Transcript for:
Lecture on Pair of Linear Equations in Two Variables

[Music] hello everyone this is sammit sankanaver i welcome you all back with my new video in today's class we'll have a detailed revision of one more important chapter from sslc mathematics that is pair of linear equations in two variables let us start this video with very basic understanding of variables and linear equations without understanding the meaning of linear equations and variables it becomes very difficult to understand this chapter so let us first understand what is the meaning of linear equations in two variables whenever you see any equation in mathematics you come across two letters of english alphabets namely x and y these two terms play an important role in forming any equation because the value of these terms depend on each other and they keep on changing according to the situation and hence these terms inside an equation are called as variables remember variables can be represented using any english alphabet and not only x and y now let us understand why these terms are called as variables exactly using an example here suppose i consider an equation here that is y is equal to 2x plus 1 in this equation as i go on changing x value y value is also going to change suppose i consider in the first situation that x equal to 1 when i put x equal to 1 in this equation i will get y value as 3 similarly in second condition if i put x equal to 2 then i will get y value as 5 in third condition if i put x equal to 3 i will get y value as 7 so in all three conditions when i go on changing x value i will get different values of y therefore y is depending on x here and we can change values of value of x to get a different values of y but one noticeable point here is i cannot change the value of 1 i cannot change the value of 2 hence these two terms are called as constant and x and y are called as variables because i can change their values now there is no limitation for how many variables can be present inside an equation suppose there is only one equation one variable is present in equation that equation is called as equation in one variable example 2x plus 3 equal to 0 in this equation there is only one variable present that is x if there are two variables present inside an equation that is called as equation in two variables in this equation you can see there are two variable that is x and y both are present if there are three variables present in any equation that is called as equation in three variables you can see in this equation x y and z all three variables are present therefore it can have multiple variables depending on the variables we can have those equations hope you have understood what is the meaning of variables now let us understand what is the meaning of linear equations linear equations are those equations upon plotting or graph if it is going to give us a straight line that equation can be termed as a linear equation now whenever you are given with a equation you know how to plot it on the graph that you have studied in nine standard if after plotting such equation on the graph you are getting a straight line those equation can be termed as linear equation now every time you cannot plot it on the graph and check whether it is linear equation or no algebraically also you can identify whether that equation is linear equation or no if power of any variable it can contain multiple variables if power of any variable is going to be equal to 1 that equation can be termed as linear equation similarly if power of any equation inside an equation is 2 that equation can be termed as quadratic equation and if power of any variable inside an equation is 3 that equation can be termed as cubic equation so all those equations that are going to come in this chapter are going to be of variables of power one though therefore in this chapter we are going to study only about linear equation we have one more chapter in your assessment syllabus that is quadratic equation wherein power of all the variables are going to be equal to 2 hope you understood what is variables and linear equations now let us go ahead to understand linear equations how they can be plotted on the graph [Music] now i am going to consider an example and show you how the graph of linear equation looks like here i have considered an equation that is 2x minus y is equal to 0 this pink line represents the equation 2x minus y is equal to 0 now we have already studied in n standard how to plot a graph using any given equation considering different values of x we obtain different values of y and that we plot on the graph same thing i have done here now all these points that i have marked are going to fall on this straight line and further whenever i substitute these values of x and y corresponding to different points on this straight line inside this equation they are going to satisfy this equation that is 2x minus y is equal to 0 further any other points of x and y any other values of x and y that are going to that are going to satisfy this equation and that are not visible right now in this graph that those two points will also follow the same straight line when this line extended further this can be considered as a very important point corresponding to linear equation i will note it for you each solution x comma y of a linear equation in two variable which is of the form ax plus b y plus c is equal to 0 corresponds to a point on the line representing the equation and vice versa just meaning of this once again i will explain any value of x and y which will satisfy this equation will fall on this straight line when extended further or any point that are present on this linear equation will satisfy this equation when they are substituted in the form of different values of x and y example when i consider this point that is which 1 comma 2 whose x value is 1 y value is 2 when i substitute this value in this equation that is 2 into 1 that is 2 minus y value is 2 2 minus 2 is going to equal to 0 means this point is satisfying this equation likewise all these points are going to satisfy this equation that is a meaning of this point and this is going to be very important point corresponding to linear equations now let us go ahead and understand why do we always consider a pair of linear equation in two variables and not a single equation of two variables this i can explain you considering an example here suppose i am going to give you an equation in single variable that is 2x minus 8 is equal to 0 and i ask you to solve the equation solving the equation is nothing but finding the value of unknown term that is present inside the equation that you can easily do by shifting minus 8 on the right side and taking 2 below 8 therefore you can find here in this equation x value as 4 but if i give you one more equation that is containing two variables and ask you to solve it becomes very difficult for you to solve and no algebraic method can help you to solve this equation alone unless and until you imagine any particular values of x and y and manually find the solution but if i give you one more equation of similar manner containing two variables then you can solve both using both the equation you can solve and find the x and y values why is it like this because in order to solve linear equation of two variables we must need one pair of two linear equation what is the reason for that this one fact you must remember every time you are solving any equation that number of equations required to find the solution will be equal to number of variables means in the first example there is only one variable or only one unknown term so one equation was sufficient to find the value of that variable here you can see in second example there are two variables so we need two equations to find the values of these two variables now suppose in any equation there are three variables then we must need three equations in order to obtain all the values of those three variables likewise number of equations required to find the value of those unknown terms will be equal to how many unknown terms are present in the equation hope you remember this one concept every time you study the study mathematics and and you are supposed to solve any given equation now let us go ahead and understand very important concept that is going to be very useful in this chapter that is general form of the linear equation in two variable general form of linear equation two variable is going to be ax plus b y plus c is equal to 0 where e is going to be coefficient of x b is nothing but coefficient of y and c is the constant term now whenever we are going to consider a pair of linear equation first equation i can write as a 1 x plus b 1 y plus c 1 is equal to 0 second equation i can write as a 2x plus b2y plus c2 is equal to 0 here remember always a 1 and a 2 are coefficient of x b 1 and b 2 are coefficient of y c 1 and c 2 are constant terms and a 1 a 2 b 1 b 2 c 1 c 2 all are going to be real numbers remember one important fact here that a 1 and b 1 cannot be equal to 0 at a time remember what happens if a 1 and b 1 both become equal to 0 at a time then a 1 is equal to 0 means this whole term becomes 0 b 1 is equal to 0 means this whole term also will be equal to 0 then c 1 will be equal to 0 any constant term suppose constant term is one here can one be equal to zero it is never possible therefore this condition is not satisfying here therefore a one and b one cannot be zero at a time either a can be zero then these two terms will be remaining then b can be 0 c can also be 0 that is also considered as linear equation in 2 variables but a 1 and b 1 cannot be 0 at a time that one fact we have to remember for example 2 can that can be written in the general form as you can see example 2 here x plus 2 i is equal to 2 this is not present in general form because in general form always we equate that to 0 so i can take this minus 12 on the left side of the equation that is x plus 2y minus 12 is equal to 0 now this is in the general form here a value that is coefficient of x is 1 nothing means we take it as 1 b value that is coefficient of y is 2 c value that is constant term is minus 12 here because in standard it should be in the form plus but here it is minus because of that constant term which is in the form of minus therefore these are the different values of a b and c in this equation likewise any equation will be given to you and you will be asked to write the values of coefficient of x or coefficient of y or value of constant term so you should always take care of the signs that are present next to them now let us understand how we can form the pair of linear equation in two variables reading the problem statement many students find it difficult to form the linear equation reading the problem statement and because of which they are not able to solve the equation which are going to be very easy equations this was actually the part of ninth standard in pair of linear equation in two variables that you have already studied but once again i am going to take that part in this video that is how we can form the equation for this i am going to consider an example here that is romela went to a stationary shop and purchased two pencils and three raises for rupees nine her friend sonali saw the new variety of pencils and erasers with rumila and she also bought five pencils and six erasers of the same kind for rupees 21. now this example should be represented in the form of pair of linear equation in two variables and further same example we are going to solve using the graph that will cover your geometrical method of solving the pair of linear equation first let us understand how we can form the equation step by step we are going to read the question and we are going to write our equations according to the statements first statement is given that romera went to a straightener shape stationary shop and purchased two pencils plus three erasers and she totally gave nine rupees that i can represent as two pencil plus three eraser is equal to nine now her friend sonali also went to the same shop and she bought five pencils and six erasers five pencils plus six erasers and totally she paid for that is 21 rupees now here what we are supposed to find cost of one pencil and cost of one eraser now they are the unknown terms for us so those things we are going to equate to our variables that is let cost of one pencil be equal to x and cost of one eraser be equal to y now when i put x x a and y value x and y in this equation it becomes 2 x plus 3 y equal to 9 and 5 x plus 6 y is equal to 21 here x is nothing but cost of one pencil when you multiply that by two pencil you will get cost of two pencil cost of one eraser is nothing but y when you multiply by three results you will get cost of three raises total cost of three razors total cost of three eraser plus total cost of two pencils is equal to nine according to the first statement similarly total cost of five pencil plus total cost of six erasers is equal to 29 according to second statement same thing we have represented in the form of equation now we have got one pair of two linear equations now let us go ahead and understand how this equation can be solved by different different methods basically you have two methods of solving the pair of linear equation first one is geometrical method second one is algebraic method geometrical method is nothing but plotting both the equations separately on the graph and wherever they are going to intersect that is going to give the value of x and y that is the solution let us understand first graphical method then we will go ahead and understand the algebraic methods now using both the equation that we just obtained in order to solve them graphically you already know how to solve these equations by plotting them on the graph that you have studied in ninth standard we are going to represent this equation in terms of y first that we can write it as y is equal to 9 minus 2x by 3 first equation second equation we can write as y is equal to 21 minus 5x upon 6 now by substituting different values of x we obtain different values of y in both the equation that i am going to take as when i put first x equal to 0 i get y value as 2 into 0 here that is 0 then i get 9 by 3 as 3 therefore for x equal to 0 i get y is equal to 3 in the first equation similarly i am going to put x equal to 3 now in this equation that is 2 into 3 is going to be 6 9 minus 6 is 3 again 3 by 3 is going to be equal to 1 then i get y value is equal to 1 now what i will do i put y value as 0 then i will get x value whenever y is equal to 0 here 3 into 0 is 0 then minus 2 is going to come on left side of equal sign then 2x will be equal to 9 therefore x value will be equal to 9 by 2 that is nothing but 4.5 now i obtained a point three two points are more than sufficient to plot a graph but for to be honest a percent will take three points similarly in this equation again i will put first y is equal to zero x is equal to zero sorry five into 0 is equal to 0 then i get 21 by 6 when i solve this i get answer as 3.5 that is y value secondly i am going to put x equal to 3 when i put x equal to 3 in this equation i will get 5 into 3 as 15 21 minus 15 is going to be equal to 6 6 by 6 is going to give me answer as 1 i am getting y value as 1 here now i will put y value as 0 then obtain x value when i put y is equal to 0 6 into 0 is 0 minus 5 x i'm going to take it on the left side of the equal sign that is going to be equal to 5 x equal to 21 then i take x value as 21 divided by 5 i get it as 4.2 now i have obtained different values of x and y for both the equation now i can easily plot it on the graph when i plot it on the graph two equations this sky blue equation sky view a blue line is representing 2x plus 3y is equal to 9 equation and purple line is representing 5x plus 6y is equal to 21 this equation now different points that we obtained here are represented on this two lines here you can see that both the lines are going to intersect at the point 3 comma 1 now this 3 comma 1 itself is going to be the solution of these two equations means 3 comma 1 that is x value is 3 and y value is 1 this point is going to satisfy both this equation together therefore answer for these two equations that we obtained graphically that is pair of linear equation is nothing but x equal to 3 and y is equal to 1 but we have substituted x as cost of 1 pencil that is equal to 3 rupees and y as cost of 1 eraser that is 1 rupee now you can substitute this equation back in the previous uh this equation that is equation 1 and equation 2 and check whether we are getting correct answer or no with this we are going to finish graphical method this is also known as geometrical method of finding solution of pair of linear equations in two variables now next we are going to understand different patterns when we solve pair of linear equations what are the different possibilities let us see whenever we consider two lines there are mainly three possibilities and that is same in case of payroll linear equation as well let us understand those three possibilities one by one the very first possibility is they intersect at only one point let us understand this possibility with an example here suppose i have considered two equations x minus two y is equal to zero and three x plus four y is equal to twenty when i plot graph of these two equations by considering different points of x and y i get this pattern you can see here these two lines are meeting at one particular point that is x equal to 4 and y is equal to 2 means x equal to 4 and y is equal to 2 is the unique solution or one common solution that can satisfy both this equation therefore for this first possibility we can conclude one point that the point four comma two is the only common point for both the equations therefore x equal to four and y equal to is the only solution for given pair of linear equation now second possibility that we have is they coincide or overlap each other let us understand this with an example here i have considered two equations 2x plus 3y is equal to 9 and 4x plus 6y is equal to 18. when we plot graph of these two equations by considering different points of x and y we can see such pattern wherein both lines are overlapping we can see only one line because another line is falling behind this line means all those points are overlapping here therefore we have many solutions that satisfy both these equations means all those points which satisfy one equation will definitely satisfy another equation that are lying on this line thus we can conclude one point for this second possibility that is all the points that satisfy both equations coincide means those pair of equations have infinitely many solutions we cannot have only one solution we have infinitely many solutions that will satisfy both these equations next third possibility is they never meet or we have parallel lines if i consider two examples that is x plus 2y is equal to 4 and 2x plus 4y is equal to 12. in these two equations when i plot them on the graph by considering different values of x and y i see such pattern wherein those two lines are never going to meet they are going to be parallel here all those equat all those points that are present in on these lines they are never going to meet means these two pair of equations are never going to have a common solution means there is no particular value of x and y which can satisfy both these equations that is the one conclusion that we can draw for the third point that i can write in the form of point as there is no such common point which can satisfy both the equations therefore those pair of equations have no common solution these are the main main three conclusions that we can draw considering geometrical pattern of pair of different kinds of linear equations now let us understand how these pair of linear equations behave algebraically and what are the different conclusions that we can draw algebraically when we consider such pair of linear equations every time we need not plot the graph of pair of linear equations in order to understand geometrical and algebraic behavior of those pair of equations but depending on the ratios of coefficient of variables and constants we can arrive at the following conclusions let us see what are those conclusions here in this table i have considered three pair of linear equations as an example in the first example we can consider the ratio of coefficient of x that is a 1 by a 2. here you can see coefficient of x is 1 equation x is 3 a 1 by a 2 is 1 by 3 and b 1 by b 2 i have considered as 1 by 2 here you can clearly see that a 1 by a 2 is not equal to b 1 by b 2 whenever we have this condition where ratio of coefficient of x is not equal to ratio of coefficient of y we will have intersecting lines this we have seen intersecting lines in the previous slide and algebraic conclusion of this type of linear equations is going to be they will have unique solution means there will be only one point of x and one point of y which will satisfy both these equations in the second example you can see a 1 by a 2 is going to be equal to b 1 by b 2 that is equal to c 1 by c 2 when we have ratio of coefficient of x is equal to ratio of coefficient of y which is equal to ratio of constant terms then when all these three ratios are going to be equal then we have coinciding lines means we have overlapping lines means algebraic conclusion of that pair of linear equation is going to be will have infinitely many solutions means there will be infinitely many such points of x and y which will satisfy these both equations condition is a 1 by 2 should be equal to b 1 by b 1 that should be equal to c 1 by c 2 all three ratios are going to be equal in the third example we have considered here you can see a 1 by a 2 is equal to b 1 by b 2 but that is not equal to c 1 by c 2 when we have this condition we will have parallel lines means algebraic conclusion of this type of pair of linear equation is will have no common solution means there will be no such point of x and y which will satisfy both these equations so what what will be the approach in order to understand algebraic and geometrical behavior first we'll calculate these all three ratios that is a 1 by a 2 b 1 by b 2 and c 1 by c 2 if a 1 by a 2 is not equal to b 1 by b 2 then these two possibilities are eliminated then we should understand that it will have intersecting lens and unique solution if a1 by a2 is equal to b1 by b2 then we should see the third condition whether it is equal to 7 by c2 if it is equal to c1 by c2 then we will have coinciding lines and infinitely many solution if c 1 by c 2 is not equal to a 1 by b 2 and that is equal to b 1 by b 2 in that case we will have parallel lines and we should conclude that there will be no common solution remember considering all these three points we can arrive at one more conclusion common conclusion that is if at least one solution is possible means in case of unique solution or infinitely many solutions in these two cases those pair of linear equations are called as consistent pair of equations because we have at least one solution in this case but in this third case where we have no common solution that pair of linear equation is called as inconsistent pair of linear equation in many examples they will be asking you find out whether these given pair of linear equations are consistent or inconsistent if it has has at least one solution means if it is having unique solution or if it is having infinitely many solution in both the cases that that pair of linear equation is considered as consistent pair or else it is taken as inconsistent pair of equation now let us go ahead and understand an important concept of this chapter that is how we can solve a pair of linear equations algebraically in the previous slides we have understood how we can solve pair of linear equations geometrically or also known as graphically in algebraic method there are mainly three methods let us understand one by one the very first method that we will consider is substitution method in order to understand this method i am going to consider an example here that is five years hence the age of vinod will be three times that of his son five years ago wino's age was seven times that of his son what are the percentages you will be able to see this kind of examples frequently in this chapter so let us understand first how we can form the equations using these statements and then we can solve them using substitution method step by step will understand this in the very first step we will understand what are the variables that we have to identify in this that the present we know is presentation and is the sun's present age these are the variables that we have to identify using this problem statement so i will assign windows percentage as x and his son's present age as y now according to the first statement that we will consider in step two five years hence means after five years his age will be for you more than whatever it is now therefore i'll consider x plus five that will be three times of that of his son if his age is being increased by five after five years his son's age will also be increased by phi after five years therefore y plus five three times of y plus five will be equal to x plus five according to the first statement means five years after windows presentation will be x plus five and according to this statement that will be 3 times of his sun is after 5 years that is 3 into 1 plus 5 when i simplify this i will get x plus 5 is equal to 3 y plus 3 into 5 that is 15 or i can simplify and keep it in standard form of linear equation that is x minus 3 y is equal to 10 this i will consider as equation number one similarly i will use second statement in order to form one more equation that is in next step that is according to second statement five years ago means x minus five if his percentage is five five years ago it will be x minus five we knows age was seven times of his sun's age five years ago is the sun's age will also be y minus five therefore x minus five will be equal to seven into y minus five according to the second statement when we simplify this we'll get x minus five is equal to 7 y into 7 into 5 will be 35 so when i keep this in the form of standard equation that is x minus 7 y is equal to minus 30 here that is second equation that i have got now i have got two equations of one pair which will make one pair and i can solve this pair of linear equations let us see how we can solve this by substitution method here i will consider or i will represent any one equation in terms of only one variable you can see here i have considered first equation that i have expressed only in terms of x means all the other terms i have sent on other side of the equals therefore x equal to 10 plus 3y i can also represent this equation only in terms of y but here i've considered in terms of x now this i will name it as equation number three now this third equation i will substitute in second equation remember this third equation i have formed using first equation so i will not substitute this third equation in one equation i will put it definitely in second equation now when i substitute this third equation in second equation it will look something like this that is in fourth step x i have substituted as 10 plus 3y here minus 7y as it is and minus 30 as it is or when i open this bracket it will be 10 plus 3y minus 7y it will be 10 minus 4y that is equal to minus 30 or i can write taking 30 on the left side of equation and sending 4y on the right side of equals and that will be 10 plus 30 that is equal to 4y therefore 10 plus 30 is equal to 40 i get 4y is equal to 40 therefore y value i get it as 10. now substituting this y is equal to 10 in any of the equation that is 1 2 or 3 i will get x value i will substitute y is equal to 10 in equation number 3 when i do it i will get answer as x equal to 10 into 10 plus 3 into y y value is 10 that is 10 plus 3 into 10 so i will get x value as 40 therefore i can write here you can see x i have ascendant we know the percentage therefore minus percentage is 40 and his son's percentage is 10. you can verify these two answers by substituting in the statement given in the problem both the statement that is first statement and second statement and verify whether we have got the correct answer now no therefore presentation of we know this 40 years and that of his son is 10 years this is how we can go ahead with the substitution method wherein one equation is represented as only one variable like this that is x and that equation is substituted in another equation therefore it is called as substitution method now let us verify whether we have got the correct answer or no through graphical solution that is these are the two equations that i had formed in the previous slide that is x minus 3 y is equal to 10 and x minus 7 y is equal to minus 30 when i plot graph of these two equations i get something like this you can see here these two lines formed by these two equations are going to intersect at one common point that is x equal to 40 and y is equal to 10 therefore x equal to 14 y is equal to 10 is the only one common solution for these two equations which means present age of we know is 40 and present age of his son is 10 years therefore in graphical solution also we are getting same answer as x equal to 14 y is equal to 10 therefore our answer is matching with graphical solution and the substitution method therefore x equal to 14 y is equal to 10 is the correct answer now the second method that we are going to see in algebraic methods is elimination method i am going to consider an example in order to explain you this elimination method let us see what is that example meena went to a bank to withdraw rupees 2000. she asked the cashier to give her rupees 50 and rupees 100 notes only meena got 25 notes in all find how many notes of rupees 50 and rupees 100 she received here we are supposed to find number of 50 rupees nodes and number of hundred rupees nodes therefore these two terms are going to act as x and y in forming the equation first in the very first step we need to form the equations depending on this problem statement then we are going to solve them using elimination method very first step is let x be the number of 50 rupees nodes and y be the number of hundred rupees nodes these things i am going to assign in this problem then in the second step i will form the equation it is given that she got 25 nodes in r means if x are number of 50 rupees notes and wire number of 100 rupees notes x and y total she has got 25 so x plus y is equal to 25 this is the first statement or first equation that we have formed using the first statement now we know that total rupees is 2 000 50 rupees notes multiplied by number of 50 rupees notes plus 100 rupees notes multiplied by number of hundred rupees notes is going to give me total 2 000 rupees so this i can represent as second equation now remember one thing in order to solve problems using elimination method coefficient of any one variable in both the equation should be equal but here if we can see coefficient of x is 1 in first equation and coefficient of y is also 1 in the first equation but in second equation coefficient of x is 50 and coefficient of y is 100 none of those coefficients are matching or they are same so in order to make coefficients same in both the equations i have to multiply throughout the equation by one particular number which will give me equal coefficients in both the equations therefore i will consider equation 1 and multiply that equation throughout by 50 so that i will get coefficient of x in equal remember whenever we add subtract multiply or divide by any particular number that has to be done throughout the equation and not only to the particular variable here multiplying equation 1 by 50 throughout therefore when i do it i am not changing value of this equation it is going to remain same therefore i will get it as 50x plus 50y is equal to 25 into 50 that will give me 1 to 5 0 so now you can see equation 3 and equation 2 here coefficient of x are same now i can subtract equation 3 in equation 2 so that i will get a proper solution when i do that i will be eliminating this x term completely and i will remain i will have only y terms remaining so next step is subtracting equation 3 from equation 2. when i do that first i will write down equation 2 as it is below right below that i will write distance 3 as it is with the minus sign i have purposely kept this third equation in a bracket so that when i open the bracket or for this minus sign all the signs are going to change therefore plus 50 x will become minus 50 x and plus 5y will become minus 5y and plus 1 to 5 will become minus 1 to 5 0 now you can see that plus 50 x and minus 58 x get cancelled and they are eliminated so by doing this i have eliminated x term in both the equation therefore this name of this method is called as what elimination method that one thing you have to remember here after that we have 100y minus 50y we will have plus 50y that is equal to two thousand minus one two five zero i have 750 by solving this i get what 50y is equal to 750 when i when i solve this y is equal to 750 upon 50 so i will get answer as y is equal to 15 therefore put y is equal to 15 in equation 1 now our equation 1 was x plus y is equal to 25 when i put y is equal to 15 in that equation i will get x is equal to 25 minus 15 therefore x is equal to 10 i got which means x we had assigned to number of 50 rupees notes and why we had assigned to 100 rupees notes therefore we can say she got 10 notes of 50 rupees and 50 notes of 100 rupees we can substitute these values in the equation 1 and equation 2 and verify whether we have got the correct answers or no now let us understand whether we have solved it correctly or no by uh verifying it considering graphical solution so we have uh two equation that is x plus y is equal to 25 and 50 x plus 100 y is equal to 2 000 when we solve this graphically by considering different values of x and y we get such pattern wherein both the lines are going to meet at one particular point that is x equal to 10 and y is equal to 15 which means for one common solution of x equal to 10 and y is equal to 15 these both equations are going to get satisfied therefore correct answer for this two equation is going to be x equal to 10 and y is equal to 15. the third method that we are going to consider for solving the pair of linear equation is cross multiplication method in order to solve this method first we need to form a certain formula using the given pair of linear equations then we can solve that method substituting the particular values in that formula let us understand how we can form the formula first by considering the equations first we have to note on the given pair of linear equations in general form we have studied how to represent given pair of linear equation in general form that is going to be a 1 x plus b 1 y plus c 1 is equal to 0 and second equation can be written as a 2 x plus b 2 y plus c 2 is equal to 0 now using these two equations we have to first formulate a formula let us understand how we can do that first we can write x as it is then in the denominator leaving the coefficient that are present in x using other coefficient that is b1 c1 and b2 c2 we have to write the denominator part for x let us understand how we can do that considering arrow from top to down in these two terms of the equations and subtracting that from the terms that is uh imagining considering the arrow from down to up that is b1 c2 minus b2 c1 you can see here b1 c2 minus b2 c1 so first part of our this big formula is over now second part we are going to consider as y y divided by in the denominator now we are going to leave all those terms uh coefficient that are present in y term means middle term we are going to leave then we are going to only consider c1 c2 and even e2 now in order to form this second part of this equation we are going to assume that again a1 x and a2 x are repeated next to c1 and c2 here so when i again consider an arrow from top to down it will be c1 down down will be a2 here if i imagine this is here that will be a 2 c 1 minus down to up that is c 2 a 1 or i can write it as even c 2 so second part of my equation is over now for formulating the third part i can consider constant term as one here here for one divided by now leaving terms present in constant term i will form the denominator part using this these terms here that is from up to down that is a one b two here minus a 2 b 1 a 2 b 1 you can see now by writing these terms i have formulated this formula now substituting the given values of a 1 a 2 b 1 b 2 and c 1 c 2 in the given pair of linear equation i can solve the equation and form value for x and y let us understand that with an example here i am considering an example that is find the values of x and y by cross multiplication method so pair of given linear equations are 2x plus y is equal to 5 and 3x plus 2y is equal to 8 first i need to write down these two equations in general form that is 2x plus y minus 5 is equal to 0 and 3x plus 2y minus 8 is equal to 0 the first these are written in general form that is this form now we have to substitute all the values in this formula so formula is x divided by b 1 c 2 here you can see in the first term that is b 1 is present nothing means 1 we are going to take it as 1 c 2 is minus 8 minus b 2 b 2 value is 2 here and c 1 c 1 value is minus 5 here so first part is over second part is y divided by a 2 c 1 that is a 2 is 3 c 1 is minus 5 so 3 into minus 5 and this minus as it is a 1 into c2 a1 is 2 c2 is minus 8 that is 2 into minus 8 in the third term 1 as it is divided by a1 b2 a1 is 2 and b2 is also 2 that is 2 into 2 this minus as it is then we have a 2 into b 1 a 2 is 3 and b 1 is 1 that i have 3 into 1 when i simplify this in the next step that i have x divided by 8 into 1 is 8 plus into minus is minus so we have minus 8 minus minus become plus here 5 into 2 is 10 that is minus 8 plus 10 here then y divided by 5 into 3 is 15 plus into minus is minus so i have minus 15 then minus minus become plus 8 into 2 is 16 then 1 divided by 2 to the 4 and 3 ones are 3 so we have simplified this in the next step further when i solve this 8 upon 10 minus 8 is going to be 2 that is x by 2 will be equal to y by 16 minus 15 that is equal to 1 y by 1 that should be equated to the both these terms are going to be equated to last term that is 1 divided by 4 minus 3 is again 1 so 1 by 1 is nothing but 1 therefore x divided by 2 is equated to 1 also y divided by 1 is equated to 1 when i solve these two finally i get x value as 2 and y value as 1 this is how if we can obtain x and y values by forming first this formula then substituting values of a 1 a 2 b 1 between c 1 c 2 in this formula and then by doing cross multiplication we are obtaining x and y values that is why name of this method is known as cross multiplication method hope you have understood all three methods of solving pair of linear equation using algebraic methods now we have arrived at the last part of this chapter that is equations reducible to pair of linear equations in two variables suppose variables are present in the denominator of any equation that cannot be considered as a linear equation therefore those equations first has to be converted in in the form of linear equations by making suitable substitution then we can solve them as the linear equations using any available algebraic method let us understand how we can convert any equation in the linear equation first and then we can solve this with an example solve the following pair of equations by reducing them to a pair of linear equations here we have two equations given you can see as i told you variables are present in the denominator in that case first we need to make suitable substitution then convert these two equations in the form of linear equation then we can solve here you can see in both the equation first term is 1 upon x and in second term 1 upon y minus 2 is present so generally i can consider 1 upon x minus 1 as p and 1 upon y minus 2 as q and after that i can form the equation as it will become put 1 upon x minus 1 is equal to p and 1 upon y minus 2 is equal to q then when i make these substitutions these equations can be reduced to for you p plus 1 by y minus 2 is q therefore 5 p plus q is equal to 2 and 6 p minus 3 q is equal to 1 now we have two linear equations here now you already know how to solve this pair of linear equations using any method of algebraic solution we can find the solution of p and q i am using elimination method here in order to solve these two equations we have understood already that very variable coefficient of any one variable should be same but here none of the variables are having same coefficients so what i will do first i will multiply equation 1 by 3 then i will subtract or add that one equation in equation 2. have minus three q in equation two so and we have plus sign here so if i have three q here then i can add these two equation in order to eliminate q term therefore what i will do is multiply equation one by 3 and add equation 1 to equation 2 when i do that i will get 6 p plus 15 p i will get it as 21 p and q term is eliminated here that is minus three q plus liquid you get cancelled and it will become zero and one plus six i'm going to get as seven therefore p value i'm going to get as one by three and when i substitute p is equal to one by three in any of these two equations i can obtain q value as 1 by 3 this you already know how to obtain p and q values uh by solving any method we have discussed about this in previous methods so i got two values of p and q now further we have assigned 1 by x equal to p and 1 by y minus 2 is equal to q here so i will put p value and key value in this substitutions then i will obtain x and y values individually so i get 1 upon x minus 1 is equal to 1 by 3 because that is equal to p and p value is 1 by 3 i will do cross multiplication here 3 will go up and x minus 1 will come here therefore 3 is equal to x minus 1 now when i solve this i get x value as 4 similarly i will equate 1 upon y minus 2 is equal to q but q value is 1 by 3 here therefore when i substitute all the values here and do cross multiplication 1 into 3 is 3 and y minus 2 into 1 is 5 minus 2 therefore i get 3 is equal to y minus 2 or y is equal to 5 like this we can obtain x and y values by solving this method by first converting it to reducing it to linear equation then solving it by any available algebraic method hope you understood all the methods that we have considered in this chapter and all the different different concepts of this chapter so i will come with the next video taking next chapter of sslc mathematics till then have a good day and take care thank you for watching the video