Students can solve the system of linear equations by using the comparison method. Here, you need to compare a pair of equations and find the value of the variables. You can find a detailed explanation of solving the simultaneous linear equations in the following sections. The comparison method is one of the easiest ways to solve a pair of equations. So, follow the below guidelines and solve the questions.<\/p>\n
Simultaneous Equations are nothing but two or more equations having two or more quantities. There are many ways to solve the system of linear equations, but we are using the comparison method to get the answer easily. Check the steps and get the solution.<\/p>\n
Example 1.<\/strong><\/p>\n Solve the pair of simultaneous equations 5x + 3y = 9, 2x – 3y = 12 using the comparison method with detailed explanation.<\/p>\n Solution:<\/strong><\/p>\n Given the system of linear equations are<\/p>\n 5x + 3y = 9 —— (i)<\/p>\n 2x – 3y = 12 ——- (ii)<\/p>\n Express of x in terms of y.<\/p>\n From Equation (i) 5x + 3y = 9, we get<\/p>\n Subtract 3y from both sides<\/p>\n 5x + 3y – 3y = 9 – 3y<\/p>\n 5x = 9 – 3y<\/p>\n Divide both sides of the equation by 5.<\/p>\n 5x\/5 = (9 – 3y)\/5<\/p>\n x = (9 – 3y)\/5<\/p>\n Therefore, x = (9 – 3y)\/5 —— (iii)<\/p>\n From equation (ii) 2x – 3y = 12, we get<\/p>\n Add 3y to both sides of the equation.<\/p>\n 2x – 3y + 3y = 12 + 3y<\/p>\n 2x = 12 + 3y<\/p>\n Divide both sides by 2.<\/p>\n 2x\/2 = (12 + 3y)\/2<\/p>\n x = (12 + 3y)\/2<\/p>\n Therefore, x = (12 + 3y)\/2 —– (iv)<\/p>\n Equate the values of x in equation (iii) and equation (iv) forming the equation in y<\/p>\n From equations (iii) and (iv), we get<\/p>\n (9 – 3y)\/5 = (12 + 3y)\/2<\/p>\n Solve the above linear equation<\/p>\n Cross multiply the fractions.<\/p>\n 2(9 – 3y) = 5(12 + 3y)<\/p>\n 18 – 6y = 60 + 15y<\/p>\n 18 – 60 = 15y + 6y<\/p>\n -42 = 21y<\/p>\n y = -42\/21<\/p>\n y = -2<\/p>\n Substituting the value of y in equation (iii) or equation (iv), find the value of x<\/p>\n Put y = -2 in equation (iii), we get<\/p>\n x = (9 – 3(-2))\/5<\/p>\n x = (9 + 6)\/5<\/p>\n = 15\/5<\/p>\n x = 3<\/p>\n Required solution of the two equations<\/p>\n Therefore, x = 3, y = -2.<\/p>\n Therefore, we have compared the values of x obtained from equation (i) and (ii) and formed an equation in y, so this method of solving simultaneous equations is known as the comparison method. Similarly, comparing the two values of y, we can form an equation in x.<\/p>\n Example 2.<\/strong><\/p>\n Solve the system of linear equations 5x + y = 10, 7x – 3y + 8 = 0 using comparison method.<\/p>\n Solution:<\/strong><\/p>\n Given pair of system of linear equations are<\/p>\n 5x + y = 10 —- (i)<\/p>\n 7x – 3y + 8 = 0 —– (ii)<\/p>\n Express x in terms of y<\/p>\n From equation (i) 5x + y = 10, we get<\/p>\n 5x = 10 – y<\/p>\n x = (10 – y)\/5<\/p>\n From equation (ii) 7x – 3y + 8 = 0, we get<\/p>\n 7x – 3y = -8<\/p>\n 7x = 3y – 8<\/p>\n x = (3y – 8)\/7<\/p>\n Equate the values of x obtained from both equations.<\/p>\n (10 – y)\/5 = (3y – 8)\/7<\/p>\n Solve the equation in y.<\/p>\n 7(10 – y) = 5(3y – 8)<\/p>\n 70 – 7y = 15y – 40<\/p>\n 70 + 40 = 15y + 7y<\/p>\n 110 = 22y<\/p>\n y = 110\/22<\/p>\n y = 5<\/p>\n Putting the value of y in equation (i)<\/p>\n Substitute y = 5 in equation (i), we get<\/p>\n 5x + 5 = 10<\/p>\n 5x = 10 – 5<\/p>\n 5x = 5<\/p>\n x = 5\/5<\/p>\n x = 1<\/p>\n Required solution of the two equations<\/p>\n Therefore, x = 1, y = 5.<\/p>\n Example 3.<\/strong><\/p>\n Solve the linear equations 2x + y = 10 and x + y = 4 by comparison method.<\/p>\n Solution:<\/strong><\/p>\n Given simultaneous linear equations are<\/p>\n 2x + y = 10 —- (i)<\/p>\n x + y = 4 —- (ii)<\/p>\n express y in terms of x.<\/p>\n From equation (i), we can write<\/p>\n y = 10 – 2x<\/p>\n From equation (ii), we can write<\/p>\n y = 4 – x<\/p>\n Equate both the values of y obtained from equations (i) and (ii)<\/p>\n 10 – 2x = 4 – x<\/p>\n 10 – 4 = 2x – x<\/p>\n x = 6<\/p>\n Substitute x = 6 in the equation (ii)<\/p>\n 6 + y = 4<\/p>\n y = 4 – 6<\/p>\n y = -2<\/p>\n Required solution of the two equations<\/p>\n Therefore, x = 6, y = -2.<\/p>\n Example 4.<\/strong><\/p>\n Find the solution to the following simultaneous equations.<\/p>\n 4x + 3y = 14 and 5x + 7y = 11.<\/p>\n Solution:<\/strong><\/p>\n Given linear equations are<\/p>\n 4x + 3y = 14 —– (i)<\/p>\n 5x + 7y = 11 —— (ii)<\/p>\n Express x in terms of y for both equations.<\/p>\n From equation (i) 4x + 3y = 14, we can write<\/p>\n 4x = 14 – 3y<\/p>\n x = (14 – 3y)\/4<\/p>\n From equation (ii) 5x + 7y = 11, we can write<\/p>\n 5x = 11 – 7y<\/p>\n x = (11 – 7y)\/5<\/p>\n Equate the values of x obtained from equations (i), (ii)<\/p>\n (14 – 3y)\/4 = (11 – 7y)\/5<\/p>\n 5(14 – 3y) = 4(11 – 7y)<\/p>\n 70 – 15y = 44 – 28y<\/p>\n 28y – 15y = 44 – 70<\/p>\n 13y = -26<\/p>\n y = -26\/13<\/p>\n y = -2<\/p>\n Substitute y = -2 in the equation (i) and simplify it to get the x value.<\/p>\n Put y = -2 in 4x + 3y = 14<\/p>\n 4x + 3(-2) = 14<\/p>\n 4x – 6 = 14<\/p>\n 4x = 14 + 6<\/p>\n 4x = 20<\/p>\n x = 20\/4<\/p>\n x = 5<\/p>\n Therefore the required solutions are x = 5, y = -2.<\/p>\n","protected":false},"excerpt":{"rendered":" Students can solve the system of linear equations by using the comparison method. Here, you need to compare a pair of equations and find the value of the variables. You can find a detailed explanation of solving the simultaneous linear equations in the following sections. The comparison method is one of the easiest ways to … Read more<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[5],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts\/35215"}],"collection":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/comments?post=35215"}],"version-history":[{"count":1,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts\/35215\/revisions"}],"predecessor-version":[{"id":138477,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts\/35215\/revisions\/138477"}],"wp:attachment":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/media?parent=35215"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/categories?post=35215"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/tags?post=35215"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}