Algebra 1

Over Lesson 6–3 Use elimination to solve the system of equations. 5x + y = 9 3x – y = 7 A. (2, –1) B. (–2, 1) C. (4, –3) D. (5, –2) Over Lesson 6–3 Use elimination to solve the system of equations. 2x + 4y = –8 2x + y = 1 A. (3, –2) B. (2, –3) C. (1, 3) D. (–1, 2) Over Lesson 6–3 Use elimination to solve the system of equations. x – 3y = 2 6x + 3y = –2 A. B. (0, 1) C. D. (–1, 3) Over Lesson 6–3 Find two numbers that have a sum of 151 and a difference of 7. A. 67, 84 B. 69, 82 C. 71, 80 D. 72, 79 Over Lesson 6–3 The student council sold pennants and pom-poms. The pom-poms cost $0.75 more than the pennants. Two pennants and two pom-poms cost $6.50. What are the prices for the pennants and the pom-poms? A. pennants = $1.50 pom poms = $2.25 B. pennants = $1.25 pom poms = $2.00 C. pennants = $1.10 pom poms = $2.10 D. pennants = $1.00 pom poms = $1.50 Over Lesson 6–3 Solve the system of equations below by using the elimination method. x – 3y = 15 4x + 3y = 15 A. (6, –3) B. (9, –2) C. (3, 1) D. (–3, 6) • Solve systems of equations by using elimination with multiplication. • Solve real-world problems involving systems of equations. Critical thinking question. How do you think you can convert the system so that it can be solved through elimination ? Solve the system by elimination. x + 2y = 11 –3x + y = –5 In some cases, you will first need to multiply one or both of the equations by a number so that one variable has opposite coefficients. Multiply One Equation to Eliminate a Variable Use elimination to solve the system of equations. 2x + y = 23 3x + 2y = 37 Multiply the first equation by –2 so the coefficients of the y terms are additive inverses. Then add the equations. 2x + y = 23 –4x – 2y = –46 Multiply by –2. 3x + 2y = 37 –x = –9 (+) 3x + equations. 2y= 37 Add the Divide each side by –1. x = Multiply One Equation to Eliminate a Variable Now substitute 9 for x in either equation to find the value of y. 2x + y = 23 First equation 2(9) + y = 23 18 + y = 23 18 + y – 18 = 23 – 18 side. x=9 Simplify. Subtract 18 from each y=5 Simplify. Answer: The solution is (9, 5). Use elimination to solve the system of equations. x + 7y = 12 3x – 5y = 10 A. (1, 5) B. (5, 1) C. (5, 5) D. (1, 1) Real-life application Numbered heards together , write the system Paige has $7.75 to buy 12 sheets of felt and card stock for her scrapbook. The felt costs $0.50 per sheet, and the card stock costs $0.75 per sheet. How many sheets of each can Paige buy? Write a system. Use f for the number of felt sheets and c for the number of card stock sheets. 0.50f + 0.75c = 7.75 f + c = 12 The cost of felt and card stock totals $7.75. The total number of sheets is 12. Step 1 0.50f + 0.75c = 7.75 Multiply the second equation by –0.50 to get + (–0.50)(f + c) = 12 opposite f-coefficients. 0.50f + 0.75c = 7.75 Add this equation to the + (–0.50f – 0.50c = –6) first equation to 0.25c = 1.75 Step 2 eliminate f. c=7 Solve for c. Step 3 f + c = 12 f + 7 = 12 –7 –7 f = 5 Write one of the original equations. Substitute 7 for c. Subtract 7 from both sides. Write the solution as an ordered pair. Paige can buy 7 sheets of card stock and 5 sheets of felt. Step 4 (7, 5) What if you cannot convert one equation to the other ? Push to zone 1/ in pairs St1,4 St 2,3 Solve the system by elimination. 2x + 5y = 26 –3x – 4y = –25 Step 1 Step 2 3(2x + 5y = 26) +(2)(–3x – 4y = –25) Multiply the first equation by 3 and the second equation by 2 to get opposite x-coefficients 6x + 15y = 78 +(–6x – 8y = –50) Add the new equations to eliminate x. 0 + 7y = 28 y =4 Solve for y. Step 3 2x + 5y = 26 2x + 5(4) = 26 2x + 20 = 26 –20 –20 2X = 6 x=3 Step 4 (3, 4) Write one of the original equations. Substitute 4 for y. Subtract 20 from both sides. Solve for x. Write the solution as an ordered pair. Multiply Both Equations to Eliminate a Variable Use elimination to solve the system of equations. 4x + 3y = 8 3x – 5y = –23 Method 1 Eliminate x. 4x + 3y = 8 3x – 5y = –23 12x + 9y = 24 Multiply by 3. (+)–12x + 20y= 29y 92Multiply by –4. = Divide each side 116 yby 29. Add the equations. = Multiply Both Equations to Eliminate a Variable Now substitute 4 for y in either equation to find x. 4x + 3y = 8 First equation 4x + 3(4) = 8 y=4 4x + 12 = 8 Simplify. 4x + 12 – 12 = 8 – 12 side. 4x = –4 Subtract 12 from each Simplify. Divide each side by 4. x Answer: The solution is (–1, 4). = –1 Multiply Both Equations to Eliminate a Variable Method 2 Eliminate y. 4x + 3y = 8 20x + 15y = 40 Multiply by 5. 3x – 5y = –23 29x (+) 9x – 15y= –69 Multiply by 3. = Divide each side –29 by 29. x = –1 Simplify. Add the Now substitute –1 for x in either equation. equations. Multiply Both Equations to Eliminate a Variable 4(–1) + 3y = 8 4x + 3y equation x = –1 –4 + 3y = 8 Simplify. –4 + 3y + 4 = 8 + 4 3y = 12 = 8 First Add 4 to each side. Simplify. Divide each side by 3. y =4 Simplify. Answer: The solution is (–1, 4), which matches the result obtained with Method 1. Use elimination to solve the system of equations. 3x + 2y = 10 2x + 5y = 3 A. (–4, 1) B. (–1, 4) C. (4, –1) D. (–4, –1)