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Solving Systems Using Elimination Section 6-3 Part 2 Goals Goal • To solve systems by adding or subtracting to eliminate a variable. Rubric Level 1 – Know the goals. Level 2 – Fully understand the goals. Level 3 – Use the goals to solve simple problems. Level 4 – Use the goals to solve more advanced problems. Level 5 – Adapts and applies the goals to different and more complex problems. Vocabulary • Elimination Method Further Elimination • In Part 1 of this lesson, you was that to eliminate a variable, its coefficients must have a sum or difference of zero. • 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, so you can add or subtract to eliminate the variable. Further Elimination Procedure Step 1: Put the equations in Standard Form. Step 2: Determine which variable to eliminate. Standard Form: Ax + By = C Look for variables that have the same coefficient. Step 3: Multiply the equations and solve. Solve for the variable. Step 4: Plug back in to find the other variable. Substitute the value of the variable into the equation. Step 5: Check your solution. Substitute your ordered pair into BOTH equations. Example: Multiplying One Equation 2x + 2y = 6 3x – y = 5 Step 1: Put the equations in Standard Form. Step 2: Determine which variable to eliminate. They already are! None of the coefficients are the same! Find the least common multiple of each variable. LCM = 6x, LCM = 2y Which is easier to obtain? 2y (you only have to multiply the bottom equation by 2) Example: Continued 2x + 2y = 6 3x – y = 5 Step 3: Multiply the equations and solve. Multiply the bottom equation by 2 2x + 2y = 6 2x + 2y = 6 (2)(3x – y = 5) (+) 6x – 2y = 10 8x = 16 x=2 Step 4: Plug back in to find the other variable. 2(2) + 2y = 6 4 + 2y = 6 2y = 2 y=1 Example: Continued 2x + 2y = 6 3x – y = 5 Step 5: Check your solution. (2, 1) 2(2) + 2(1) = 6 3(2) - (1) = 5 Solving with multiplication adds one more step to the elimination process. Example: Multiplying One Equation x + 4y = 7 4x – 3y = 9 Step 1: Put the equations in Standard Form. Step 2: Determine which variable to eliminate. They already are! Find the least common multiple of each variable. LCM = 4x, LCM = 12y Which is easier to obtain? 4x (you only have to multiply the top equation by -4 to make them inverses) Example: Continued x + 4y = 7 4x – 3y = 9 Step 3: Multiply the equations and solve. Multiply the top equation by -4 (-4)(x + 4y = 7) -4x – 16y = -28 4x – 3y = 9) (+) 4x – 3y = 9 -19y = -19 y=1 Step 4: Plug back in to find the other variable. x + 4(1) = 7 x+4=7 x=3 Example: Continued x + 4y = 7 4x – 3y = 9 Step 5: Check your solution. (3, 1) (3) + 4(1) = 7 4(3) - 3(1) = 9 Your Turn: Solve the system by elimination. x + 2y = 11 –3x + y = –5 x + 2y = 11 –2(–3x + y = –5) x + 2y = 11 +(6x –2y = +10) 7x + 0 = 21 7x = 21 x=3 Multiply each term in the second equation by –2 to get opposite y-coefficients. Add the new equation to the first equation. Simplify and solve for x. Continued x + 2y = 11 3 + 2y = 11 –3 –3 2y = 8 y=4 (3, 4) Write one of the original equations. Substitute 3 for x. Subtract 3 from each side. Simplify and solve for y. Write the solution as an ordered pair. Your Turn: Solve the system by elimination. 3x + 2y = 6 –x + y = –2 3x + 2y = 6 3(–x + y = –2) 3x + 2y = 6 +(–3x + 3y = –6) 0 + 5y = 0 5y = 0 y=0 Multiply each term in the second equation by 3 to get opposite x-coefficients. Add the new equation to the first equation. Simplify and solve for y. Continued –x + y = –2 –x + 3(0) = –2 –x + 0 = –2 –x = –2 x=2 (2, 0) Write one of the original equations. Substitute 0 for y. Simplify and solve for x. Write the solution as an ordered pair. Example: Multiplying Both Equations 3x + 4y = -1 4x – 3y = 7 Step 1: Put the equations in Standard Form. Step 2: Determine which variable to eliminate. They already are! Find the least common multiple of each variable. LCM = 12x, LCM = 12y Which is easier to obtain? Either! I’ll pick y because the signs are already opposite. Example: Continued 3x + 4y = -1 4x – 3y = 7 Step 3: Multiply the equations and solve. Multiply both equations (3)(3x + 4y = -1) 9x + 12y = -3 (4)(4x – 3y = 7) (+) 16x – 12y = 28 25x = 25 x=1 Step 4: Plug back in to find the other variable. 3(1) + 4y = -1 3 + 4y = -1 4y = -4 y = -1 Example: Continued 3x + 4y = -1 4x – 3y = 7 Step 5: Check your solution. (1, -1) 3(1) + 4(-1) = -1 4(1) - 3(-1) = 7 Your Turn: Solve the system by elimination. –5x + 2y = 32 2x + 3y = 10 2(–5x + 2y = 32) 5(2x + 3y = 10) –10x + 4y = 64 +(10x + 15y = 50) 19y = 114 y=6 Multiply the first equation by 2 and the second equation by 5 to get opposite x-coefficients Add the new equations. Simplify and solve for y. Continued 2x + 3y = 10 2x + 3(6) = 10 2x + 18 = 10 –18 –18 2x = –8 x = –4 (–4, 6) Write one of the original equations. Substitute 6 for y. Subtract 18 from both sides. Simplify and solve for x. Write the solution as an ordered pair. Your Turn: Solve the system by elimination. 2x + 5y = 26 –3x – 4y = –25 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. 0 + 7y = 28 Simplify and solve for y. y =4 Your Turn: 2x + 5y = 26 2x + 5(4) = 26 2x + 20 = 26 –20 –20 2x = 6 x=3 (3, 4) Write one of the original equations. Substitute 4 for y. Subtract 20 from both sides. Simplify and solve for x. Write the solution as an ordered pair. IDENTIFYING THE NUMBER OF SOLUTIONS CONCEPT NUMBER OF SOLUTIONS OF A LINEAR SYSTEM SUMMARY y y y x x x Lines intersect Lines are parallel Lines coincide one solution no solution infinitely many solutions Identifying The Number of Solutions • If both variable terms are eliminated as you solve a system of equations, the answer is either no solution or infinite solutions. – No solution: get a false statement when solving the system. – Infinite solutions: get a true statement when solving the system. A Linear System with Infinite Solutions Show that this linear system M ETHOD: Elimination has infinitely many solutions. –2x y 3 – 4 x 2y 6 Equation 1 Equation 2 You can multiply Equation 1 by –2. 4x – 2y – 6 – 4 x 2y 6 0 0 Multiply Equation 1 by –2. Write Equation 2. Add Equations. True statement! The variables are eliminated and you are left with a statement that is true regardless of the values of x and y. This result tells you that the linear system has infinitely many solutions. A Linear System with No Solution Show that this linear system METHOD: Elimination has no solution. 2x y 5 2x y 1 Equation 1 Equation 2 You can multiply Equation 1 by –1. -2x - y -5 2x y 1 0 -4 Multiply Equation 1 by –1. Write Equation 2. Add Equations. False statement! The variables are eliminated and you are left with a statement that is false regardless of the values of x and y. This result tells you that the linear system has no solution. Your Turn: Solve the systems using elimination. 2 x 5 y 7 1) 2 12 y 5 x 5 False Statement No Solution 12x 8 y 20 2) 3x 2 y 5 True Statement Infinite Solutions Summary 7.3 The Elimination Method Summary of Methods for Solving Systems Example 6x + y = 10 y=5 Suggested Method Substitution Why The value of one variable is known and can easily be substituted into the other equation. 7.3 The Elimination Method Summary of Methods for Solving Systems Example Suggested Method Why 2x – 5y = –20 4x + 5y = 14 Elimination eliminate ‘y’ 5 Add the two equations 7.3 The Elimination Method Summary of Methods for Solving Systems Example Suggested Method Why 9a – 2b = –11 8a + 4b = 25 Elimination eliminate ‘b’ 4 Multiply first equation by 2 Add the equations Joke Time • How does an octopus go to war? • Well-Armed! • Why does a Moo-rock taste better than an Earth-rock? • Because it’s a little meteor! • What did the elder chimney say to the younger chimney? • You’re too young to smoke! Assignment • 6-3 Part 2 Exercises Pg. 402 - 404: #6 – 34 even