{"id":491,"date":"2023-05-04T10:00:55","date_gmt":"2023-05-04T04:30:55","guid":{"rendered":"https:\/\/www.aplustopper.com\/?p=491"},"modified":"2023-05-05T09:16:35","modified_gmt":"2023-05-05T03:46:35","slug":"graphical-method","status":"publish","type":"post","link":"https:\/\/www.aplustopper.com\/graphical-method\/","title":{"rendered":"Graphical Method Of Solving Linear Equations In Two Variables"},"content":{"rendered":"<h2><strong>Graphical Method Of Solving Linear Equations In Two Variables<\/strong><\/h2>\n<p>Let the system of pair of linear equations be<br \/>\na<sub>1<\/sub>x + b<sub>1<\/sub>y = c<sub>1<\/sub>\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0\u2026.(1)<br \/>\na<sub>2<\/sub>x + b<sub>2<\/sub>y = c<sub>2<\/sub>\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u2026.(2)<br \/>\nWe know that given two lines in a plane, only one of the following three possibilities can happen &#8211;<br \/>\n(i) The two lines will intersect at one point.<br \/>\n(ii) The two lines will not intersect, however far they are extended, i.e., they are parallel.<br \/>\n(iii) The two lines are coincident lines.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"\" src=\"https:\/\/cbsenotesblog.files.wordpress.com\/2016\/08\/graphical-method-1.png\" alt=\"graphical-method-1.png\" width=\"457\" height=\"173\" \/><br \/>\n<strong>Types Of Solutions:<\/strong><br \/>\nThere are three types of solutions<\/p>\n<ol>\n<li>Unique solution.<\/li>\n<li>Infinitely many solutions<\/li>\n<li>No solution.<\/li>\n<\/ol>\n<p><strong>Read More:<\/strong><\/p>\n<ul>\n<li><span style=\"font-weight: 400;\"><a href=\"https:\/\/www.aplustopper.com\/what-is-a-linear-equation\/\">What is a Linear Equation<\/a><\/span><\/li>\n<li><a href=\"https:\/\/www.aplustopper.com\/solution-of-linear-equation-two-variables\/\"><span style=\"font-weight: 400;\">Linear Equations In Two Variables<\/span><\/a><\/li>\n<li><a href=\"https:\/\/www.aplustopper.com\/linear-equations-in-one-variable\/\"><span style=\"font-weight: 400;\">Linear Equations In One Variable<\/span><\/a><\/li>\n<li><a href=\"https:\/\/www.aplustopper.com\/lesson\/rs-aggarwal-class-10-solutions-linear-equations-in-two-variables\/\"><span style=\"font-weight: 400;\">RS Aggarwal Class 10 Solutions Linear Equations In Two Variables<\/span><\/a><\/li>\n<li><a href=\"https:\/\/www.aplustopper.com\/lesson\/rs-aggarwal-class-9-solutions-linear-equations-in-two-variables\/\"><span style=\"font-weight: 400;\">RS Aggarwal Class 9 Solutions Linear Equations In Two Variables<\/span><\/a><\/li>\n<li><a href=\"https:\/\/www.aplustopper.com\/linear-equations-rs-aggarwal-class-8-maths-solutions-ex-8a\/\"><span style=\"font-weight: 400;\">RS Aggarwal Class 8 Solutions Linear Equations<\/span><\/a><\/li>\n<li><a href=\"https:\/\/www.aplustopper.com\/linear-equation-one-variable-rs-aggarwal-class-7-solutions\/\"><span style=\"font-weight: 400;\">RS Aggarwal Class 7 Solutions Linear Equations in One Variable<\/span><\/a><\/li>\n<li><span style=\"font-weight: 400;\"><a href=\"https:\/\/www.aplustopper.com\/linear-equation-one-variable-rs-aggarwal-class-6-maths-solutions-ex-9a\/\">RS Aggarwal Class 6 Solutions Linear Equation In One Variable<\/a><\/span><\/li>\n<\/ul>\n<p><strong>(A) Consistent:<\/strong> If a system of simultaneous linear equations has at least one solution then the system is said to be consistent.<br \/>\n<strong>(i) Consistent equations with unique solution:<\/strong> The graphs of two equations intersect at a unique point.<br \/>\n<strong>For Example<\/strong>\u00a0Consider<br \/>\nx + 2y = 4<br \/>\n7x + 4y = 18<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"\" src=\"https:\/\/cbsenotesblog.files.wordpress.com\/2016\/08\/graphical-method-21.png\" alt=\"graphical-method-2.png\" width=\"189\" height=\"200\" \/><br \/>\nThe graphs (lines) of these equations intersect each other at the point (2, 1) i.e., x = 2, y = 1.<br \/>\nHence, the equations are consistent with unique solution.<br \/>\n<strong>(ii) Consistent equations with infinitely many solutions:\u00a0<\/strong>The graphs (lines) of the two equations will be coincident.<br \/>\n<strong>For Example<\/strong>\u00a0Consider\u00a02x + 4y = 9 \u00a0 \u21d2 \u00a0 3x + 6y = 27\/2<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"\" src=\"https:\/\/cbsenotesblog.files.wordpress.com\/2016\/08\/graphical-method-3.png\" alt=\"graphical-method-3.png\" width=\"195\" height=\"188\" \/><br \/>\nThe graphs of the above equations coincide. Coordinates of every point on the lines are the solutions of the equations. Hence, the given equations are consistent with infinitely many solutions.<br \/>\n<strong>(B) Inconsistent Equation:\u00a0<\/strong>If a system of simultaneous linear equations has no solution, then the system is said to be inconsistent.<br \/>\n<strong>No Solution:<\/strong> The graph (lines) of the two equations are parallel.<br \/>\n<strong>For Example<\/strong>\u00a0Consider<br \/>\n4x + 2y = 10<br \/>\n6x + 3y = 6<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"\" src=\"https:\/\/cbsenotesblog.files.wordpress.com\/2016\/08\/graphical-method-4.png\" alt=\"graphical-method-4.png\" width=\"193\" height=\"200\" \/><br \/>\nThe graphs (lines) of the given equations are parallel. They will never meet at a point. So, there is no solution. Hence, the equations are inconsistent.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"\" src=\"https:\/\/cbsenotesblog.files.wordpress.com\/2016\/08\/graphical-method-5.png\" alt=\"graphical-method-5.png\" width=\"413\" height=\"116\" \/><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"\" src=\"https:\/\/cbsenotesblog.files.wordpress.com\/2016\/08\/graphical-method-6.png\" alt=\"graphical-method-6.png\" width=\"415\" height=\"228\" \/><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"\" src=\"https:\/\/cbsenotesblog.files.wordpress.com\/2016\/08\/graphical-method-7.png\" alt=\"graphical-method-7.png\" width=\"273\" height=\"176\" \/><br \/>\nFrom the table above you can observe that if the line a<sub>1<\/sub>x + b<sub>1<\/sub>y + c<sub>1<\/sub> = 0 and a<sub>2<\/sub>x + b<sub>2<\/sub>y + c<sub>2\u00a0<\/sub>= 0 are<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"\" src=\"https:\/\/cbsenotesblog.files.wordpress.com\/2016\/08\/graphical-method-8.png\" alt=\"graphical-method-8.png\" width=\"407\" height=\"162\" \/><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-12434\" src=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2016\/08\/graphical-method.jpg\" alt=\"\" width=\"680\" height=\"392\" srcset=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2016\/08\/graphical-method.jpg 680w, https:\/\/www.aplustopper.com\/wp-content\/uploads\/2016\/08\/graphical-method-300x173.jpg 300w\" sizes=\"(max-width: 680px) 100vw, 680px\" \/><\/p>\n<h2>Graphical Method Examples<\/h2>\n<p><iframe loading=\"lazy\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/mo8qxxrRce4?feature=oembed\" frameborder=\"0\" allowfullscreen><\/iframe><\/p>\n<p><strong>Example 1:<\/strong> \u00a0\u00a0\u00a0The path of highway number 1 is given by the equation x + y = 7 and the highway number 2 is given by the equation 5x + 2y = 20. Represent these equations geometrically.<br \/>\n<strong>Sol.<\/strong>\u00a0 \u00a0 We have, x + y = 7<br \/>\n\u21d2 y = 7 \u2013 x \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0&#8230;.(1)<br \/>\nIn tabular form<\/p>\n<table border=\"2\">\n<tbody>\n<tr>\n<td style=\"text-align: center;\" width=\"50\">x<\/td>\n<td style=\"text-align: center;\" width=\"49\">1<\/td>\n<td style=\"text-align: center;\" width=\"49\">4<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\" width=\"50\">y<\/td>\n<td style=\"text-align: center;\" width=\"49\">6<\/td>\n<td style=\"text-align: center;\" width=\"49\">3<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\" width=\"50\">Points<\/td>\n<td style=\"text-align: center;\" width=\"49\">A<\/td>\n<td style=\"text-align: center;\" width=\"49\">B<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>and 5x + 2y = 20<br \/>\n\u21d2 y = \\(\\frac { 20-5x }{ 2 }\\) \u00a0 \u00a0 \u00a0 \u00a0 &#8230;.(2)<br \/>\nIn tabular form<\/p>\n<table style=\"height: 92px;\" border=\"2\" width=\"156\">\n<tbody>\n<tr>\n<td style=\"text-align: center;\" width=\"50\">x<\/td>\n<td style=\"text-align: center;\" width=\"49\">2<\/td>\n<td style=\"text-align: center;\" width=\"49\">4<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\" width=\"50\">y<\/td>\n<td style=\"text-align: center;\" width=\"49\">5<\/td>\n<td style=\"text-align: center;\" width=\"49\">0<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\" width=\"50\">Points<\/td>\n<td style=\"text-align: center;\" width=\"49\">C<\/td>\n<td style=\"text-align: center;\" width=\"49\">D<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Plot the points A (1, 6), B(4, 3) and join them to form a line AB.<br \/>\nSimilarly, plot the points C(2, 5). D (4, 0) and join them to get a line CD. Clearly, the two lines intersect at\u00a0the point C. Now, every point on the line AB gives us a solution of equation (1). Every point on CD gives us\u00a0a solution of equation (2).<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-134825\" src=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2018\/08\/Graphical-Method-Of-Solving-Linear-Equations-In-Two-Variables-1.png\" alt=\"Graphical Method Of Solving Linear Equations In Two Variables 1\" width=\"279\" height=\"284\" \/><\/p>\n<p><strong>Example 2:<\/strong> \u00a0\u00a0\u00a0 A father tells his daughter, \u201c Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be.\u201d Represent this situation algebraically and graphically.<br \/>\n<strong>Sol.<\/strong>\u00a0 \u00a0Let the present age of father be x years and that of daughter = y years<br \/>\nSeven years ago father\u2019s age = (x \u2013 7) years<br \/>\nSeven years ago daughter\u2019s age = (y \u2013 7) years<br \/>\nAccording to the problem<br \/>\n(x \u2013 7) = 7(y \u2013 7) \u00a0or \u00a0x \u2013 7y = \u2013 42 \u00a0 \u00a0 \u00a0&#8230;.(1)<br \/>\nAfter 3 years father\u2019s age = (x + 3) years<br \/>\nAfter 3 years daughter\u2019s age = (y + 3) years<br \/>\nAccording to the condition given in the question<br \/>\nx + 3 = 3(y + 3) \u00a0or \u00a0x \u2013 3y = 6 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0&#8230;.(2)<br \/>\nx \u2013 7y = \u201342 \u00a0\u00a0\u21d2 \u00a0 \\(y=\\frac { x+42 }{ 7 }\\)<\/p>\n<table style=\"height: 92px;\" border=\"2\" width=\"219\">\n<tbody>\n<tr>\n<td style=\"text-align: center;\" width=\"50\">x<\/td>\n<td style=\"text-align: center;\" width=\"49\">0<\/td>\n<td style=\"text-align: center;\" width=\"49\">7<\/td>\n<td style=\"text-align: center;\" width=\"45\">14<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\" width=\"50\">y<\/td>\n<td style=\"text-align: center;\" width=\"49\">6<\/td>\n<td style=\"text-align: center;\" width=\"49\">7<\/td>\n<td style=\"text-align: center;\" width=\"45\">8<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\" width=\"50\">Points<\/td>\n<td style=\"text-align: center;\" width=\"49\">A<\/td>\n<td style=\"text-align: center;\" width=\"49\">B<\/td>\n<td style=\"text-align: center;\" width=\"45\">C<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>x \u2013 3y = 6 \u00a0\u00a0\u21d2 \u00a0 \\(y=\\frac { x-6 }{ 3 }\\)<\/p>\n<table border=\"2\">\n<tbody>\n<tr>\n<td style=\"text-align: center;\" width=\"50\">x<\/td>\n<td style=\"text-align: center;\" width=\"49\">6<\/td>\n<td style=\"text-align: center;\" width=\"49\">12<\/td>\n<td style=\"text-align: center;\" width=\"45\">18<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\" width=\"50\">y<\/td>\n<td style=\"text-align: center;\" width=\"49\">0<\/td>\n<td style=\"text-align: center;\" width=\"49\">2<\/td>\n<td style=\"text-align: center;\" width=\"45\">4<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\" width=\"50\">Points<\/td>\n<td style=\"text-align: center;\" width=\"49\">D<\/td>\n<td style=\"text-align: center;\" width=\"49\">E<\/td>\n<td style=\"text-align: center;\" width=\"45\">F<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Plot the points A(0, 6), B(7, 7), C(14, 8) and join them to get a straight line ABC. Similarly plot the points D(6, 0), E(12, 2) and F(18, 4) and join them to get a straight line DEF.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-134827\" src=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2018\/08\/Graphical-Method-Of-Solving-Linear-Equations-In-Two-Variables-2.png\" alt=\"Graphical Method Of Solving Linear Equations In Two Variables 2\" width=\"281\" height=\"179\" \/><\/p>\n<p><iframe loading=\"lazy\" width=\"500\" height=\"375\" src=\"https:\/\/www.youtube.com\/embed\/ZRXRkXO-2K0?feature=oembed\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture\" allowfullscreen><\/iframe><\/p>\n<p><strong>Example 3:<\/strong> \u00a0\u00a0\u00a0 10 students of class X took part in a Mathematics quiz. If the number of girls is 4 more than the number of\u00a0boys, find the number of boys and girls who took part in the quiz.<br \/>\n<strong>Sol.<\/strong>\u00a0 \u00a0 Let the number of boys be x and the number of girls be y.<br \/>\nThen the equations formed are<br \/>\nx + y = 10 \u00a0 \u00a0 \u00a0&#8230;.(1)<br \/>\nand y = x + 4 \u00a0 \u00a0 &#8230;.(2)<br \/>\nLet us draw the graphs of equations (1) and (2) by finding two solutions for each of the equations. The<br \/>\nsolutions of the equations are given.<br \/>\nx + y = 10 \u00a0\u00a0\u21d2 \u00a0y = 10 \u2013\u00a0x<\/p>\n<table border=\"2\">\n<tbody>\n<tr>\n<td style=\"text-align: center;\" width=\"50\">x<\/td>\n<td style=\"text-align: center;\" width=\"49\">0<\/td>\n<td style=\"text-align: center;\" width=\"49\">8<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\" width=\"50\">y<\/td>\n<td style=\"text-align: center;\" width=\"49\">10<\/td>\n<td style=\"text-align: center;\" width=\"49\">2<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\" width=\"50\">Points<\/td>\n<td style=\"text-align: center;\" width=\"49\">A<\/td>\n<td style=\"text-align: center;\" width=\"49\">B<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>y = x + 4<\/p>\n<table border=\"2\">\n<tbody>\n<tr>\n<td style=\"text-align: center;\" width=\"50\">x<\/td>\n<td style=\"text-align: center;\" width=\"49\">0<\/td>\n<td style=\"text-align: center;\" width=\"49\">1<\/td>\n<td style=\"text-align: center;\" width=\"45\">3<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\" width=\"50\">y<\/td>\n<td style=\"text-align: center;\" width=\"49\">4<\/td>\n<td style=\"text-align: center;\" width=\"49\">5<\/td>\n<td style=\"text-align: center;\" width=\"45\">7<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\" width=\"50\">Points<\/td>\n<td style=\"text-align: center;\" width=\"49\">C<\/td>\n<td style=\"text-align: center;\" width=\"49\">D<\/td>\n<td style=\"text-align: center;\" width=\"45\">E<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Plotting these points we draw the lines AB and CE passing through them to represent the equations. The two lines AB and Ce intersect at the point E (3, 7). So, x = 3 and y = 7 is the required solution of the pair of linear equations.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-134828\" src=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2018\/08\/Graphical-Method-Of-Solving-Linear-Equations-In-Two-Variables-3.png\" alt=\"Graphical Method Of Solving Linear Equations In Two Variables 3\" width=\"341\" height=\"265\" srcset=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2018\/08\/Graphical-Method-Of-Solving-Linear-Equations-In-Two-Variables-3.png 341w, https:\/\/www.aplustopper.com\/wp-content\/uploads\/2018\/08\/Graphical-Method-Of-Solving-Linear-Equations-In-Two-Variables-3-300x233.png 300w\" sizes=\"(max-width: 341px) 100vw, 341px\" \/><br \/>\ni.e. Number of boys = 3<br \/>\nNumber of girls = 7.<br \/>\n<strong>Verification:<\/strong><br \/>\nPutting x = 3 and y = 7 in (1), we get<br \/>\nL.H.S. = 3 + 7 = 10 = R.H.S.,\u00a0(1) is verified.<br \/>\nPutting x = 3 and y = 7 in (2), we get<br \/>\n7 = 3 + 4 = 7, (2) is verified.<br \/>\nHence, both the equations are satisfied.<\/p>\n<p><strong>Example 4:<\/strong> \u00a0\u00a0\u00a0 Half the perimeter of a garden, whose length is 4 more than its width is 36m. Find the dimensions of the garden.<br \/>\n<strong>Sol.<\/strong>\u00a0 \u00a0 Let the length of the garden be x and width of the garden be y.<br \/>\nThen the equation formed are<br \/>\nx = y + 4 \u00a0 \u00a0 &#8230;.(1)<br \/>\nHalf perimeter = 36<br \/>\nx + y = 36 \u00a0 \u00a0 \u00a0 &#8230;.(2)<br \/>\ny = x &#8211; 4<\/p>\n<table border=\"2\">\n<tbody>\n<tr>\n<td style=\"text-align: center;\" width=\"50\">x<\/td>\n<td style=\"text-align: center;\" width=\"49\">0<\/td>\n<td style=\"text-align: center;\" width=\"49\">4<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\" width=\"50\">y<\/td>\n<td style=\"text-align: center;\" width=\"49\">-4<\/td>\n<td style=\"text-align: center;\" width=\"49\">0<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\" width=\"50\">Points<\/td>\n<td style=\"text-align: center;\" width=\"49\">A<\/td>\n<td style=\"text-align: center;\" width=\"49\">B<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>y = 36 &#8211; x<\/p>\n<table border=\"2\">\n<tbody>\n<tr>\n<td style=\"text-align: center;\" width=\"50\">x<\/td>\n<td style=\"text-align: center;\" width=\"49\">10<\/td>\n<td style=\"text-align: center;\" width=\"49\">20<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\" width=\"50\">y<\/td>\n<td style=\"text-align: center;\" width=\"49\">26<\/td>\n<td style=\"text-align: center;\" width=\"49\">16<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\" width=\"50\">Points<\/td>\n<td style=\"text-align: center;\" width=\"49\">C<\/td>\n<td style=\"text-align: center;\" width=\"49\">D<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Plotting these points we draw the lines AB and CD passing through them to represent the equations.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-134829\" src=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2018\/08\/Graphical-Method-Of-Solving-Linear-Equations-In-Two-Variables-4.png\" alt=\"Graphical Method Of Solving Linear Equations In Two Variables 4\" width=\"343\" height=\"322\" srcset=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2018\/08\/Graphical-Method-Of-Solving-Linear-Equations-In-Two-Variables-4.png 343w, https:\/\/www.aplustopper.com\/wp-content\/uploads\/2018\/08\/Graphical-Method-Of-Solving-Linear-Equations-In-Two-Variables-4-300x282.png 300w\" sizes=\"(max-width: 343px) 100vw, 343px\" \/><br \/>\nThe two lines AB and CD intersect at the point (20, 16), So, x = 20 and y = 16 is the required solution of the pair of linear equations i.e. length of the garden is 20 m and width of the garden is 16 m.<br \/>\n<strong>Verification:<\/strong><br \/>\nPutting x = 20 and y = 16 in (1). We get<br \/>\n20 = 16 + 4 = 20, (1) is verified.<br \/>\nPutting x = 20 and y = 16 in (2). we get<br \/>\n20 + 16 = 36<br \/>\n36 = 36, (2) is verified.<br \/>\nHence, both the equations are satisfied.<\/p>\n<p><strong>Example 5:<\/strong> \u00a0\u00a0\u00a0Draw the graphs of the equations x \u2013 y + 1 = 0 and 3x + 2y \u2013 12 = 0. Determine the coordinates of the vertices of the triangle formed by these lines and the x-axis, and shade the triangular region.<br \/>\n<strong>Sol.<\/strong>\u00a0 \u00a0Pair of linear equations are:<br \/>\nx \u2013 y + 1 = 0 \u00a0 \u00a0 \u00a0 \u00a0 &#8230;.(1)<br \/>\n3x + 2y \u2013 12 = 0 \u00a0 \u00a0 \u00a0&#8230;.(2)<br \/>\nx \u2013 y + 1 = 0 \u00a0\u00a0\u21d2 \u00a0y = x + 1<\/p>\n<table border=\"2\">\n<tbody>\n<tr>\n<td style=\"text-align: center;\" width=\"50\">x<\/td>\n<td style=\"text-align: center;\" width=\"49\">0<\/td>\n<td style=\"text-align: center;\" width=\"49\">4<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\" width=\"50\">y<\/td>\n<td style=\"text-align: center;\" width=\"49\">1<\/td>\n<td style=\"text-align: center;\" width=\"49\">5<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\" width=\"50\">Points<\/td>\n<td style=\"text-align: center;\" width=\"49\">A<\/td>\n<td style=\"text-align: center;\" width=\"49\">B<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>3x + 2y \u2013 12 = 0 \u00a0 \u00a0\u21d2 \u00a0\u00a0y = \\(\\frac { 12-3x }{ 2 }\\)<\/p>\n<table border=\"2\">\n<tbody>\n<tr>\n<td style=\"text-align: center;\" width=\"50\">x<\/td>\n<td style=\"text-align: center;\" width=\"49\">0<\/td>\n<td style=\"text-align: center;\" width=\"49\">2<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\" width=\"50\">y<\/td>\n<td style=\"text-align: center;\" width=\"49\">6<\/td>\n<td style=\"text-align: center;\" width=\"49\">3<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\" width=\"50\">Points<\/td>\n<td style=\"text-align: center;\" width=\"49\">C<\/td>\n<td style=\"text-align: center;\" width=\"49\">D<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Plot the points A(0, 1), B(4, 5) and join them to get a line AB. Similarly, plot the points C(0, 6), D(2, 3) and join them to form a line CD.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-134830\" src=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2018\/08\/Graphical-Method-Of-Solving-Linear-Equations-In-Two-Variables-5.png\" alt=\"Graphical Method Of Solving Linear Equations In Two Variables 5\" width=\"326\" height=\"310\" srcset=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2018\/08\/Graphical-Method-Of-Solving-Linear-Equations-In-Two-Variables-5.png 326w, https:\/\/www.aplustopper.com\/wp-content\/uploads\/2018\/08\/Graphical-Method-Of-Solving-Linear-Equations-In-Two-Variables-5-300x285.png 300w\" sizes=\"(max-width: 326px) 100vw, 326px\" \/><br \/>\nClearly, the two lines intersect each other at the point D(2, 3). Hence x = 2 and y = 3 is the solution of the<br \/>\ngiven pair of equations.<br \/>\nThe line CD cuts the x-axis at the point<br \/>\nE (4, 0) and the line AB cuts the x-axis at the point F(\u20131, 0).<br \/>\nHence, the coordinates of the vertices of the triangle are; D(2, 3), E(4, 0), F(\u20131, 0).<br \/>\n<strong>Verification:<\/strong><br \/>\nBoth the equations (1) and (2) are satisfied by x = 2 and y = 3. Hence, Verified.<\/p>\n<p><strong>Example 6:<\/strong> \u00a0\u00a0\u00a0Show graphically that the system of equations<br \/>\nx \u2013 4y + 14 = 0; 3x + 2y \u2013 14 = 0 is consistent with unique solution.<br \/>\n<strong>Sol.<\/strong>\u00a0 \u00a0 The given system of equations is<br \/>\nx \u2013 4y + 14 = 0 \u00a0 \u00a0 \u00a0 \u00a0&#8230;.(1)<br \/>\n3x + 2y \u2013 14 = 0 \u00a0 \u00a0\u00a0&#8230;.(2)<br \/>\nx \u2013 4y + 14 = 0 \u00a0\u00a0\u21d2 \u00a0 \u00a0y = \u00a0\u00a0\\(\\frac { x + 14 }{ 4 }\\)<\/p>\n<table border=\"2\">\n<tbody>\n<tr>\n<td style=\"text-align: center;\" width=\"50\">x<\/td>\n<td style=\"text-align: center;\" width=\"49\">6<\/td>\n<td style=\"text-align: center;\" width=\"49\">-2<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\" width=\"50\">y<\/td>\n<td style=\"text-align: center;\" width=\"49\">5<\/td>\n<td style=\"text-align: center;\" width=\"49\">3<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\" width=\"50\">Points<\/td>\n<td style=\"text-align: center;\" width=\"49\">A<\/td>\n<td style=\"text-align: center;\" width=\"49\">B<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>3x + 2y \u2013 14 = 0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u21d2 \u00a0y = \u00a0\\(\\frac { -3x + 14 }{ 2 }\\)<\/p>\n<table border=\"2\">\n<tbody>\n<tr>\n<td style=\"text-align: center;\" width=\"50\">x<\/td>\n<td style=\"text-align: center;\" width=\"49\">0<\/td>\n<td style=\"text-align: center;\" width=\"49\">4<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\" width=\"50\">y<\/td>\n<td style=\"text-align: center;\" width=\"49\">7<\/td>\n<td style=\"text-align: center;\" width=\"49\">1<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\" width=\"50\">Points<\/td>\n<td style=\"text-align: center;\" width=\"49\">C<\/td>\n<td style=\"text-align: center;\" width=\"49\">D<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-134833\" src=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2018\/08\/Graphical-Method-Of-Solving-Linear-Equations-In-Two-Variables-6.png\" alt=\"Graphical Method Of Solving Linear Equations In Two Variables 6\" width=\"253\" height=\"245\" \/><br \/>\nThe given equations representing two lines, intersect each other at a unique point (2, 4). Hence, the equations are consistent with unique solution.<\/p>\n<p><strong>Example 7:<\/strong> \u00a0\u00a0\u00a0Show graphically that the system of equations<br \/>\n2x + 5y = 16; \u00a0\\(3x+\\frac { 15 }{ 2 }=24\\) \u00a0has infinitely many solutions.<br \/>\n<strong>Sol.<\/strong>\u00a0 \u00a0 The given system of equations is<br \/>\n2x + 5y = 16 \u00a0 \u00a0 \u00a0&#8230;.(1)<br \/>\n\\(3x+\\frac { 15 }{ 2 }=24\\) \u00a0\u00a0\u00a0 \u00a0&#8230;.(2)<br \/>\n2x + 5y = 16 \u00a0 \u00a0\u21d2 \u00a0y =\u00a0\u00a0\\(\\frac { 16-2x }{ 5 }\\)<\/p>\n<table border=\"2\">\n<tbody>\n<tr>\n<td style=\"text-align: center;\" width=\"50\">x<\/td>\n<td style=\"text-align: center;\" width=\"49\">-2<\/td>\n<td style=\"text-align: center;\" width=\"49\">3<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\" width=\"50\">y<\/td>\n<td style=\"text-align: center;\" width=\"49\">4<\/td>\n<td style=\"text-align: center;\" width=\"49\">2<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\" width=\"50\">Points<\/td>\n<td style=\"text-align: center;\" width=\"49\">A<\/td>\n<td style=\"text-align: center;\" width=\"49\">B<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>\\(3x+\\frac { 15 }{ 2 }=24\\) \u00a0\u21d2 y =\u00a0\\(\\frac { 48-6x }{ 15 }\\)<\/p>\n<table border=\"2\">\n<tbody>\n<tr>\n<td style=\"text-align: center;\" width=\"50\">x<\/td>\n<td style=\"text-align: center;\" width=\"49\">1\/2<\/td>\n<td style=\"text-align: center;\" width=\"49\">11\/2<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\" width=\"50\">y<\/td>\n<td style=\"text-align: center;\" width=\"49\">3<\/td>\n<td style=\"text-align: center;\" width=\"49\">1<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\" width=\"50\">Points<\/td>\n<td style=\"text-align: center;\" width=\"49\">C<\/td>\n<td style=\"text-align: center;\" width=\"49\">D<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-134852\" src=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2018\/08\/Graphical-Method-Of-Solving-Linear-Equations-In-Two-Variables-7.png\" alt=\"Graphical Method Of Solving Linear Equations In Two Variables 7\" width=\"295\" height=\"278\" \/><br \/>\nThe lines of two equations are coincident. Coordinates of every point on this line are the solution.<br \/>\nHence, the given equations are consistent with infinitely many solutions.<\/p>\n<p><strong>Example 8:<\/strong>\u00a0 \u00a0 \u00a0Show graphically that the system of equations<br \/>\n2x + 3y = 10, 4x + 6y = 12 \u00a0has no solution.<br \/>\n<strong>Sol.<\/strong>\u00a0 \u00a0 The given equations are<br \/>\n2x + 3y = 10 \u00a0 \u00a0\u00a0\u21d2 \u00a0 \u00a0 \u00a0y = \u00a0\u00a0\\(\\frac { 10-2x }{ 3 }\\)<\/p>\n<table border=\"2\">\n<tbody>\n<tr>\n<td style=\"text-align: center;\" width=\"50\">x<\/td>\n<td style=\"text-align: center;\" width=\"49\">-4<\/td>\n<td style=\"text-align: center;\" width=\"49\">2<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\" width=\"50\">y<\/td>\n<td style=\"text-align: center;\" width=\"49\">6<\/td>\n<td style=\"text-align: center;\" width=\"49\">2<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\" width=\"50\">Points<\/td>\n<td style=\"text-align: center;\" width=\"49\">A<\/td>\n<td style=\"text-align: center;\" width=\"49\">B<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>4x + 6y = 12 \u00a0 \u00a0\u00a0\u21d2 \u00a0 \u00a0 \u00a0y = \u00a0\u00a0\\(\\frac { 12-4x }{ 6 }\\)<\/p>\n<table border=\"2\">\n<tbody>\n<tr>\n<td style=\"text-align: center;\" width=\"50\">x<\/td>\n<td style=\"text-align: center;\" width=\"49\">-3<\/td>\n<td style=\"text-align: center;\" width=\"49\">3<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\" width=\"50\">y<\/td>\n<td style=\"text-align: center;\" width=\"49\">4<\/td>\n<td style=\"text-align: center;\" width=\"49\">0<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\" width=\"50\">Points<\/td>\n<td style=\"text-align: center;\" width=\"49\">C<\/td>\n<td style=\"text-align: center;\" width=\"49\">D<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Plot the points A (\u20134, 6), B(2, 2) and join them to form a line AB. Similarly, plot the points C(\u20133, 4), D(3, 0) and join them to get a line CD.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-134856\" src=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2018\/08\/Graphical-Method-Of-Solving-Linear-Equations-In-Two-Variables-8.png\" alt=\"Graphical Method Of Solving Linear Equations In Two Variables 8\" width=\"319\" height=\"340\" srcset=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2018\/08\/Graphical-Method-Of-Solving-Linear-Equations-In-Two-Variables-8.png 319w, https:\/\/www.aplustopper.com\/wp-content\/uploads\/2018\/08\/Graphical-Method-Of-Solving-Linear-Equations-In-Two-Variables-8-281x300.png 281w\" sizes=\"(max-width: 319px) 100vw, 319px\" \/><br \/>\nClearly, the graphs of the given equations are parallel lines. As they have no common point, there is no common solution. Hence, the given system of equations has no solution.<\/p>\n<p><strong>Example 9:<\/strong>\u00a0 \u00a0 Given the linear equation 2x + 3y \u2013 8 = 0, write another linear equation in two variables such that the<br \/>\ngeometrical representing of the pair so formed is :<br \/>\n(i) intersecting lines<br \/>\n(ii) parallel lines<br \/>\n(iii) coincident lines<br \/>\n<strong>Sol.<\/strong>\u00a0 \u00a0 We have, \u00a02x + 3y \u2013 8 = 0<br \/>\n(i) Another linear equation in two variables such that the geometrical representation of the pair so formed is intersecting lines is<br \/>\n3x \u2013 2y \u2013 8 = 0<br \/>\n(ii) Another parallel lines to above line is<br \/>\n4x + 6y \u2013 22 = 0<br \/>\n(iii) Another coincident line to above line is<br \/>\n6x + 9y \u2013 24 = 0<\/p>\n<p><strong>Example 10:\u00a0<\/strong>\u00a0 \u00a0Solve the following system of linear equations graphically;<br \/>\n3x + y \u2013 11 = 0 ; x \u2013 y \u2013 1 = 0<br \/>\nShade the region bounded by these lines and also y-axis. Then, determine the areas of the region bounded by these lines and y-axis.<br \/>\n<strong>Sol.<\/strong>\u00a0\u00a0\u00a0\u00a0\u00a0 We have,<br \/>\n3x + y \u2013 11 = 0 and x \u2013 y \u2013 1 = 0<br \/>\n<strong>(a)<\/strong> Graph of the equation 3x + y \u2013 11 = 0<br \/>\nWe have, 3x + y \u2013 11 = 0<br \/>\n\u21d2 \u00a0y = \u2013 3x + 11<br \/>\nWhen, x = 2, \u00a0 \u00a0 y = \u20133 \u00d7 2 + 11 = 5<br \/>\nWhen, x = 3, \u00a0 \u00a0 y = \u2013 3 \u00d7 3 + 11 = 2<br \/>\nPlotting the points P (2, 5) and Q(3, 2) on the graph paper and drawing a line joining between them, we get the graph of the equation 3x + y \u2013 11 = 0 as shown in fig.<br \/>\n<strong>(b)\u00a0<\/strong>Graph of the equation x \u2013 y \u2013 1 = 0<br \/>\nWe have,<br \/>\nx \u2013 y \u2013 1 = 0<br \/>\ny = x \u2013 1<br \/>\nWhen, x = \u2013 1, y = \u20132<br \/>\nWhen, x = 3, y = 2<br \/>\nPlotting the points R(\u20131, \u20132) and S(3, 2) on the same graph paper and drawing a line joining between them, we get the graph of the equation x \u2013 y \u2013 1 = 0 as shown in fig.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-134857\" src=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2018\/08\/Graphical-Method-Of-Solving-Linear-Equations-In-Two-Variables-9.png\" alt=\"Graphical Method Of Solving Linear Equations In Two Variables 9\" width=\"373\" height=\"348\" srcset=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2018\/08\/Graphical-Method-Of-Solving-Linear-Equations-In-Two-Variables-9.png 373w, https:\/\/www.aplustopper.com\/wp-content\/uploads\/2018\/08\/Graphical-Method-Of-Solving-Linear-Equations-In-Two-Variables-9-300x280.png 300w\" sizes=\"(max-width: 373px) 100vw, 373px\" \/><br \/>\nYou can observe that two lines intersect at<br \/>\nQ(3, 2). So, x = 3 and y = 2. The area enclosed by the lines represented by the given equations and also the y-axis is shaded.<br \/>\nSo, the enclosed area = Area of the shaded portion<br \/>\n= Area of \u2206QUT = 1\/2 \u00d7 base \u00d7 height<br \/>\n= 1\/2 \u00d7 (TU \u00d7 VQ) = 1\/2 \u00d7 (TO + OU) \u00d7 VQ<br \/>\n= 1\/2 (11 + 1) 3 = 1\/2 \u00d7 12 \u00d7 3 = 18 sq.units.<br \/>\nHence, required area is 18 sq. units.<\/p>\n<p><strong>Example 11:\u00a0<\/strong>\u00a0 \u00a0Draw the graphs of the following equations<br \/>\n2x \u2013 3y = \u2013 6; 2x + 3y = 18; y = 2<br \/>\nFind the vertices of the triangles formed and also find the area of the triangle.<br \/>\n<strong>Sol.<\/strong>\u00a0\u00a0\u00a0\u00a0\u00a0<strong> (a) \u00a0<\/strong>Graph of the equation 2x \u2013 3y = \u2013 6;<br \/>\nWe have, 2x \u2013 3y = \u2013 6 \u00a0\u00a0\u21d2 \u00a0 \u00a0 \u00a0y = \u00a0\u00a0\\(\\frac { 2x+6 }{ 3 }\\)<br \/>\nWhen, x = 0, y = 2<br \/>\nWhen, x = 3, y = 4<br \/>\nPlotting the points P(0, 2) and Q(3, 4) on\u00a0the graph paper and drawing a line joining between them we get the graph of the equation 2x \u2013 3y = \u2013 6 as\u00a0shown in fig.<br \/>\n<strong>(b) \u00a0<\/strong>Graph of the equation 2x + 3y = 18;<br \/>\nWe have 2x + 3y = 18 \u00a0\u00a0\u21d2 \u00a0 \u00a0 \u00a0y = \u00a0\u00a0\\(\\frac { -2x+18 }{ 3 }\\)<br \/>\nWhen, x = 0, \u00a0y = 6<br \/>\nWhen, x = \u2013 3, \u00a0y = 8<br \/>\nPlotting the points R(0, 6) and S(\u20133, 8) on the same graph paper and drawing a line joining between them, we\u00a0get the graph of the equation 2x + 3y = 18 as shown in fig.<br \/>\n<strong>(c) \u00a0<\/strong>Graph of the equation y = 2<br \/>\nIt is a clear fact that y = 2 is for every value of x. We may take the points T (3, 2), U(6, 2) or any other values.<br \/>\nPlotting the points T(3, 2) and U(6, 2) on the same graph paper and drawing a line joining between them, we get the graph of the equation y = 2 as shown in fig.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-134858\" src=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2018\/08\/Graphical-Method-Of-Solving-Linear-Equations-In-Two-Variables-10.png\" alt=\"Graphical Method Of Solving Linear Equations In Two Variables 10\" width=\"367\" height=\"327\" srcset=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2018\/08\/Graphical-Method-Of-Solving-Linear-Equations-In-Two-Variables-10.png 367w, https:\/\/www.aplustopper.com\/wp-content\/uploads\/2018\/08\/Graphical-Method-Of-Solving-Linear-Equations-In-Two-Variables-10-300x267.png 300w\" sizes=\"(max-width: 367px) 100vw, 367px\" \/><br \/>\nFrom the fig., we can observe that the lines taken in pairs intersect each other at points\u00a0Q(3, 4), U (6, 2) and P(0, 2). These form the three vertices of the triangle PQU.<br \/>\nTo find area of the triangle so formed<br \/>\nThe triangle is so formed is PQU (see fig.)<br \/>\nIn the \u2206PQU<br \/>\nQT (altitude) = 2 units<br \/>\nand PU (base) = 6 units<br \/>\nso, area of \u2206PQU = (base \u00d7 height)<br \/>\n= 1\/2 (PU \u00d7 QT) = 1\/2 \u00d7 6 \u00d7 2 sq. untis<br \/>\n= 6 sq. units.<\/p>\n<p><strong>Example 12:\u00a0<\/strong>\u00a0 \u00a0On comparing the ratios \\(\\frac{{{a}_{1}}}{{{a}_{2}}},\\frac{{{b}_{1}}}{{{b}_{2}}}and\\frac{{{c}_{1}}}{{{c}_{2}}}\\) \u00a0and without drawing them, find out whether the lines representing\u00a0the following pairs of linear equations intersect at a point, are parallel or coincide.<br \/>\n(i) 5x \u2013 4y + 8 = 0, 7x + 6y \u2013 9 = 0<br \/>\n(ii) 9x + 3y + 12 = 0, 18x + 6y + 24 = 0<br \/>\n(iii) 6x \u2013 3y + 10 = 0, 2x \u2013 y + 9 = 0<br \/>\n<strong>Sol.<\/strong>\u00a0\u00a0\u00a0\u00a0\u00a0 Comparing the given equations with standard forms of equations a<sub>1<\/sub>x + b<sub>1<\/sub>y + c<sub>1<\/sub> = 0\u00a0 and<br \/>\na<sub>2<\/sub>x + b<sub>2<\/sub>y + c<sub>2<\/sub> = 0 \u00a0we have,<br \/>\n<strong>(i)<\/strong> \u00a0a1\u00a0= 5, b<sub>1<\/sub> = \u2013 4, c<sub>1<\/sub> = 8;<br \/>\na<sub>2<\/sub> = 7, b<sub>2<\/sub> = 6, c<sub>2<\/sub> = \u2013 9<br \/>\n\\( \\frac{{{a}_{1}}}{{{a}_{2}}}=\\frac{5}{7},\\frac{{{b}_{1}}}{{{b}_{2}}}=\\frac{-4}{6} \\)<br \/>\n\\(\\Rightarrow \\frac{{{a}_{1}}}{{{a}_{2}}}\\ne \\frac{{{b}_{1}}}{{{b}_{2}}} \\)<br \/>\nThus, the lines representing the pair of linear equations are intersecting.<br \/>\n<strong>(ii)<\/strong> \u00a0a1\u00a0= 9, b<sub>1<\/sub> = 3, c<sub>1<\/sub> = 12;<br \/>\na<sub>2<\/sub> = 18, b<sub>2<\/sub> = 6, c<sub>2<\/sub> = 24<br \/>\n\\( \\frac{{{a}_{1}}}{{{a}_{2}}}=\\frac{9}{18}=\\frac{1}{2},\\frac{{{b}_{1}}}{{{b}_{2}}}=\\frac{3}{6}=\\frac{1}{2}and\\frac{{{c}_{1}}}{{{c}_{2}}}=\\frac{12}{24}=\\frac{1}{2}\\)<br \/>\n\\(\\Rightarrow \\frac{{{a}_{1}}}{{{a}_{2}}}=\\frac{{{b}_{1}}}{{{b}_{2}}}=\\frac{{{c}_{1}}}{{{c}_{2}}} \\)<br \/>\nThus, the lines representing the pair of linear equation coincide.<br \/>\n<strong>(iii)\u00a0<\/strong>\u00a0a1\u00a0= 6, b<sub>1<\/sub> = \u2013 3, c<sub>1<\/sub> = 10;<br \/>\na<sub>2<\/sub> = 2, b<sub>2<\/sub> = \u2013 6, c<sub>2<\/sub> = 9<br \/>\n\\( \\frac{{{a}_{1}}}{{{a}_{2}}}=\\frac{6}{2}=3,\\frac{{{b}_{1}}}{{{b}_{2}}}=\\frac{-3}{-1}=3and\\frac{{{c}_{1}}}{{{c}_{2}}}=\\frac{10}{9} \\)<br \/>\n\\(\\Rightarrow \\frac{{{a}_{1}}}{{{a}_{2}}}=\\frac{{{b}_{1}}}{{{b}_{2}}}\\ne \\frac{{{c}_{1}}}{{{c}_{2}}} \\)<br \/>\nThus, the lines representing the pair of linear equations are parallel.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Graphical Method Of Solving Linear Equations In Two Variables Let the system of pair of linear equations be a1x + b1y = c1\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0\u2026.(1) a2x + b2y = c2\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u2026.(2) We know that given two [&hellip;]<\/p>\n","protected":false},"author":3,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_genesis_hide_title":false,"_genesis_hide_breadcrumbs":false,"_genesis_hide_singular_image":false,"_genesis_hide_footer_widgets":false,"_genesis_custom_body_class":"","_genesis_custom_post_class":"","_genesis_layout":"","footnotes":""},"categories":[5],"tags":[53,54,52],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v17.8 (Yoast SEO v17.8) - 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