# D3notes - Supplementary Course Module D: Solving Equations...

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Unformatted text preview: Supplementary Course Module D: Solving Equations and Inequalities Assignment 3 & Systems of Equations; Inequalities Solving Systems of Equations & To solve a system of two equations in two variables, say x and y , we are looking for all pairs of values ( x; y ) which simultaneously satisfy both equations. & There are two common methods which are used for such problems: elimination and substitution. example: Solve the system 2 x ¡ 3 y = 1 1 8 y ¡ 6 x = 4 2 Substitution: From 1 : x = 1 2 (1 + 3 y ) Sub into 2 : ¡ 6 & 1 2 (1 + 3 y ) ¡ + 8 y = 4 y = ¡ 7 Sub y = ¡ 7 into 1 : x = 1 2 (1 ¡ 21) = ¡ 10 ) there is one solution: ( ¡ 10 ; ¡ 7) Elimination: 1 ¢ 3 6 x ¡ 9 y = 3 2 ¡ 6 x + 8 y = 4 Add: ¡ y = 7 y = ¡ 7 & Sub y = ¡ 7 into 1 : 2 x + 21 = 1 2 x = ¡ 20 x = ¡ 10 ) there is one solution: ( ¡ 10 ; ¡ 7) example: Solve the system 3 x ¡ 4 y = 8 1 ¡ 6 x + 8 y = ¡ 16 2 1 ¢ 2 6 x ¡ 8 y = 16 2 ¡ 6 x + 8 y = ¡ 16 Add: y = 0 y 2 R ) there are in&nite solutions (such a system is called dependent). example: Solve the system 6 x ¡ 8 y = 16 1 ¡ 6 x + 8 y = 16 2 Add: y = 32- > no solution exists ) the system has no solution (such a system is called inconsistent). & A system of equations involving three unknowns may also be solved if three equations linking the variables are given. (And a system of equations involving four unknowns may be solved if 4 equations are known, etc.) & such systems may require various applications of both the substitution and elimination methods example: Solve the system 2 x + y + z = 7 1 x ¡ y + 2 z = 11 2 5 x + y ¡ 2 z = 1 3 Add 2 + 3 : 6 x = 12 () x = 2 Sub into...
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## This note was uploaded on 11/25/2010 for the course MATH MA110 taught by Professor Hu during the Fall '10 term at Wilfred Laurier University .

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D3notes - Supplementary Course Module D: Solving Equations...

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