Check out the derivation to get more details about the Cross-Multiplication Method.<\/p>\n
General Solution Using the Method of Cross-Multiplication<\/h3>\n
Let the pair of linear equations in two variables are<\/p>\n
a\u2081x + b\u2081y + c\u2081 = 0<\/p>\n
a\u2082x + b\u2082y + c\u2082 = 0<\/p>\n
For solving these linear equations follow the below included steps<\/p>\n
Given simultaneous linear equations in two variables are<\/p>\n
a\u2081x + b\u2081y + c\u2081 = 0 ——- (i)<\/p>\n
a\u2082x + b\u2082y + c\u2082 = 0 ——– (ii)<\/p>\n
Multiply equation (i) by b\u2082 and equation (ii) by b\u2081.<\/p>\n
b\u2082 (a\u2081x + b\u2081y + c\u2081 = 0)<\/p>\n
b\u2082 (a\u2081x + b\u2081y + c\u2081) = b\u2082 x 0<\/p>\n
a\u2081b\u2082x + b\u2081b\u2082y + b\u2082c\u2081 = 0 ——- (iii)<\/p>\n
b\u2081 (a\u2082x + b\u2082y + c\u2082 = 0)<\/p>\n
b\u2081 (a\u2082x + b\u2082y + c\u2082) = b\u2081 x 0<\/p>\n
a\u2082b\u2081x + b\u2082b\u2081y + c\u2082b\u2081 = 0 —- (iv)<\/p>\n
Subtract equation (iv) from equation (iii)<\/p>\n
[a\u2081b\u2082x + b\u2081b\u2082y + b\u2082c\u2081 = 0] – [a\u2082b\u2081x + b\u2082b\u2081y + c\u2082b\u2081 = 0]<\/p>\n
[a\u2081b\u2082x + b\u2081b\u2082y + b\u2082c\u2081] – [a\u2082b\u2081x + b\u2082b\u2081y + c\u2082b\u2081] = 0<\/p>\n
a\u2081b\u2082x + b\u2081b\u2082y + b\u2082c\u2081 – a\u2082b\u2081x – b\u2082b\u2081y – c\u2082b\u2081 = 0<\/p>\n
x(a\u2081b\u2082 – a\u2082b\u2081) + b\u2082c\u2081 – c\u2082b\u2081 = 0<\/p>\n
x(a\u2081b\u2082 – a\u2082b\u2081) = c\u2082b\u2081 – b\u2082c\u2081<\/p>\n
x = (c\u2082b\u2081 – b\u2082c\u2081) \/ (a\u2081b\u2082 – a\u2082b\u2081)<\/p>\n
Substitute x = (c\u2082b\u2081 – b\u2082c\u2081) \/ (a\u2081b\u2082 – a\u2082b\u2081) in equation (i)<\/p>\n
a\u2081(c\u2082b\u2081 – b\u2082c\u2081) \/ (a\u2081b\u2082 – a\u2082b\u2081) + b\u2081y + c\u2081 = 0<\/p>\n
b\u2081y = -a\u2081(c\u2082b\u2081 – b\u2082c\u2081) \/ (a\u2081b\u2082 – a\u2082b\u2081) – c\u2081<\/p>\n
y = (-a\u2081(c\u2082b\u2081 – b\u2082c\u2081) \/ (a\u2081b\u2082 – a\u2082b\u2081) – c\u2081) \/ b\u2081<\/p>\n
y = (c\u2081a\u2082 – c\u2082a\u2081) \/ (b\u2082a\u2081 – b\u2081a\u2082)<\/p>\n
The solution of the equations is<\/p>\n
x \/ (c\u2082b\u2081 – b\u2082c\u2081) \/ (a\u2081b\u2082 – a\u2082b\u2081) = y \/ (c\u2081a\u2082 – c\u2082a\u2081) \/ (b\u2082a\u2081 – b\u2081a\u2082) = 1\/(b\u2082a\u2081 – b\u2081a\u2082)<\/p>\n
The above method is called Cross-Multiplication Methos as it uses cross-multiplication technique.<\/p>\n
You can also represent the cross-multiplication as the following diagram.<\/p>\n
![\"\"](\"https:\/\/www.worksheetsbuddy.com\/wp-content\/uploads\/2021\/01\/cross-multiplication.png\")
<\/p>\n
Cross Multiplication Method Examples<\/h3>\n
Example 1.<\/strong><\/p>\n Solve the linear equations x + y = 5, x – y = 3 by using the cross multiplication method?<\/p>\n Solution:<\/strong><\/p>\n Given system of linear equations are<\/p>\n x + y = 5<\/p>\n x – y = 3<\/p>\n On transposition, we get<\/p>\n x + y – 5 = 0<\/p>\n x – y – 3 = 0<\/p>\n Writing the co-efficient in the following way, we get:<\/p>\n By cross-multiplication method:<\/p>\n x\/[(1(-3) – (-1)(-5)] = y\/[(-5)1 – 1(-3)] = 1\/\/[1(-1) – 1(1)]<\/p>\n x\/(-3 – 5) = y\/(-5 + 3) = 1\/(-1 – 1)<\/p>\n x\/(-8) = y\/(-2) = 1\/(-2)<\/p>\n x\/(-8) = 1\/(-2), y\/(-2) = 1\/(-2)<\/p>\n x\/8 = 1\/2, y\/2 = 1\/2<\/p>\n x = 8\/2, y = 2\/2<\/p>\n x = 4, y = 1<\/p>\n Therefore, required solution is x = 4, y = 1.<\/p>\n Example 2.<\/strong><\/p>\n Solve the simultaneous linear equations 3x – 4y = 0, 9x – 8y = 12 by the method of cross-multiplication?<\/p>\n Solution:<\/strong><\/p>\n Given simultaneous equations are,<\/p>\n 3x – 4y = 0<\/p>\n 9x – 8y = 12<\/p>\n On transposition, we get<\/p>\n 3x – 4y = 0 ——– (i)<\/p>\n 9x – 8y – 12 = 0 ———- (ii)<\/p>\n Multiply equation (i) by -8 and equation (ii) by -4.<\/p>\n -8(3x – 4y) = 0 x -8<\/p>\n -24x + 32y = 0 —— (iii)<\/p>\n -4(9x – 8y – 12) = 0 x -4<\/p>\n -36x + 32y + 48 = 0 ——- (iv)<\/p>\n Subtract equation (iv) from equation (iii)<\/p>\n [-24x + 32y = 0] – [-36x + 32y + 48 = 0]<\/p>\n -24x + 32y – (-36x + 32y + 48) = 0<\/p>\n -24x + 32y + 36x – 32y – 48 = 0<\/p>\n 12x – 48 = 0<\/p>\n 12x = 48<\/p>\n x = 48\/12<\/p>\n x = 4<\/p>\n Substitute x = 4 in equation (ii)<\/p>\n 9(4) – 8y – 12 = 0<\/p>\n 36 – 8y – 12 = 0<\/p>\n -8y + 24 = 0<\/p>\n -8y = -24<\/p>\n y = 24\/8<\/p>\n y = 3<\/p>\n Therefore, required solution is x = 4, y = 3.<\/p>\n Example 3.<\/strong><\/p>\n Solve the linear equations 2x + 3y = 5, x – 4y = 8 by using the cross multiplication method?<\/p>\n Solution:<\/strong><\/p>\n Given linear equations are 2x + 3y = 5, x – 4y = 8<\/p>\n On transposing, we get<\/p>\n 2x + 3y – 5 = 0 ——- (i)<\/p>\n x – 4y – 8 = 0 ——- (ii)<\/p>\n Multiply equation (ii) by 2, equation (i) by 1.<\/p>\n From equation (i), we get<\/p>\n 2x + 3y – 5 = 0 —– (iii)<\/p>\n From equation (ii), we get<\/p>\n 2(x – 4y – 8) = 0<\/p>\n 2x – 8y – 16 = 0 —— (iv)<\/p>\n Subtract equation (iii) from equation (iv).<\/p>\n [2x – 8y – 16] – [2x + 3y – 5] = 0<\/p>\n 2x – 8y – 16 – 2x – 3y + 5 = 0<\/p>\n -11y – 11 = 0<\/p>\n -11y = 11<\/p>\n y = -11\/11<\/p>\n y = -1<\/p>\n Substitute y = -1 in equation (i)<\/p>\n 2x + 3(-1) – 5 = 0<\/p>\n 2x – 3 – 5 = 0<\/p>\n 2x – 8 = 0<\/p>\n 2x = 8<\/p>\n x = 8\/2<\/p>\n x = 4<\/p>\n Therefore, the required solution is x = 4, y = -1.<\/p>\n Example 4.<\/strong><\/p>\n Solve the system of linear equations 2x + 5y = 20, x + 2y = 4?<\/p>\n Solution:<\/strong><\/p>\n Given simultaneous linear equations are 2x + 5y = 20, x + 2y = 4<\/p>\n On transposing, we get<\/p>\n 2x + 5y – 20 = 0 —— (i)<\/p>\n x + 2y – 4 = 0 —— (ii)<\/p>\n Multiply equation (i) by 2.<\/p>\n 2(2x + 5y – 20) = 2 x 0<\/p>\n 4x + 10y – 40 = 0 ——- (iii)<\/p>\n Multiply equation (ii) by 5.<\/p>\n 5(x + 2y – 4) = 0 x 5<\/p>\n 5x + 10y – 20 = 0 —– (iv)<\/p>\n Subtract equation (iii) from equation (iv)<\/p>\n [5x + 10y – 20] – [4x + 10y – 40] = 0<\/p>\n 5x + 10y – 20 – 4x – 10y + 40 = 0<\/p>\n x + 20 = 0<\/p>\n x = -20<\/p>\n Put x = -20 in equation (iii)<\/p>\n 4(-20) + 10y – 40 = 0<\/p>\n -80 + 10y – 40 = 0<\/p>\n 10y – 120 = 0<\/p>\n 10y = 120<\/p>\n y = 120\/10<\/p>\n y = 12<\/p>\n Therefore the required solution is x = -20, y = 12.<\/p>\n","protected":false},"excerpt":{"rendered":" Cross multiplication method is used to solve a pair of simultaneous linear equations in two variables. This method helps you to find the accurate values of variables easily. It is applicable only when we have two linear equations in two variables. Get the steps to solve the system of linear equations with the cross multiplication … Read more<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[5],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/posts\/35224"}],"collection":[{"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/comments?post=35224"}],"version-history":[{"count":1,"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/posts\/35224\/revisions"}],"predecessor-version":[{"id":92650,"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/posts\/35224\/revisions\/92650"}],"wp:attachment":[{"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/media?parent=35224"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/categories?post=35224"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/bigideasmathanswers.com\/wp-json\/wp\/v2\/tags?post=35224"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}![\"\"](\"https:\/\/www.worksheetsbuddy.com\/wp-content\/uploads\/2021\/01\/cross-multiplication-example-1.jpg\")
<\/p>\n