{"id":370,"date":"2022-12-15T10:00:27","date_gmt":"2022-12-15T04:30:27","guid":{"rendered":"https:\/\/www.aplustopper.com\/?p=370"},"modified":"2022-12-16T09:26:01","modified_gmt":"2022-12-16T03:56:01","slug":"elimination-method","status":"publish","type":"post","link":"https:\/\/www.aplustopper.com\/elimination-method\/","title":{"rendered":"Solving Systems Of Equations By Elimination Method"},"content":{"rendered":"

Solving Systems Of Equations By Elimination Method<\/strong><\/h2>\n

Step I:<\/strong> Let the two equations obtained be
\na1<\/sub>x + b1<\/sub>y + c1<\/sub> = 0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u2026.(1)
\na2<\/sub>x + b2<\/sub>y + c2<\/sub> = 0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0\u2026.(2)
\nStep II:<\/strong> Multiplying the given equation so as to make the co-efficients of the variable to be eliminated equal.
\nStep III:<\/strong> Add or subtract the equations so obtained in Step II, as the terms having the same coefficients may be either of opposite or the same sign.
\nStep IV :<\/strong> Solve the equations in one varibale so obtained in Step III.
\nStep V:<\/strong> Substitute the value found in Step IV in any one of the given equations and then copmpute the value of the other variable.<\/p>\n

Elimination Method Examples<\/strong><\/h2>\n

Example 1:<\/strong> \u00a0 \u00a0Solve the following system of linear equations by applying the method of elimination by equating the coefficients :
\n(i) 4x \u2013 3y = 4 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0(ii) 5x \u2013 6y = 8
\n2x + 4y = 3 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 3x + 2y = 6
\nSol.<\/strong>\u00a0 \u00a0(i)<\/strong>\u00a0\u00a0 We have,
\n4x \u2013 3y = 4 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 ….(1)
\n2x + 4y = 3 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 ….(2)
\nLet us decide to eliminate x from the given equation. Here, the co-efficients of x are 4 and 2 respectively. We find the L.C.M. of 4 and 2 is 4. Then, make the co-efficients of x equal to 4 in the two equations.
\nMultiplying equation (1) with 1 and equation (2) with 2, we get
\n4x \u2013 3y = 4 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 ….(3)
\n4x + 8y = 6 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0….(4)
\nSubtracting equation (4) from (3), we get
\n\u201311y = \u20132 \u00a0\u21d2 \u00a0\u00a0 y = \u00a0\\(\\frac { 2 }{ 11 }\\)
\nSubstituting y = 2\/11 in equation (1), we get
\n\u21d2 4x \u2013 3 \u00d7 \\(\\frac { 2 }{ 11 }\\) = 4
\n\u21d2 4x \u2013 \\(\\frac { 6 }{ 11 }\\) = 4
\n\u21d2 4x = 4 +\u00a0\\(\\frac { 6 }{ 11 }\\)
\n\u21d2 4x =\u00a0\\(\\frac { 50 }{ 11 }\\)
\n\u21d2 x = \\(\\frac { 50 }{ 44 }\\)\u00a0=\u00a0\\(\\frac { 25 }{ 22 }\\)
\nHence, solution of the given system of equation is :
\nx = \\(\\frac { 25 }{ 22 }\\), \u00a0y =\u00a0\\(\\frac { 2 }{ 11 }\\)<\/p>\n

(ii)\u00a0<\/strong>We have;
\n5x \u2013 6y = 8 \u00a0 \u00a0 \u00a0 ….(1)
\n3x + 2y = 6 \u00a0 \u00a0 \u00a0 ….(2)
\nLet us eliminate y from the given system of equations. The co-efficients of y in the given equations are 6 and 2 respectively. The L.C.M. of 6 and 2 is 6. We have to make the both coefficients equal to 6. So, multiplying both sides of equation (1) with 1 and equation (2) with 3, we get
\n5x \u2013 6y = 8 \u00a0 \u00a0 \u00a0 ….(3)
\n9x + 6y = 18 \u00a0 \u00a0 ….(4)
\nAdding equation (3) and (4), we get
\n14x = 26 \u00a0 \u00a0 \u00a0\u21d2 \u00a0 x =\u00a0\\(\\frac { 13 }{ 7 }\\)
\nPutting x = \\(\\frac { 13 }{ 7 }\\)\u00a0in equation (1), we get
\n5 \u00d7 \\(\\frac { 13 }{ 7 }\\)\u00a0\u2013 6y = 8 \u00a0 \u00a0 \u21d2 \\(\\frac { 65 }{ 7 }\\)\u00a0\u2013 6y = 8
\n\u21d2 6y = \\(\\frac { 65 }{ 7 }\\)\u00a0\u2013 8 = \u00a0\\(\\frac { 65-56 }{ 7 }\\) = \\(\\frac { 9 }{ 7 }\\)
\n\u21d2 y = \\(\\frac { 9 }{ 42 }\\)\u00a0= \\(\\frac { 3 }{ 14 }\\)
\nHence, the solution of the system of equations is x = \\(\\frac { 13 }{ 7 }\\)\u00a0, y =\u00a0\\(\\frac { 3 }{ 14 }\\)<\/p>\n

Example 2:<\/strong>\u00a0 \u00a0\u00a0Solve the following system of linear equations by usnig the method of elimination by equating the coefficients:
\n3x + 4y = 25 ; \u00a0 \u00a0 \u00a05x \u2013 6y = \u2013 9
\nSol.<\/strong>\u00a0 \u00a0 The given system of equation is
\n3x + 4y = 25 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0….(1)
\n5x \u2013 6y = \u2013 9 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0….(2)
\nLet us eliminate y. The coefficients of y are 4 and \u2013 6. The LCM of 4 and 6 is 12.\u00a0So, we make the coefficients of y as 12 and\u00a0\u2013 12.
\nMultiplying equation (1) by 3 and equation (2) by 2, we get
\n9x + 12y = 75 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0….(3)
\n10x \u2013 12y = \u2013 18 \u00a0 \u00a0 \u00a0 …(4)
\nAdding equation (3) and equation (4), we get
\n19x = 57 \u00a0 \u21d2 \u00a0 x = 3.
\nPutting x = 3 in (1), we get,
\n3 \u00d7 3 + 4y = 25
\n\u21d2 4y = 25 \u2013 9 = 16 \u00a0\u21d2 \u00a0 y = 4
\nHence, the solution is x = 3, y = 4.
\nVerification:<\/strong> Both the equations are satisfied by x = 3 and y = 4, which shows that the solution is correct.<\/p>\n

Example 3:<\/strong>\u00a0 \u00a0\u00a0Solve the following system of equations:
\n15x + 4y = 61; \u00a0 4x + 15y = 72
\nSol.<\/strong>\u00a0 \u00a0The given system of equation is
\n15x + 4y = 61 \u00a0 \u00a0 \u00a0 \u00a0 ….(1)
\n4x + 15y = 72 \u00a0 \u00a0 \u00a0 \u00a0….(2)
\nLet us eliminate y. The coefficients of y are 4 and 15. The L.C.M. of 4 and 15 is 60. So, we make the coefficients of y as 60. Multiplying (1) by 15 and (2) by 4, we get
\n225x + 60y = 915 \u00a0 \u00a0 ….(3)
\n16x + 60y = 288\u00a0 \u00a0 \u00a0\u00a0….(4)
\nSubstracting (4) from (3), we get
\n209x = 627 \u00a0 \u00a0\u21d2 \u00a0x = 3
\nPutting x = 3 in (1), we get
\n15 \u00d7 3 + 4y = 61 45 + 4y = 61
\n4y = 61 \u2013 45 = 16 \u00a0 \u21d2 \u00a0 y = 4
\nHence, the solution is x = 3, y = 4.
\nVerification:<\/strong> On putting x = 3 and y = 4 in the given equations, they are satisfied. Hence, the solution is correct.<\/p>\n

Example 4:<\/strong>\u00a0 \u00a0\u00a0Solve the following system of equations by using the method of elimination by equating the co-efficients.
\n\\(\\frac { x }{ y }\\) + \\(\\frac { 2y }{ 5 }\\)\u00a0+ 2 = 10; \u00a0\\(\\frac { 2x }{ 7 }\\) \u2013 \\(\\frac { 5 }{ 2 }\\)\u00a0+ 1 = 9
\nSol.<\/strong>\u00a0 \u00a0 The given system of equation is
\n\\(\\frac { x }{ y }\\) + \\(\\frac { 2y }{ 5 }\\)\u00a0+ 2 = 10 \u00a0\u21d2 \u00a0 \\(\\frac { x }{ y }\\) + \\(\\frac { 2y }{ 5 }\\)\u00a0= 8 …(1)
\n\\(\\frac { 2x }{ 7 }\\) \u2013 \\(\\frac { 5 }{ 2 }\\)\u00a0+ 1 = 9 \u00a0\u00a0\u21d2 \u00a0 \\(\\frac { 2x }{ 7 }\\) \u2013 \\(\\frac { 5 }{ 2 }\\)\u00a0= 8 ….(2)
\nThe equation (1) can be expressed as :
\n\\(\\frac { 5x+4y }{ 10 }\\)\u00a0= 8\u00a0 \u00a0\u00a0\u21d2 \u00a05x + 4y = 80 ….(3)
\nSimilarly, the equation (2) can be expressed as :
\n\\(\\frac { 4x-7y }{ 14 }\\)\u00a0= 8\u00a0 \u00a0\u00a0\u21d2 \u00a04x \u2013 7y = 112 ….(4)
\nNow the new system of equations is
\n5x + 4y = 80 ….(5)
\n4x \u2013 7y = 112 ….(6)
\nNow multiplying equation (5) by 4 and equation (6) by 5, we get
\n20x \u2013 16y = 320 ….(7)
\n20x + 35y = 560 ….(8)
\nSubtracting equation (7) from (8), we get ;
\ny =\u00a0\\(-\\frac { 240 }{ 51 }\\)
\nPutting y = \\(-\\frac { 240 }{ 51 }\\) in equation (5), we get ;
\n5x + 4 \u00d7 \\(\\frac { -240 }{ 51 }\\) = 80 \u00a0 \u21d2 \u00a0 \u00a05x \u2013 \\(\\frac { 960 }{ 51 }\\)\u00a0= 80
\n\u21d2 \u00a05x = 80 + \\(\\frac { 960 }{ 51 }\\)\u00a0= \\(\\frac { 4080+960 }{ 51 }\\)\u00a0=\u00a0\\(\\frac { 5040 }{ 51 }\\)
\n\u21d2 \u00a0x = \\(\\frac { 5040 }{ 255 }\\)\u00a0= \\(\\frac { 1008 }{ 51 }\\)= \\(\\frac { 336 }{ 17 }\\) \u00a0 \u21d2 \u00a0x =\u00a0\\(\\frac { 336 }{ 17 }\\)
\nHence, the solution of the system of equations is, x = \\(\\frac { 336 }{ 17 }\\), \u00a0 y = \\(-\\frac { 80 }{ 17 }\\)\u00a0.<\/p>\n

Example 5:<\/strong>\u00a0 \u00a0\u00a0Solve the following system of linear equatoins by using the method of elimination by equating the coefficients
\n\u221a3x \u2013 \u221a2y = \u221a3 = ; \u221a5x \u2013 \u221a3y = \u221a2
\nSol.<\/strong> \u00a0 \u00a0The given equations are
\n\u221a3x \u2013 \u221a2y = \u221a3 \u00a0 \u00a0 \u00a0….(1)
\n\u221a5x \u2013 \u221a3y = \u221a2 \u00a0 \u00a0 \u00a0 ….(2)
\nLet us eliminate y. To make the coefficients of equal, we multiply the equation (1) by \u221a3 and equation (2) by \u221a2 to get
\n3x \u2013 \u221a6y = 3 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 ….(3)
\n\u221a10x + \u221a6y = 2 \u00a0 \u00a0 \u00a0 \u00a0….(4)
\nAdding equation (3) and equation (4), we get
\n3x + \u221a10x = 5 \u21d2 \u00a0(3 + \u221a10) x = 5
\n\\( \\Rightarrow \\text{x}=\\frac{5}{3+\\sqrt{10}}=\\left( \\frac{5}{\\sqrt{10}+3} \\right)\\times \\left( \\frac{\\sqrt{10}-3}{\\sqrt{10}-3} \\right)\\)
\n\\(=5\\left( \\sqrt{10}-3 \\right)\\)
\nPutting x = 5( \u221a10\u2013 3) in (1) we get
\n\u221a3 \u00d7 5(\u221a10 \u2013 3) \u2013\u221a2 y =\u00a0\u221a3
\n\u21d2 5\u221a30 \u2013 15\u221a3 \u2013 \u221a2y =\u00a0\u221a3
\n\u21d2 \u221a2y = 5\u221a30 \u2013 15\u221a3 \u2013\u00a0\u221a3
\n\u21d2 \u221a2y = 5\u221a30 \u2013 16\u221a3
\n\u21d2 \\(y=\\frac{5\\sqrt{30}}{\\sqrt{2}}-\\frac{16\\sqrt{3}}{\\sqrt{2}}\\)
\n\u21d2 y = 5\u221a15\u00a0\u2013 8\u221a6
\nHence, the solution is x = 5( \u221a10\u2013 3) and y = 5\u221a15\u00a0\u2013 8\u221a6<\/p>\n

Example 6: \u00a0\u00a0<\/strong> Solve for x and y :
\n\\(\\frac { ax }{ b }\\) \u2013 \\(\\frac { by }{ a }\\)\u00a0= a + b ; ax \u2013 by = 2ab
\nSol. \u00a0<\/strong> The given system of equations is
\n\\(\\frac { ax }{ b }\\) \u2013 \\(\\frac { by }{ a }\\)\u00a0= a + b \u00a0 \u00a0 ….(1)
\nax \u2013 by = 2ab \u00a0 \u00a0 \u00a0….(2)
\nDividing (2) by a, we get
\nx \u2013 \\(\\frac { by }{ a }\\)\u00a0= 2b ….(3)
\nOn subtracting (3) from (1), we get
\n\\(\\frac { ax }{ b }\\) \u2013 x = a \u2013 b \u00a0 \u00a0\u21d2 \u00a0 \u00a0\\(x\\left( \\frac{a}{b}-1 \\right)\\) = a \u2013 b
\n\u21d2 x = \\(\\frac{(a-b)b}{a-b}\\)\u00a0= b \u00a0 \u00a0 \u00a0\u21d2 x = b
\nOn substituting the value of x in (3), we get
\nb \u2013 \u00a0\\(\\frac { by }{ a }\\)\u00a0 = 2b \u00a0\u21d2 \\(b\\left( 1-\\frac{y}{a} \\right)\\)\u00a0= 2b
\n\u21d2 \u00a01 \u2013 \u00a0\\(\\frac { y }{ a }\\)\u00a0= 2 \u00a0 \u00a0\u21d2 \u00a0 \u00a0\u00a0\\(\\frac { y }{ a }\\) = 1 \u2013 2
\n\u21d2 \u00a0\\(\\frac { y }{ a }\\) = \u20131 \u00a0 \u00a0 \u00a0\u21d2 \u00a0 y = \u2013a
\nHence, the solution of the equations is
\nx = b, y = \u2013 a<\/p>\n

Example 7: \u00a0 \u00a0<\/strong>Solve the following system of linear equations :
\n2(ax \u2013 by) + (a + 4b) = 0
\n2(bx + ay) + (b \u2013 4a) = 0
\nSol.<\/strong>\u00a0 \u00a0 2ax \u2013 2by + a + 4b = 0 …. (1)
\n2bx + 2ay + b \u2013 4a = 0 …. (2)
\nMultiplyng (1) by b and (2) by a and subtracting, we get
\n2(b2<\/sup> + a2<\/sup>) y = 4 (a2<\/sup> + b2<\/sup>) \u21d2 \u00a0\u00a0y = 2
\nMultiplying (1) by a and (2) by b and adding, we get
\n2(a2<\/sup> + b2<\/sup>) x + a2<\/sup> + b2<\/sup>\u00a0= 0
\n2(a2<\/sup> + b2<\/sup>) x = \u2013 (a2<\/sup> + b2<\/sup>) \u00a0\u21d2 \u00a0\u00a0x = \u2013 1\/2
\nHence x = \u20131\/2, and y = 2<\/p>\n

Example 8: \u00a0 \u00a0<\/strong>Solve (a \u2013 b) x + (a + b) y = a2<\/sup>\u00a0\u2013 2ab \u2013 b2<\/sup>
\n(a + b) (x + y) = a2<\/sup> + b2<\/sup>
\nSol.<\/strong>\u00a0 \u00a0The given system of equation is
\n(a \u2013 b) x + (a + b) y = a2<\/sup>\u00a0\u2013 2ab \u2013 b2<\/sup>\u00a0….(1)
\n(a + b) (x + y) = a2<\/sup> + b2<\/sup>\u00a0….(2)
\n\u21d2 \u00a0(a + b) x + (a + b) y = a2<\/sup> + b2<\/sup>\u00a0….(3)
\nSubtracting equation (3) from equation (1), we get
\n(a \u2013 b) x \u2013 (a + b) x = (a2<\/sup> \u2013 2ab\u2013 b2<\/sup>) \u2013 (a2<\/sup> + b2<\/sup>)
\n\u21d2 \u00a0\u20132bx = \u2013 2ab \u2013 2b2<\/sup>
\n\u21d2 \u00a0\\(\\text{x}=\\frac{-2ab}{-2b}-\\frac{2{{b}^{2}}}{-2b}=a+b\\)
\nPutting the value of x in (1), we get
\n\u21d2 \u00a0(a \u2013 b) (a + b) + (a + b) y = a2<\/sup>\u00a0 \u2013 2ab \u2013\u00a0b2<\/sup>
\n\u21d2 \u00a0(a + b) y = a2<\/sup>\u00a0 \u2013 2ab \u2013 b2<\/sup> \u2013 (a2<\/sup>\u00a0 \u2013 b2<\/sup>\u00a0)
\n\u21d2 \u00a0(a + b) y = \u2013 2ab
\n\u21d2 \u00a0y = \\(\\frac { -2ab }{ a+b }\\)
\nHence, the solution is x = a + b,
\ny =\u00a0\\(\\frac { -2ab }{ a+b }\\)<\/p>\n","protected":false},"excerpt":{"rendered":"

Solving Systems Of Equations By Elimination Method Step I: Let the two equations obtained be a1x + b1y + c1 = 0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u2026.(1) a2x + b2y + c2 = 0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0\u2026.(2) Step II: Multiplying the given equation so as to make the co-efficients of the variable to […]<\/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":[45,46,44],"yoast_head":"\nSolving Systems Of Equations By Elimination Method - A Plus Topper<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/www.aplustopper.com\/elimination-method\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Solving Systems Of Equations By Elimination Method\" \/>\n<meta property=\"og:description\" content=\"Solving Systems Of Equations By Elimination Method Step I: Let the two equations obtained be a1x + b1y + c1 = 0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u2026.(1) a2x + b2y + c2 = 0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0\u2026.(2) Step II: Multiplying the given equation so as to make the co-efficients of the variable to […]\" \/>\n<meta property=\"og:url\" content=\"https:\/\/www.aplustopper.com\/elimination-method\/\" \/>\n<meta property=\"og:site_name\" content=\"A Plus Topper\" \/>\n<meta property=\"article:publisher\" content=\"https:\/\/www.facebook.com\/aplustopper\/\" \/>\n<meta property=\"article:published_time\" content=\"2022-12-15T04:30:27+00:00\" \/>\n<meta property=\"article:modified_time\" content=\"2022-12-16T03:56:01+00:00\" \/>\n<meta name=\"twitter:card\" content=\"summary\" \/>\n<meta name=\"twitter:label1\" content=\"Written by\" \/>\n\t<meta name=\"twitter:data1\" content=\"Veerendra\" \/>\n\t<meta name=\"twitter:label2\" content=\"Est. reading time\" \/>\n\t<meta name=\"twitter:data2\" content=\"6 minutes\" \/>\n<script type=\"application\/ld+json\" class=\"yoast-schema-graph\">{\"@context\":\"https:\/\/schema.org\",\"@graph\":[{\"@type\":\"Organization\",\"@id\":\"https:\/\/www.aplustopper.com\/#organization\",\"name\":\"Aplus Topper\",\"url\":\"https:\/\/www.aplustopper.com\/\",\"sameAs\":[\"https:\/\/www.facebook.com\/aplustopper\/\"],\"logo\":{\"@type\":\"ImageObject\",\"@id\":\"https:\/\/www.aplustopper.com\/#logo\",\"inLanguage\":\"en-US\",\"url\":\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2018\/12\/Aplus_380x90-logo.jpg\",\"contentUrl\":\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2018\/12\/Aplus_380x90-logo.jpg\",\"width\":1585,\"height\":375,\"caption\":\"Aplus Topper\"},\"image\":{\"@id\":\"https:\/\/www.aplustopper.com\/#logo\"}},{\"@type\":\"WebSite\",\"@id\":\"https:\/\/www.aplustopper.com\/#website\",\"url\":\"https:\/\/www.aplustopper.com\/\",\"name\":\"A Plus Topper\",\"description\":\"Improve your Grades\",\"publisher\":{\"@id\":\"https:\/\/www.aplustopper.com\/#organization\"},\"potentialAction\":[{\"@type\":\"SearchAction\",\"target\":{\"@type\":\"EntryPoint\",\"urlTemplate\":\"https:\/\/www.aplustopper.com\/?s={search_term_string}\"},\"query-input\":\"required name=search_term_string\"}],\"inLanguage\":\"en-US\"},{\"@type\":\"WebPage\",\"@id\":\"https:\/\/www.aplustopper.com\/elimination-method\/#webpage\",\"url\":\"https:\/\/www.aplustopper.com\/elimination-method\/\",\"name\":\"Solving Systems Of Equations By Elimination Method - 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