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.

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.

Step 1: Put the equations in Standard Form. Step 2: Determine which variable to eliminate. Step 3: Multiply the equations and add the equations. Step 4: Plug back in to find the other variable. Step 5: Check your solution. Standard Form: Ax + By = C Look for variables that have the same coefficient. Solve for the variable. Substitute the value of the variable into the equation. Substitute your ordered pair into BOTH equations. Elimination Procedure

x + y = 5 3x – y = 7 Step 1: Put the equations in Standard Form. Step 2: Determine which variable to eliminate. They already are! The y’s have the same coefficient. Step 3: Add the equations. Add to eliminate y. x + y = 5 (+) 3x – y = 7 4x = 12 x = 3 Example:

Step 4: Plug back in to find the other variable. x + y = 5 (3) + y = 5 y = 2 Step 5: Check your solution. (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? x + y = 5 3x – y = 7 Example: Continued

4x + y = 7 4x – 2y = -2 Step 1: Put the equations in Standard Form. They already are! Step 2: Determine which variable to eliminate. The x’s have the same coefficient. Step 3: Multiply the equations and add the equations. Multiple the 2 nd equation by (-1) and add to eliminate x. 4x + y = 7 - 4x + 2y = 2 3y = 9 y = 3 Example:

Step 4: Plug back in to find the other variable. 4x + y = 7 4x + (3) = 7 4x = 4 x = 1 Step 5: Check your solution. (1, 3) 4(1) + (3) = 7 4(1) - 2(3) = -2 4x + y = 7 4x – 2y = -2 Example: Continued

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: Multiplying One Equation

Step 4: Plug back in to find the other variable. 2(2) + 2y = y = 6 2y = 2 y = 1 2x + 2y = 6 3x – y = 5 Step 3: Multiply the equations and solve. Multiply the bottom equation by 2 2x + 2y = 6 (2)(3x – y = 5) 8x = 16 x = 2 2x + 2y = 6 (+) 6x – 2y = 10 Example: Continued

Step 5: Check your solution. (2, 1) 2(2) + 2(1) = 6 3(2) - (1) = 5 2x + 2y = 6 3x – y = 5 Solving with multiplication adds one more step to the elimination process. Example: Continued

3x + 4y = -1 4x – 3y = 7 Step 1: Put the equations in Standard Form. They already are! Step 2: Determine which variable to eliminate. 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: Multiplying Both Equations

3x + 4y = -1 4x – 3y = 7 Step 4: Plug back in to find the other variable. 3(1) + 4y = y = -1 4y = -4 y = -1 Step 3: Multiply the equations and solve. Multiply both equations (3)(3x + 4y = -1) (4)(4x – 3y = 7) x = 1 9x + 12y = -3 (+) 16x – 12y = 28 25x = 25 Example: Continued

Step 5: Check your solution. (1, -1) 3(1) + 4(-1) = -1 4(1) - 3(-1) = 7 3x + 4y = -1 4x – 3y = 7 Example: Continued

3x – 4y = 10 x + 4y = –2 Solve by elimination. 3x – 4y = 10 Write the system so that like terms are aligned. Add the equations to eliminate the y-terms. 4x = 8 Simplify and solve for x. x + 4y = –2 4x + 0 = 8 Divide both sides by 4. 4x = 8 4 x = 2 Your Turn:

x + 4y = –2 Write one of the original equations y = –2 Substitute 2 for x. –2 4y = –4 4y –4 4 y = –1 (2, –1) Subtract 2 from both sides. Divide both sides by 4. Write the solution as an ordered pair. Continued

3x + 3y = 15 –2x + 3y = –5 Solve by elimination. 3x + 3y = 15 –(–2x + 3y = –5) 3x + 3y = x – 3y = +5 Add the opposite of each term in the second equation. Eliminate the y term. Simplify and solve for x. 5x + 0 = 20 5x = 20 x = 4 Your Turn:

Write one of the original equations. Substitute 4 for x. Subtract 12 from both sides. 3x + 3y = 15 3(4) + 3y = y = 15 –12 3y = 3 y = 1 Simplify and solve for y. Write the solution as an ordered pair. (4, 1) Continued

x + 2y = 11 –3x + y = –5 Solve the system by elimination. Multiply each term in the second equation by –2 to get opposite y-coefficients. x + 2y = 11 –2(–3x + y = –5) x + 2y = 11 +(6x –2y = +10) Add the new equation to the first equation. 7x + 0 = 21 7x = 21 x = 3 Simplify and solve for x. Your Turn:

Write one of the original equations. x + 2y = 11 Substitute 3 for x y = 11 Subtract 3 from each side. –3 2y = 8 y = 4 Simplify and solve for y. Write the solution as an ordered pair. (3, 4) Continued

–5x + 2y = 32 2x + 3y = 10 Solve the system by elimination. 2(–5x + 2y = 32) 5(2x + 3y = 10) Multiply the first equation by 2 and the second equation by 5 to get opposite x-coefficients –10x + 4y = 64 +(10x + 15y = 50) Add the new equations. Simplify and solve for y. 19y = 114 y = 6 Your Turn:

Write one of the original equations. 2x + 3y = 10 Substitute 6 for y. 2x + 3(6) = 10 Subtract 18 from both sides. –18 2x = –8 2x + 18 = 10 x = –4 Simplify and solve for x. Write the solution as an ordered pair. (–4, 6) Continued

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. Simplify and solve for y. y = y = 28 Your Turn:

Write one of the original equations. 2x + 5y = 26 Substitute 4 for y. 2x + 5(4) = 26 Simplify and solve for x. Write the solution as an ordered pair. (3, 4) x = 3 2x + 20 = 26 –20 2x = 6 Subtract 20 from both sides. Your Turn:

Show that this linear system has infinitely many solutions. – 2 x  y  3 Equation 1 – 4 x  2y  6 Equation 2 M ETHOD : Elimination You can multiply Equation 1 by –2. 4x – 2y  – 64x – 2y  – 6 Multiply Equation 1 by –2. – 4 x  2y  6 Write Equation 2. 0  0 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 Infinite Solutions

Show that this linear system has no solution. 2 x  y  5 Equation 1 2 x  y  1 Equation 2 You can multiply Equation 1 by –1. -2 x - y  -5 Multiply Equation 1 by –1. 2 x  y  1 Write Equation 2. 0  - 4 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. A Linear System with No Solution M ETHOD : Elimination

Your Turn: Solve the systems using elimination. 1) 2) False Statement No Solution True Statement Infinite Solutions

Summary

Summary of Methods for Solving Systems Substitution The value of one variable is known and can easily be substituted into the other equation. 6x + y = 10 y = 5 Example Suggested Method Why Substitution

Summary of Methods for Solving Systems Elimination eliminate ‘y’  5 Add the two equations 2x – 5y = –20 4x + 5y = 14 Example Suggested Method Why Elimination

Summary of Methods for Solving Systems Elimination 9a – 2b = –11 8a + 4b = 25 Example Suggested Method Why eliminate ‘b’  4 Multiply first equation by 2 Add the equations Elimination

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.