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Intermediate Algebra Worksheets By O. Pauline Chow HACC Math 051 Intermediate Algebra Lecture Notes for chapter 3.1-3.5 (audio dated 8/17/07 at 10:03 pm, 29 minutes long) Solve 2x2 Systems of Linear Equations by Graphing Solve 2x2 Systems of Linear Equations by Substitution Solve 2x2 Systems of Linear Equations by Addition Methods Applications of Systems of Linear Equations Solve 3x3 Systems of Linear Equations Solve 2x2 Systems of Linear Equations by Graphing Ex 1. Solve the system by graphing. 1 y = x+2 2 5 y = x−2 2 For each equation, we set up a table of values and graph. x y= 1 x+2 2 0 2 –4 0 4 4 x y= 5 x−2 2 0 –2 0.8 0 2 3 The solution to the system is the point of intersection of the two lines: (2, 3). Ex 2. Solve the system by graphing. Intermediate Algebra, by Miller, O’Neill, Hyde, 2nd edition, McGraw Hill Intermediate Algebra Worksheets By O. Pauline Chow HACC x=4 y = 2x − 3 For each equation, we set up a table of values and graph. x y 4 2 4 0 x 0 1.5 2 Y = 2x – 3 –3 0 1 The solution to the system is the point of intersection of the two lines: (4, 5). Solve 2x2 Systems of Linear Equations by Subsitution Ex 1. Solve the system using substitution. x = 2 y − 10 2x + 3 y = 8 Using equation 1, we substitute 2y – 10 for x into equation 2. 2(2y – 10) + 3y = 8 Now we solve for y. 4y – 20 + 3y = 8 7y – 20 = 8 7y = 28 y=4 We substitute 4 for y into equation 1 and solve for x. Intermediate Algebra, by Miller, O’Neill, Hyde, 2nd edition, McGraw Hill Intermediate Algebra Worksheets By O. Pauline Chow HACC x = 2(4) – 10 = 8 – 10 = –2 So, the solution to the system is (–2, 4). Ex 2. Solve the system using substitution. 5x – y = 19 2x + 7y = –22 Before we can use the substitution, we shall solve for y in equation 1. 5x – y = 19 –y = –5x + 19 y = 5x – 19 (3) Now we substitute 5x – 19 for y into equation 2. 2x + 7(5x – 19) = –22 2x + 35x – 133 = –22 37x = 111 x=3 Next we substitute 3 for x in equation 3 to solve for y. y = 5(3) – 19 = –4 The solution to the system is (3, –4). Ex 3. Solve the system using substitution. 4x + 5y = 3 3x – 2y = 8 We use equation 1 to solve for x. 4x + 5y = 3 4x = –5y + 3 5 3 x=− y+ 4 4 Now substitute − (3) 5 3 y + for x into equation 2 and solve for y. 4 4 Intermediate Algebra, by Miller, O’Neill, Hyde, 2nd edition, McGraw Hill Intermediate Algebra Worksheets By O. Pauline Chow HACC 3 5 3 − y + − 2 y = 8 4 4 15 9 − y + − 2y = 8 4 4 −15 y + 9 − 8 y = 32 We multiply 4 to the equation. −23 y + 9 = 32 −23 y = 23 y = −1 Now we substitute –1 for y into equation 3 to solve for x. 5 3 8 x = − (−1) + = = 2 4 4 4 The solution to the system is (2, –1) Solve 2x2 Systems of Linear Equations by Addition Ex 1: Solve the system using Addition. 5x – 3y = 15 2x + 7y = 56 We need to multiply each equation by an appropriate constant to make the coefficients of x or y opposite to each other. We shall multiply equation 1 by –2 and equation 2 by 5. –2(5x – 3y = 15) 5(2x + 7y = 47) –10x + 6y = –30 10x + 35y = 235 Now add the two equations. –10x + 6y = –30 10x + 35y = 235 41y = 205 y=5 Now substitute 5 for y in equation 1 to solve for x. 5x – 3(5) = 15 5x – 15 = 15 5x = 30 x=6 Intermediate Algebra, by Miller, O’Neill, Hyde, 2nd edition, McGraw Hill Intermediate Algebra Worksheets By O. Pauline Chow HACC The solution to the system is (6, 5). Ex 2. Solve the system using Addition. x – 3y = 15 2x – 6y = 31 We multiply equation 1 by –2. –2x + 6y = –30 2x – 6y = 31 0 = 1 No solution! There is no solution to the system. This means the lines are parallel to each other. Ex 3. Solve the system using Addition. 2x – 3y = 4 4x – 6y = 8 We multiply equation 1 by –2. –4x + 6y = –8 4x – 6y = 8 0 = 0 An identity! This means the lines are the same and the equations are equivalent. So, we write the solution in set notation, {(x, y) | 2x – 3y = 4}. Applications of Systems of Linear Equations Ex 1. A fruit punch that contains 25% fruit juice is combined with a fruit drink that contains 10% fruit juice. How many ounces of each should be used to make 48oz of a mixture that is 15% fruit juice? To solve this problem, we let x and y be the number of ounces of 25% and 10% fruit juice, respectively. Since the total number of ounces of the mixture is 48, x + y = 48. We use the percent to find the number of ounces of fruit in each juice. Juice I % of fruit 25% # of oz of juice x # of oz of fruit in each juice (25%)(x) Juice II 10% y (10%)(y) Mixture 15% 48 x + y = 48 (15%)(48) 0.25x + 0.10y = 7.2 Now we solve the 2x2 system using substitution method. We shall solve the equation 1 for x and substitute into equation 2. Intermediate Algebra, by Miller, O’Neill, Hyde, 2nd edition, McGraw Hill Intermediate Algebra Worksheets By O. Pauline Chow HACC x = 48 – y 0.25(48 – y) + 0.10y = 7.2 12 – 0.25y + 0.10y = 7.2 12 – 0.15y = 7.2 –0.15y = –4.8 y = 32 x = 48 – 32 = 16 16 oz of 25% fruit juice and 32 oz of 10% fruit juice. Ex 2. Jack cruised 42 miles downriver in 45 minutes. His return trip upriver took 1 hour. Determine the speed of his motorboat in still water and the speed of the river’s current. We let x mph be the speed of the motorboat in still water and y mph be the speed of the river’s current. When the boat is going upriver, its speed is (x – y) mph. When the boat is going downriver, its speed is (x + y) mph. distance Speed time d = st Downriver 42 miles (x + y) mph 42 = 0.75(x + y) 45 45 minutes = = 0.75 hour 60 Upriver 42 miles (x – y) mph 1 hour 42 = 1(x – y) Now we have a 2x2 system. We will divide 0.75 into equation 1 to make the system easier. x + y = 56 x – y = 42 Now we just add the equations and solve for x. 2x = 98 x = 49 We substitute 49 for x into equation 2 to solve for y. 49 – y = 42 –y = –7 y=7 Speed of the boat in still water is 49 mph and current is 7 mph. Ex 3. A rectangle has the perimeter of 42 m. The length is 1 m longer than the width. Find the dimensions of the rectangle. Let x m be the width and y m be the length. We set up the two equations: length is 1 m longer than the width y = 1 + x Intermediate Algebra, by Miller, O’Neill, Hyde, 2nd edition, McGraw Hill Intermediate Algebra Worksheets By O. Pauline Chow HACC Perimeter is 42 m 2x + 2y = 42 y=1+x 2x + 2y = 42 We will substitute 1 + x for y into equation 2. 2x + 2(1 + x) = 42 2x + 2 + 2x = 42 4x = 40 x = 10 y = 11 The dimensions of the rectangle are 10 m by 11 m. Solve 3x3 Systems of Linear Equations Ex 1 Solve the 3x3 system. 2x + 3y – z = –6 3x – y + 2z = 11 x + 4y – 5z = –15 We shall apply the addition method. First we use equations 1 and 2 to eliminate x. 3(2x + 3y – z = –6) –2(3x – y + 2z = 11) 6x + 9y – 3z = –18 –6x + 2y – 4z = –22 11y – 7z = –40 (4) Next we use equations 1 and 3 to eliminate x. 2x + 3y – z = –6 –2(x + 4y – 5z = –15) 2x + 3y – z = –6 –2x – 8y +10z = 30 –5y + 9z = 24 (5) Now we have a 2x2 system. We shall apply the addition method. 5(11y – 7z = -40) 11(–5y + 9z = 24) 55y – 35z = -200 –55y + 99z = 264 64z = 64, z = 1 We will use back substitution to solve for y and x. Substitute 1 for z into equation 4 to solve for y. 11y – 7(1) = –40 11y = –33 y = –3 We will substitute –3 for y and 1 for z into equation 3 to solve for x. Intermediate Algebra, by Miller, O’Neill, Hyde, 2nd edition, McGraw Hill Intermediate Algebra Worksheets By O. Pauline Chow HACC x + 4(–3) – 5(1) = –15 x – 12 – 5 = –15 x – 17 = –15 x=2 The solution to the system is (2, –3, 1). Intermediate Algebra, by Miller, O’Neill, Hyde, 2nd edition, McGraw Hill