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Algebra 2 Solving Systems Algebraically Lesson 3-2 Part 2 Goals Goal • To solve a linear systems using elimination. 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 • Equivalent Systems Essential Question Big Idea: Solving Equations and Inequalities • When can you use elimination to solve a system of linear equation? Definition • You can also solve systems of linear equations with the elimination method. • With elimination, you get rid of one of the variables by adding the two equations of the system. • You may have to multiply one or both equations by a number to create variable terms that can be eliminated. • The elimination method is sometimes called the addition method or linear combination. Elimination Method • Like substitution, the goal of elimination is to get one equation that has only one variable. To do this by elimination, you add the two equations in the system together. • To eliminate a variable, the coefficients of that variable in the two equations must be additive inverses. To achieve this, we use properties of algebra to change the system to an equivalent system, one with the same solution set. The three transformations that produce an equivalent system are listed next. Elimination Method Transformations of a Linear System 1. Interchange any two equations of the system. 2. Multiply or divide any equation of the system by a nonzero real number. 3. Replace any equation of the system by the sum of that equation and a multiple of another equation in the system. Elimination Method • When adding the two equations in the system together to eliminate one of the variables, remember that an equation stays balanced if you add equal amounts to both sides. • So, if 5x + 2y = 1, you can add 5x + 2y to one side of an equation and 1 to the other side and the balance is maintained. Elimination Method Since –2y and 2y have opposite coefficients (additive inverses), the y-term is eliminated. The result is one equation that has only one variable: 6x = –18. When you use the elimination method to solve a system of linear equations, align all like terms in the equations. Then determine whether any like terms can be eliminated because they have opposite coefficients. Elimination Method • 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 to eliminate the variable. • Elimination is easiest when the equations are in standard form. 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 add the equations. 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: x+y=5 3x – y = 7 Step 1: Put the equations in Standard Form. Step 2: Determine which variable to eliminate. Step 3: Add the equations. They already are! The y’s have the same coefficient. Add to eliminate y. x+ y=5 (+) 3x – y = 7 4x = 12 x=3 Example: Continued x+y=5 3x – y = 7 Step 4: Plug back in to find the other variable. Step 5: Check your solution. x+y=5 (3) + y = 5 y=2 (3, 2) (3) + (2) = 5 3(3) - (2) = 7 The solution is (3, 2). What do you think the answer would be if you solved using substitution? Example: 4x + y = 7 4x – 2y = -2 Step 1: Put the equations in Standard Form. Step 2: Determine which variable to eliminate. Step 3: Multiply the equations and add the equations. They already are! The x’s have the same coefficient. Multiple the 2nd equation by (-1) and add to eliminate x. 4x + y = 7 - 4x + 2y = 2 3y = 9 y=3 Example: Continued 4x + y = 7 4x – 2y = -2 Step 4: Plug back in to find the other variable. Step 5: Check your solution. 4x + y = 7 4x + (3) = 7 4x = 4 x=1 (1, 3) 4(1) + (3) = 7 4(1) - 2(3) = -2 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 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 3x – 4y = 10 by elimination. x + 4y = –2 3x – 4y = 10 x + 4y = –2 4x + 0 = 8 4x = 8 4x = 8 4 4 x=2 Write the system so that like terms are aligned. Add the equations to eliminate the y-terms. Simplify and solve for x. Divide both sides by 4. Continued x + 4y = –2 2 + 4y = –2 –2 –2 4y = –4 4y –4 4 4 y = –1 (2, –1) Write one of the original equations. Substitute 2 for x. Subtract 2 from both sides. Divide both sides by 4. Write the solution as an ordered pair. Your Turn: Solve 3x + 3y = 15 by elimination. –2x + 3y = –5 3x + 3y = 15 –(–2x + 3y = –5) 3x + 3y = 15 + 2x – 3y = +5 5x + 0 = 20 5x = 20 x=4 Add the opposite of each term in the second equation. Eliminate the y term. Simplify and solve for x. Continued 3x + 3y = 15 3(4) + 3y = 15 12 + 3y = 15 –12 –12 3y = 3 y=1 (4, 1) Write one of the original equations. Substitute 4 for x. Subtract 12 from both sides. Simplify and solve for y. Write the solution as an ordered pair. 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. –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. 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 Substitution 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. Elimination Summary of Methods for Solving Systems Example Suggested Method Why 2x – 5y = –20 4x + 5y = 14 Elimination eliminate ‘y’ 5 Add the two equations Elimination 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 Essential Question Big Idea: Solving Equations and Inequalities • When can you use elimination to solve a system of linear equation? • Use elimination when the system contains a pair of additive inverses. Then add to eliminate a variable. You can also multiply one or both equations by nonzero numbers to make an equivalent system with additive inverses. Assignment • Section 3-2 Part 2, Pg. 159 – 161; #1 – 6 all, 8 – 48 even.