Lesson 5.2 Solving Systems of Linear Equations Using Algebraic Methods
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1 Name: Date: Lesson 5. Solving Systems of Linear Equations Using Algebraic Methods Solve the systems of linear equations using the elimination method. 1. 3y x 5. x 5y y 1 x y x q 1 p w 3v 5 4 p q 5 5 w 1 3v a b n m 5 3 3a 1 b m 6n x 1 6y p 1 5q x 1 3y 5 6 3p 10q Chapter 5 Lesson 5.
2 Name: Date: Solve the systems of linear equations using the substitution method. 9. 3a b p 1 3q 5 7 b 5 a 7 q 5 p c b y x 5 3 b c 5 5 y x h 1 k x 1 y 5 36 h 1 k 5 7 5y x t 1 s x 1 4y 5 6 7t s x 5 6y Extra Practice Course 3A 85
3 Name: Date: Solve the systems of linear equations using the substitution method or the elimination method. Explain why you choose the method x 1 5y m n x 4y 5 8 n 5m m 1 4n w 4v 5 1 m 5 5n 1 v 5 6w h 1 9k y x 5h 5k 5 0 3x y b 1 4c p q b 1 16c p 1 4q Chapter 5 Lesson 5.
4 8. 3x 1 4y 5 15 x 0 1 y x 1 3 y x 5 y x 0 1 y Only the pair of values x 5 1 and y 5 3 appear in both tables. So the solution of the system of equations is x 5 1 and y x 1 4y 5 64 x y x 1 8y 5 7 x y 7 4 Only the pair of values x 5 8 and y 5 6 appear in both tables. So the solution of the system of equations is x 5 8 and y 5 6. Jolene took 8 minutes to fold a paper airplane and 6 minutes to fold a paper star. 10. x 5 9y 51 8 x y 0 x 1 3y x y 8 Only the pair of values x 5 9 and y 5 appear in both tables. So the solution of the system of equations is x 5 9 and y 5. x y At the present time, Janice is 18 years old and Jennifer is 6 years old. 11. x 1 y x 1 3 y x 1 3y Only the pair of values x 5 3 and y 5 1 appear in both tables. So the solution of the system of equations is x 5 3 and y 5 1. Difference The difference between Jack s walking speed and cycling speed is 9 mi/h. Lesson y x 5 Equation 1 3y 1 x 5 16 Equation (3y x) 1 (3y 1 x) y 1 3y x 1 x y 5 18 y 5 3 Substitute y 5 3 into Equation 1: 3(3) x 5 9 x 5 x 5 7 equations is x 5 7 and y x 5y 5 13 Equation 1 9y x 5 17 Equation (x 5y) 1 (9y x) (17) x x 5y 1 9y y 5 4 y 5 1 Substitute y 5 1 into Equation 1: x 5(1) 5 13 x x 5 8 equations is x 5 8 and y q 1 p 5 9 Equation 1 p q 5 5 Equation Subtract Equation from Equation 1: (7q 1 p) (p q) q 1 q 1 p p 5 4 8q 5 4 q 5 3 Substitute q 5 3 into Equation 1: 7(3) 1 p p 5 9 p 5 8 p 5 4 equations is p 5 4 and q 5 3. Extra Practice Course 3A 175
5 4. w 3v 5 4 Equation 1 w 1 3v 5 9 Equation (w 3v) 1 (w 1 3v) w 1 w 3v 1 3v w 5 33 w 5 11 Substitute w 5 11 into Equation 1: (11) 3v 5 4 3v 5 4 3v 5 18 v 5 6 equations is v 5 6 and w a b 5 6 Equation 1 3a 1 b 5 19 Equation (a b) 1 (3a 1 b) a 1 3a b 1 b 5 5 5a 5 5 a 5 5 Substitute a 5 5 into Equation 1: (5) b b 5 6 b 5 4 equations is a 5 5 and b n m 5 3 Equation 1 3m 6n 5 15 Equation (6n m) 1 (3m 6n) n 6n m 1 3m 5 18 m 5 18 m 5 9 Substitute m 5 9 into Equation 1: 6n n 5 1 n 5 equations is m 5 9 and n x 1 6y 5 14 Equation 1 6x 1 3y 5 6 Equation Multiply Equation by : (6x 1 3y) 5 (6) 1x 1 6y 5 1 Equation 3 Subtract Equation 3 from Equation 1: (8x 1 6y) (1x 1 6y) x 1x 1 6y 6y 5 4x 5 x 5 1 Substitute x 5 1 into Equation 1: 8 1 6y y y 5 18 y 5 3 equations is x 5 1 and y p 1 5q 5 18 Equation 1 3p 10q 5 69 Equation (4p 1 5q) 5 (18) 8p 1 10q 5 36 Equation 3 Add Equation and Equation 3: (3p 10q) 1 (8p 1 10q) (36) 3p 1 8p 10q 1 10q p 5 33 p 5 3 Substitute p 5 3 into Equation 1: 4(3) 1 5q q q 5 30 q 5 6 equations is p 5 3 and q a b 5 13 Equation 1 b 5 a 7 Equation Substitute Equation into Equation 1: 3a (a 7) a a a a 5 6 Substitute a 5 6 into Equation : b 5 (6) 7 b b 5 5 equations is a 5 6 and b p 1 3q 5 7 Equation 1 q 5 p 1 5 Equation Substitute Equation into Equation 1: 5p 1 3(p 1 5) 5 7 5p 6p p 5 p 5 Substitute p 5 into Equation : q 5 () 1 5 q q 5 39 equations is p 5 and q Answers
6 11. 6c b 5 5 Equation 1 b c 5 5 Equation Use Equation to solve for b in terms of c: b c 5 5 b c Equation 3 6c (5 1 c) 5 5 6c 5 c 5 5 5c 5 10 c 5 Substitute c 5 into Equation 3: b equations is b 5 7 and c y x 5 3 Equation 1 y x 5 4 Equation Use Equation to solve for y in terms of x: y x 5 4 y x Equation 3 (4 1 x) x x x x 5 3 x 5 5 Substitute x 5 5 into Equation 3: y (5) equations is x 5 5 and y h 1 k 5 7 Equation 1 h 1 k 5 7 Equation Use Equation to solve for h in terms of k: h 1 k 5 7 h 5 7 k Equation 3 4(7 k) 1 k k 1 k k 5 7 7k 5 1 k 5 3 Substitute k 5 3 into Equation 3: h 5 7 (3) 5 1 equations is h 5 1 and k x 1 y 5 36 Equation 1 5y x 5 39 Equation Use Equation to solve for x in terms of y: 5y x 5 39 x 5 5y 39 Equation 3 3(5y 39) 1 y y y y y y 5 9 Substitute y 5 9 into Equation 3: x 5 5(9) equations is x 5 6 and y t 1 s 5 3 Equation 1 7t s 5 15 Equation Use Equation to solve for s in terms of t: 7t s 5 15 s 5 7t 15 Equation 3 5t 1 7t t t 5 1 t 5 1 Substitute t 5 1 into Equation 3: s 5 7(1) 15 s 5 8 s 5 4 equations is s 5 4 and t x 1 4y 5 6 Equation 1 5 x 5 6y Equation Use Equation to solve for x in terms of y: 5 x 5 6y x 5 6y 1 5 Equation 3 5(6y 1 5) 1 4y y y y 5 51 y 5 3 Substitute y 5 3 into Equation 3: x equations is x 5 4 and y x 1 5y 5 35 Equation 1 6x 4y 5 8 Equation (3x 1 5y) 5 (35) 6x 1 10y 5 70 Equation 3 Subtract Equation 3 from Equation : (6x 4y) (6x 1 10y) x 6x 4y 10y y 5 98 Extra Practice Course 3A 177
7 Substitute y 5 7 into Equation 1: 3x 1 5(7) x x 5 0 equations is x 5 0 and y 5 7. substitution method will result in an algebraic fraction that will make the steps complicated m n 5 13 Equation 1 n 5m 5 11 Equation (7m n) 1 (n 5m) m 5m n 1 n 5 m 5 m 5 1 Substitute m 5 1 into Equation 1: 7(1) n n 5 13 n 5 6 n 5 3 equations is m 5 1 and n 5 3. substitution method will result in an algebraic fraction that will make the steps complicated m 1 4n 5 38 Equation 1 m 5 5n 1 Equation (9m 1 4n) 5 (38) 18m 1 8n 5 76 Equation 3 Multiply Equation by 9: 9(m) 5 9(5n 1) 18m 5 45n 189 Equation 4 Subtract Equation 4 from Equation 3: (18m 1 8n) 18m 5 76 (45n 189) 8n n n 1 45n n 5 65 n 5 5 Substitute n 5 5 into Equation 1: 9m 1 4(5) m m 5 18 m 5 equations is m 5 and n 5 5. substitution method will result in an algebraic fraction that will make the steps complicated. 0. 5w 4v 5 1 Equation 1 v 5 6w 1 14 Equation Substitute Equation into Equation 1: 5w 4(6w 1 14) Answers 5w 4w w 5 57 w 5 3 Substitute w 5 3 into Equation : v 5 6(3) equations is w 5 3 and v 5 4. Substitution method is used as v is already expressed in terms of w. 1. h 1 9k 5 19 Equation 1 5h 5k 5 0 Equation Multiply Equation 1 by 5: 5(h 1 9k) 5 5(19) 10h 1 45k 5 95 Equation 3 Multiply Equation by : (5h 5k) 5 (0) 10h 10k 5 40 Equation 4 Subtract Equation 4 from Equation 3: (10h 1 45k) (10h 10k) h 10h 1 45k 1 10k k 5 55 k 5 1 Substitute k 5 1 into Equation 1: h 1 9(1) 5 19 h h 5 10 h 5 5 equations is h 5 5 and k 5 1. substitution method will result in an algebraic fraction that will make the steps complicated.. 5y x Equation 1 3x y 5 18 Equation Substitute Equation 1 into Equation : 5y 1 9 y y y y 5 9 y 5 3 Substitute y 5 3 into Equation 1: 5(3) x 3x 5 4 x 5 8 equations is x 5 8 and y 5 3. Substitution method is used as 3x is already expressed in terms of y. 3. 3b 1 4c 5 6 Equation 1 7b 1 16c 5 34 Equation Multiply Equation 1 by 4: 4(3b 1 4c) 5 4(6) 1b 1 16c 5 4 Equation 3 Subtract Equation 3 from Equation :
8 (7b 1 16c) (1b 1 16c) 5 34 (4) 7b 1b 1 16c 16c b 5 10 b 5 Substitute b 5 into Equation 1: 3() 1 4c c 5 6 4c c 5 1 c 5 3 equations is b 5 and c 5 3. substitution method will result in an algebraic fraction that will make the steps complicated. 4. 7p q 5 18 Equation 1 3p 1 4q 5 1 Equation Use Equation 1 to solve for q in terms of p: 7p q 5 18 q 5 7p 18 Equation 3 Substitute Equation 3 into Equation : 3p 1 4(7p 18) 5 1 3p 1 8p p p 5 93 p 5 3 Substitute p 5 3 into Equation 3: q 5 7(3) 18 q 5 3 equations is p 5 3 and q 5 3. Substitution method is used as q can easily be expressed in terms of p. Lesson Let the number of art magazines be x and the number of science magazines be y. x 1 y 5 6 Equation 1 4x 1 7y Equation Solve Equation 1 for x in terms of y: x 5 6 y Equation 3 Substitute Equation 3 into Equation : 4(6 y) 1 7y y 1 7y y y 5 30 y 5 10 Substitute y 5 10 into Equation 3: x x 5 16 Jenny purchased 16 art magazines and 10 science magazines.. Let the number of adult tickets be x and the number of children s tickets be y. x 1 y 5 95 Equation 1 1x 1 9y Equation Solve Equation 1 for x in terms of y: x 5 95 y Equation 3 Substitute Equation 3 into Equation : 1(95 y) 1 9y ,140 1y 1 9y ,140 3y y 5 1, y 5 60 Substitute y 5 60 into Equation 3: x x 5 35 There were 35 adult tickets and 60 children s tickets sold. 3. Let the number of packets of roasted peanuts be x and the number of packets of beef jerky be y. 5x 1 3y Equation 1 3x 1 y Equation (5x 1 3y) 5 (37.80) 10x 1 6y Equation 3 Multiply Equation by 3: 3(3x 1 y) 5 3(3.87) 9x 1 6y Equation 4 Subtract Equation 4 from Equation 3: (10x 1 6y) (9x 1 6y) x 9x 1 6y 6y x Substitute x into Equation 3: 10(3.99) 1 6y y y y The cost of a packet of roasted peanuts is $3.99 and that of a packet of beef jerky is $ Let the number of wheat crackers a glass container can hold be x and the number of wheat crackers a plastic container can hold be y. 6x 1 y Equation 1 4x y Equation (6x 1 y) 5 (180) 1x 1 4y Equation 3 Multiply Equation by 3: 3(4x) 5 3(5 1 5y) 1x y Equation 4 Subtract Equation 4 from Equation 3: (1x 1 4y) 1x ( y) 4y y 19y 5 85 y 5 15 Substitute y 5 15 into Equation 3: Extra Practice Course 3A 179
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