tailieunhanh - Engineering Mathematics 4 Episode 3

Tham khảo tài liệu 'engineering mathematics 4 episode 3', kỹ thuật - công nghệ, cơ khí - chế tạo máy phục vụ nhu cầu học tập, nghiên cứu và làm việc hiệu quả | SIMULTANEOUS EQUATIONS 69 More difficult worked problems on simultaneous equations Problem 8. Solve 2 3 C - 7 1 x y 1 4 - - -2 2 x y In this type of equation the solution is easier if a substitution is initially made. Let - a and - b x y Thus equation 1 becomes 2a C 3b 7 3 and equation 2 becomes a 4b 2 4 Multiplying equation 4 by 2 gives 2a - 8b -4 5 Subtracting equation 5 from equation 3 gives 0 C 11b 11 . b 1 Substituting b 1 in equation 3 gives 2a C 3 7 2a 7 - 3 4 . a 2 Checking substituting a 2 and b 1 in equation 4 gives LHS 2 - 4 1 2 - 4 -2 RHS Hence a 2 and b 1 However since - a x and since - b y Hence the solution is x which may be checked in 1 then x - a 1 then y - b 1 y 1 2 he original eq _ 1 2 1 4 1 1 uations. Problem 9. Solve 1 C ẳ 4 2a 5b 1 4 C 1 a 2b 2 1 1 Let - x and - y a b x3 then 2 C 5y 4 3 1 4x C 1y 4 To remove fractions equation 3 is multiplied by 10 giving 10 xì C 10 3y 10 4 25 . 5x C 6y 40 Multiplying equation 4 by 2 gives 5 8x C y 21 6 Multiplying equation 6 by 6 gives 48x C 6y 126 7 Subtracting equation 5 from equation 7 gives 43x C 0 86 86 43 2 x Substituting x 2 into equation 3 gives 2 C 3 y 4 2 5 3 5 y 4 - 1 3 y 3 3 5 1 1 _ 1 Since - x then a - a x 2 1 1 1 and since - y then b - - b y 5 Hence the solution is a b Ệ which may be checked in the original equations. Problem 10. Solve 1 4 xCy 27 1 4 2x y 33 1 2 70 ENGINEERING MATHEMATICS To eliminate fractions both sides of equation 1 are multiplied by 27 x C y giving 27 x C y D 27 x C y A x C y 27 . 27 1 D 4 x C y 27 D 4x C 4y 3 Similarly in equation 2 33 D 4 2x y . 33 D 8x - 4y 4 Equation 3 equation 4 gives 60 D 12x . x D D 5 12 Substituting x D 5 in equation 3 gives 27 D 4 5 C 4y from which 4y D 27 - 20 D 7 and y D 7 D 13 44 Hence x 5 J 1 is the required solution which may be checked in the original equations. Now try the following exercise Exercise 34 Further more difficult problems on simultaneous equations 4. c 1 d C 2 1 1 n 4 3 1 c 3 - d 13 5 1 4 20 c D 3

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