to c1 by c and after writing that we will replace the value so we will replace the value 3 by 2k minus 1 and that is equal to 1 here nothing is there so coefficient is 1 and plus k minus 1 is equal to minus 1 and why minus because it is on the right side equal to sin so minus 2k plus 1 negative negative cancelled I have made my habit we will cancel it there is no problem many times we write it here it is difficult to make and if we are not going according to sin convention we do not see it we have made mistakes so what we will do in this I do not know after k key value we will use it to remove k key value so we have multiplied cross so after multiplying cross what we have said k minus 1 is equal to 1 times 2k minus 1 so we will write 3k minus 3 and that is equal to 2k minus 1k 3k, we will write the variables on the left hand side this is negative on the left hand side minus 3 will be positive so we will write k is equal to 2 so the answer is k is equal to after k is equal to 2 so 3 2k 2 into 2 is 4 k minus 1 is 3 so 3 by 3 this is 1 by 2 minus 1 so 3 by 3 is anyway 1 by 1 not equal to 1 by 2 into 2 plus 1 5 so you will see that our condition is also being fulfilled so our option D came out condition is fulfilled and how easily we have asked this so I am telling you to remember the conditions after remembering the conditions you will not have any problem this in this question I have already given a condition if I read this question for you if am is not equal to d by m then the system of equations ax plus by is equal to c and lx plus my is equal to m now if I get this question in the exam and if I remember those 3 conditions in my mind so I will think about 2 conditions I will give you time to think about 5 seconds so those 2 conditions will be no solution because in no solution there is not equal to solution or that condition will be unique solution because in unique solution there is not equal to solution difference is that in unique solution it will not go towards c1 c2 and in no solution a1 by a2 b1 by b2 was equal to but it was not equal to c1 by c2 we will see which case is being fulfilled so here our a1 by a2 is a by l and b1 by b2 is b by m so a by l is b by m so a1 by a2 b1 by b2 is talking about c1 by b2 c2 which is cnn nothing is happening in this condition this means that there is only one case of unique solution how? I will tell you so am our is not equal to bl so cross multiply or you will see l is left hand side so this is in denominator and m should be in right hand side so a by l is not equal to b by m so from here I will write a1 by a2 is not equal to b1 by b2 because a1 by a2 is a by l and b1 by b2 is b by m so from here which condition came out? unique solution so in this question he gave you a solution in other questions he gave you a question and then he applied the condition and gave you a solution so in both ways the solution is possible you have to pay attention to what you have to do and especially in this type of question you have to pay more attention because here two cases are possible here a unique solution and a no solution is possible so here we saw because a1 by a2 is not equal to b1 by a2 or b1 by b2 then the solution is unique it is possible that he used c and n and gave you a question and here the answer is no solution so here there are conditions you have to check all the conditions so while doing this question you need to pay more attention so that we can take the right answers after doing this we move on to the last question and we will see that it says if the system of equations 2x plus 3y is equal to 7 and 2ax plus a plus by is equal to 28 has infinitely been used again the same question I give you some time think about the condition till the time I am writing the condition is very simple a1 by a2 sorry equal to b1 by b2 and that is equal to c1 by c2 3 over a2 so we write 2 a2 2a then we write b 3 by a plus b and this is 7 by 28 with a negative because it is here so it went 1 by 4 so my 2 by 2a is 1 by 4 so 2a8 and a4 if a is 4 so 3 by 4 plus b because I have replaced a it goes 1 by 4 so 4 plus b12 and b8 both are positive the solution will never be negative so option c and b will never be my answer now a is equal to 2b 4 is equal to 2 into 8 not possible b is equal to 2 into 4 so we can double b so in this video option b is correct in this video you have seen that the question of inconsistent or consistent pairs and the three conditions related questions can be approached so easily if we know the definition of consistent and inconsistent pairs we know the three conditions a1 by a2 not equal to b1 by b2 which is for unique solution a1 by a2 equal to b1 by b2 equal to c1 by c2 which is for infinitely many solutions and a1 by a2 equal to b1 by b2 not equal to c1 by c2 which is for no solution if we know the three conditions and definition then we can easily answer such questions and you will get to know how to solve these questions you will not face any difficulty you will correct these questions if you have done any calculation mistakes I hope you will avoid these questions because these are short short questions we should not make any mistake with this I am ending today's video I hope you liked this video with more videos we will meet soon we will solve the questions of linear equations in the next video till then thank you so much I hope you will be good wish you all the best and thank you