Linear equations in two variables, explain the geometry of lines or the graph of two lines, plotted to solve the given equations. As we already know, the linear equation represents a straight line. The plotting of these graphs will help us to solve the equations, which consist of unknown variables. Previously we have learned to solve linear equations in one variable, here we will find the solutions for the equations having two variables. Show  DefinitionAn equation is said to be linear equation in two variables if it is written in the form of ax + by + c=0, where a, b & c are real numbers and the coefficients of x and y, i.e a and b respectively, are not equal to zero. For example, 10x+4y = 3 and -x+5y = 2 are linear equations in two variables. The solution for such an equation is a pair of values, one for x and one for y which further makes the two sides of an equation equal. Also, read: Solution of Linear Equations in Two VariablesThe solution of linear equations in two variables, ax+by = c, is a particular point in the graph, such that when x-coordinate is multiplied by a and y-coordinate is multiplied by b, then the sum of these two values will be equal to c. Basically, for linear equation in two variables, there are infinitely many solutions. ExampleIn order to find the solution of Linear equation in 2 variables, two equations should be known to us. Consider for Example: 5x + 3y = 30 The above equation has two variables namely x and y. Graphically this equation can be represented by substituting the variables to zero. The value of x when y=0 is 5x + 3(0) = 30 ⇒ x = 6 and the value of y when x = 0 is, 5 (0) + 3y = 30 ⇒ y = 10 It is now understood that to solve linear equation in two variables, the two equations have to be known and then the substitution method can be followed. Let’s understand this with a few example questions. Unique SolutionFor the given linear equations in two variables, the solution will be unique for both the equations, if and only if they intersect at a single point. The condition to get the unique solution for the given linear equations is, the slope of the line formed by the two equations, respectively, should not be equal. Consider, m1 and m2 are two slopes of equations of two lines in two variables. So, if the equations have a unique solution, then: m1 ≠ m2 No SolutionIf the two linear equations have equal slope value, then the equations will have no solutions. m1 = m2 This is because the lines are parallel to each other and do not intersect. System of Linear Equations in Two VariablesInstead of finding the solution for a single linear equation in two variables, we can take two sets of linear equations, both having two variables in them and find the solutions. So, basically the system of linear equations is defined when there is more than one linear equation. For example, a+b = 15 and a-b = 5, are the system of linear equations in two variables. Because, the point a = 10 and b = 5 is the solution for both equations, such as: a+b=10 + 5 = 15 a-b=10-5 = 5 Hence, proved point (10,5) is solution for both a+b=15 and a-b=5. Problems and SolutionsQuestion: Find the value of variables which satisfies the following equation: 2x + 5y = 20 and 3x+6y =12. Solution: Using the method of substitution to solve the pair of linear equation, we have: 2x + 5y = 20…………………….(i) 3x+6y =12……………………..(ii) Multiplying equation (i) by 3 and (ii) by 2, we have: 6x + 15y = 60…………………….(iii) 6x+12y = 24……………………..(iv) Subtracting equation (iv) from (iii) 3y = 36 ⇒ y = 12 Substituting the value of y in any of the equation (i) or (ii), we have 2x + 5(12) = 20 ⇒ x = −20 Therefore, x=-20 and y =12 is the point where the given equations intersect. Now, it is important to know the situational examples which are also known as word problems from linear equations in 2 variables. Check: Linear Equations Calculator Word ProblemsQuestion 1: A boat running downstream covers a distance of 20 km in 2 hours while for covering the same distance upstream, it takes 5 hours. What is the speed of the boat in still water? Solution: These types of questions are the real-time examples of linear equations in two variables. In water, the direction along the stream is called downstream. And, the direction against the stream is called upstream. Let us consider the speed of a boat is u km/h and the speed of the stream is v km/h, then: Speed Downstream = (u + v) km/h Speed Upstream = (u – v) km/h We know that, Speed = Distance/Time So, the speed of boat when running downstream = (20⁄2) km/h = 10 km/h The speed of boat when running upstream = (20⁄5) km/h = 4 km/h From above, u + v = 10…….(1) u – v = 4 ………. (2) Adding equation 1 and 2, we get: 2u = 1 u = 7 km/h Also, v = 3 km/h Therefore, the speed of the boat in still water = u = 7 km/h Question 2: A boat running upstream takes 6 hours 30 minutes to cover a certain distance, while it takes 3 hours to cover the same distance running downstream. What is the ratio between the speed of the boat and speed of the water current, respectively? Solution: If the speed downstream is a km/hr and the speed upstream is b km/hr, then Speed in still water = (a + b)/2 km/h Rate of stream = ½ (a − b) kmph Let the Boat’s rate upstream be x kmph and that downstream be y kmph. Then, distance covered upstream in 6 hrs 30 min = Distance covered downstream in 3 hrs. ⇒ x × 6.5 hrs = y × 3hrs ⇒ 13/2x = 3y ⇒ y = 13x/6 \(\begin{array}{l}The\ required\ ratio\ is\ \frac{y + x}{2}~ :~ \frac{y – x}{2}\end{array} \) \(\begin{array}{l}\Rightarrow~\frac{\frac{13x}{6}~+~x}{2}~:~\frac{\frac{13x}{6}~-~x}{2}\end{array} \) \(\begin{array}{l}\Rightarrow~\frac{\frac{19x}{6}}{2}~:~\frac{\frac{7x}{6}}{2}\end{array} \) = 19:7 For a system of linear equations in two variables, we can find the solutions by the elimination method. For linear equations in two variables, there are infinitely many solutions. A linear equation in two variables is an equation which has two different solutions. The coefficient of x is 3 and the coefficient of y is -6. The constant of the equation 3x-6y=-13 is -13.
3.1 Introduction Linear equations in two variables are equations where we have two variables. We require two equations to find the solution of linear equations in two variables. Let’s understand the basic concept of linear equations in two variables. 3.2 lINEAR EQUATIONS IN TWO VARIABLES In case of a pair of linear equations, the values of x and y, which satisfy both the equations, are the solutions of the equation. Geometric representation of a Linear EquationsGeometrically, a linear equation represents a straight line. Every solution of an equation is a point on the line represented by the equation.Likewise, a pair of linear equations represents two straight lines. When their graphs are drawn, there are three possibilities. (i) If the lines intersect at a point, then that point gives the unique solution of the two equations. In this case, the pair of equations is consistent.(ii) If the lines coincide, then there are infinitely many solutions — each point on the line being a solution. In this case, the pair of equations is dependent (consistent).(iii) If the lines are parallel, then the pair of equations has no solution. In this case, the pair of equations is inconsistent. Example 1. Find whether the given pairs of linear equations are consistent or inconsistent? 4x + y = 87x – 2y = 1 Solution: 4x + y = 87x – 2y = 1 Here, a1 = 4 b1 = 1 c1 = –8 a2 = 7 b2 = –2 c2 = –1 ∴ a1a2=47;b1 b2=1–2=–12;c1c2=–8–1=8 Therefore, according to the above rule, this system of equations has a unique solution. Example 2. There are two lines drawn in each of the following graphs. Each line represents a linear equation in two variables. State whether the pair of linear equations represented in each graph is consistent, inconsistent, or dependent. (b) In the given graph, the lines representing the two equations intersect at a point. (c) In the given graph, the lines are coincident i.e., the same line represents the two equations. 3.3 Graphical Solution Of A Pair Of Linear Equations In Two Variables Suppose you go to a stationary shop to buy some registers and pens. If you buy two registers and two pens, then you have to pay Rs 30. However, if you buy two registers and four pens, then you have to pay Rs 40.Now, how can we represent this information algebraically, i.e., in the form of equations?Let us denote the cost of each register by Rs x and that of each pen by Rs y.Now, it is given to us that two registers and two pens cost Rs 30. ∴ 2x + 2y = 30 ⇒ x + y – 15 = 0 _______________ (1)We also know that two registers and four pens cost Rs 40. ∴ 2x + 4y = 40 ⇒ x + 2y – 20 = 0 ______________ (2)Thus, we now have two linear equations in two variables from the given situation.x + y – 15 = 0 ______________ (1)x + 2y – 20 = 0 ______________ (2) These two equations are known as a pair of linear equations in two variables. If we try to individually solve each equation, we will be unable to do so. c2We can arrive at an answer only if we try and solve both these equations together.The general form of a pair of linear equations in two variables is written as a1x + b1y +c1 = 0 where a1, b1, c1 are real numbers, and a12+b12≠0 (i.e., a1, b1≠0) The lines represented by a pair of linear equations in two variables, a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0, (a) have a unique solution (or are consistent) and represent intersecting lines if a1a2≠b1 b2 (b) have infinite solutions (or are dependent) and represent coincident lines if a1a2=b1b2=c1c2 (c) have no solution (or are inconsistent) and represent parallel lines if a1a2=b1b2≠c1c2A pair of linear equations represents two straight lines. When their graphs are drawn, there are three possibilities.(i) The two lines may intersect at one point.(ii) The two lines may be parallel.(iii) They may represent the same line, i.e. they may be coincident.The solution of a pair of linear equations is the point of intersection of their graphs.Now, according to this, a pair of dependent equations is represented by coincident lines. Example 3. Anshuman and Vikram have 10 marbles with them. Twice the number of marbles with Anshuman is equal to three times the number of marbles with Vikram. Solve this question graphically. Solution : Let us suppose that Anshuman has x marbles and Vikram has y marbles. From the given information, we can form two equations as x + y = 10 x + y – 10 = 0 –––––––– (i)2x = 3y 2x – 3y = 0 –––––––– (ii) A table can be drawn for the corresponding values of x and y for equation (i) as
Similarly, a table can be drawn for the corresponding values of x and y for equation (ii) as
The two equations can be plotted on a graph as As seen in the graph, the two lines intersect at the point (6, 4).This implies that the solution for the pair of linear equations is x = 6 and y = 4. Thus, Anshuman has 6 marbles while Vikram has 4 marbles. Example 4 Find graphically whether the following pairs of linear equations are consistent or inconsistent? If the equations are consistent, then find their solutions as well.(a) 2x + 3y = 24x + 3y = 4(b) x + y = 52x + 2y = 10 Solution (a) 2x + 3y = 2; 4x + 3y = 4A table can be drawn for the corresponding values of x and y for the equation 2x + 3y = 2 as
Similarly, a table can be drawn for the corresponding values of x and y for the equation 4x + 3y = 4 as
These points can now be plotted and joined to obtain the graphs of the equations as Thus, the given pair of linear equations is consistent with its solution as x = 1 and y = 0. (b) x + y = 5 2x + 2y = 10
Similarly, a table can be drawn for the corresponding values of x and y for the equation 2x + 2y = 10 as
These points can now be plotted and joined to obtain the graphs of the lines as As seen in the graph, the two equations are represented by the same line. This means that the given pair of linear equations is dependent. Any point lying on this line is a solution to the pair of equations. 3.4 Expressing Given Situations MathematicallyWe come across many situations in real life when it is easy to find the solution, if we express them mathematically.Let us see such a situation. The coach of the school cricket team buys 5 bats and 20 leather balls for Rs 3500. After some time, some more boys joined the team for practice, so he buys another 4 bats and 15 balls for Rs 2750. Suppose that the price of bat and ball does not change in the time period.Can we express this situation mathematically to find out the individual prices of a ball and a bat?Let the price of a bat be Rs x and that of a ball be Rs y.It is given that 5 bats and 20 balls cost Rs 3500. ∴ Cost of 5 bats = 5x And cost of 20 balls = 20y⇒ Cost of 5 bats and 20 balls = 5x + 20y ⇒ 5x + 20y = 3500Similarly, it is also given that 4 bats and 15 balls cost Rs 2750. ⇒ 4x + 15y = 2750 Therefore, the algebraic representation of the given situation is5x + 20y = 3500 _______________ (1)4x + 15y = 2750 _______________ (2)After solving these equations, we can find out the individual prices of the ball and the bat. Example 5. Aman and Yash each have certain number of chocolates. Aman says to Yash, if you give me 10 of your chocolates, I will have twice the number of chocolates left with you. Yash replies, if you give me 10 of your chocolates, I will have the same number of chocolates as left with you. Write this situation mathematically? ⇒ x + 10 = 2(y – 10) x +10 = 2y – 20x – 2y + 30 = 0 _______________ (1)According to the second condition, Aman gives 10 chocolates to Yash such that both have equal number of chocolates.⇒ y +10 = x – 10 Thus, the algebraic representation of the given situation is x – y – 20 = 0 __________ (2) Example 6. Cadets in the military academy are made to stand in rows. If one cadet is extra in each row, then there would be 2 rows less. If one cadet is less in each row, then there would be 3 more rows. Express the given situation mathematically? ∴ xy = (y – 2) (x + 1) xy = xy – 2x + y – 22x – y = – 2 _________________ (1)It is also given to us that if one cadet is less in each row, then there are 3 more rows.∴ xy = (y + 3) (x – 1) xy = xy + 3x – y – 33x – y = 3 _______________ (2)Thus, the algebraic representation of the given situation is2x – y = – 2 _______________ (1)3x – y = 3 _______________ (2) 3.5. Substitution Method Of Solving Pairs Of Linear EquationsIn a class, the number of boys is 7 more than twice the number of girls.Can you find the number of boys and girls in the class?Let the number of boys be x and the number of girls be y.Now, according to the given condition, x – 2y = 7We cannot find the unique values of x and y by solving this equation because there are multiple values of x and y for which this equation holds true.We can write the above equation as x = 2y + 7. Thus, by taking different values of y, we will obtain different values of x.However, if we are given one more condition, then the values of x and y can be evaluated.To solve a linear equation in two variables, two linear equations in the same two variables are required.Let us consider one more condition. Suppose the total number of students in the class is 52. Now, can you find the number of boys and girls in the class? According to this condition, we can write x + y = 52Therefore, we obtain a pair of linear equations in two variables.x – 2y = 7x + y = 52Let us solve this pair of linear equations by Substitution method.x – 2y = 7 ____ (1)x + y = 52 ____ (2)y = 52 – xsubstitute this value in equation (1) we getx – 2(52 – x) = 7x – 104 + 2x = 73x = 104 + 73x = 111x = 37put this value in (2) we gety = 52 – 37y = 15 Therefore, number of boys = 37 and number of girls = 15 In this method, we have substituted the value of one variable by expressing it in terms of the other variable to solve the pair of linear equations. Example 7. Solve the following pair of linear equations.7x – 5y + 12 = 03x + 8y – 5 = 0 Solution Given, 7x – 5y + 12 = 0 ––––––––– (i)3x + 8y – 5 = 0 ––––––––– (ii)From equation (i), we obtain7x – 5y + 12 = 0y=7x+125 ––––––––– (iii) On substituting the value of y in equation (ii), we obtain 3x+87x+125–5=0 ⇒ 3x+56x+965–5=0 ⇒ 15x+56x+96–255=0 ⇒ 15x+56x+96–25=0 ⇒ 71x+71=0⇒71x=–71 3.6 Elimination method to solve a pair of linear equationsThere are many real life situations which can be represented in the form of linear equations. Let us begin with such a situation.Suppose you go to the market with your friend to buy clothes. You buy 3 shirts and 6 pairs of jeans from a shop and the total cost of your purchase comes out to be Rs 7000. Your friend, on the other hand buys 2 shirts and 3 pairs of jeans and the total cost of his purchase comes out to be Rs 3855. Now, we can represent this information mathematically as pair of linear equations in two variables. 3x + 6y = 7000 ______________ (i)2x + 3y = 3855 ______________ (ii) We can then solve the equations and get the cost of each shirt and each pair of jeans. Let’s solve them using the elimination method. On multiplying equation (ii) by 2 we obtain2(2x + 3y = 3855)4x + 6y = 7710 ______________ (iii)On subtracting equation (i) from (ii), we obtainx = 7710 – 7000x = 710 Substitute the above value in (i) we get3 × 710 + 6y = 70002130 + 6y = 70006y = 7000 – 2130y = 811.67Cost of one shirt = Rs 710 Cost of 1 pair of jeans = Rs 811.67 Example 8. Solve the following equations using elimination method.3x + 4y = 04x – 3y = 50 Solution Given, 3x + 4y = 0 ______________ (i)4x – 3y = 50 ______________ (ii)On multiplying equation (i) by 3 and equation (ii) by 4, we obtain9x + 12y = 0 ______________ (iii)16x – 12y = 200 _________ (iv)On adding equation (iii) and (iv), we obtain25x = 200⇒x=20025=8 Putting the value of x in equation (i), we obtain3 × 8 + 4y = 0⇒4y = – 24 ⇒ y=–244=–6 Thus, x = 8 and y = – 6. 3.7 Cross-Multiplication Method Of Solving Pairs Of Linear EquationsWe can represent many situations in real life as linear equation in two variables. Let us consider such a situation.Suppose Samay is older than Sumit by 30 years. After 5 years, Samay will be thrice as old as Sumit.Can we find the present ages of Samay and Sumit?Yes, we can find the present ages. However, before that, we have to represent this situation in the form of linear equations in two variables. Let the present age of Samay be x years and the present age of Sumit be y years.It is given that Samay is older than Sumit by 30 years. Thus, according to this condition, we havex – y = 30After 5 years, the age of Samay will be(x + 5) years and the age of Sumit will be (y + 5) years. However, it is given that after 5 years, Samay will be thrice as old as Sumit. Thus, we have x + 5 = 3 (y + 5) ⇒x + 5 = 3y + 15 (iii) Graphical method Now, we will follow another method to solve linear equations in two variables. Cross-multiplication method If a1x + b1y + c1 = 0, a1, b1≠ 0 and a2x + b2y +c2 = 0,a2, b2 ≠0, then xb1c2–b2c1=yc1a2–c2a1=1a1 b2–a2 b1 x – y = 30 ____________ (1)x – 3y = 10 ____________ (2) a1 = 1, b1 = –1, c1 = – 30 and a2 = 1, b2 = –3, c2 = – 10 x10–90=y–30+10=1–3+1 x–80=y–20=1–2 x–80=1–2⇒x=40 y–23=1–2⇒y=10 In this way, we can solve a pair of linear equations in two variables by cross-multiplication method.We could solve these linear equations in two variables by other methods also. But why should we use cross-multiplication method to solve linear equations? The cross-multiplication method is easier as compared to the other methods if the coefficients of variables in the equations are large numbers or numbers that look complex. From the above two equations, we obtain ⇒xb×(–2a)–1×–a2+b2 =y–a2+b2×1–(–2a)×a=1a×1–1×b ⇒ x–2ab+a2+b2=y–a2–b2+2a2=1a–b ⇒x(a–b)2=ya2–b2=1a–b Therefore, x(a–b)2=1a–b and ya2–b2=1a–b ⇒ x=(a–b)2a–b and y=a2–b2a–b x = a – b and y = (a–b)(a+b)a–b 3.8 Equations Reducible To A Pair Of Linear Equations In Two VariablesConsider the following equations. x = y ; 57x = 21y + 15; m + 20n = –13; etc. We can see that the above equations are linear equations in two variables.However, there are many equations, which do not appear to be linear equations at first glance such as 2x+3y=20;1x–1y=5; etc Example10.Solve the following pair of linear equations. 5x+1y=7 2x–3y=–4 Solution: 5x+1y=7 ––––––––––– (1) 51x+1y=7 ––––––––––– (3) From equation (2), we can write, Let 1x=a and 1y=bNow, equation (3) becomes5a + b = 7 ___________ (5)and equation (4) becomes2a – 3b = – 4 ___________ (6)On multiplying (5) by 3, we obtain15a + 3b = 21 ___________ (7)On adding equations (6) and (7), we obtain 17a = 17 ⇒a = 1 On putting the value of a in equation (5), we obtain5 × 1 + b = 7⇒5 + b = 7 ⇒b = 7 – 5 = 2 Since 1x=a=1 ⇒ y=12 Example11. Solve the following pair of linear equations. 2x–y+5x+y=73x–y+10x+y=9 Solution: The given equations can be written as and Let 1x–y=a and 1x+y=bThus, the equations become2a + 5b = 7 _________ (3)and, 3a + 10b = 9 _________ (4)Now, on multiplying equation (3) by 2, we obtain4a + 10b = 14 _____________ (5)On subtracting equation (4) from equation (5), we obtaina = 14 – 9 ⇒a = 5 Putting the value of a in equation (3), we obtain10 + 5b = 7 ⇒5b = 7 – 10 ⇒x–y=15 and x+y=–53 On adding these two equations, we obtain 2x=–2215 ⇒ x=–1115 and y=x–15 ⇒ y=–1115–15 ⇒ y=–11–315⇒ y=–1415 Thus, x=–1115 and y=–1415 Updated on April 21, 2021 |
How many solutions do two linear equations in two variables have if their lines in graph intersect at one point?
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