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Objective. You will be able to: Solve systems of equations using elimination with addition or (subtraction) and multiplication. UNIT 6 Concept 3. Solving Systems of Equations.

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    1. Objective You will be able to: Solve systems of equations using elimination with addition or (subtraction) and multiplication. UNIT 6 Concept 3

    2. Solving Systems of Equations • So far, we have solved systems using graphing and substitution. These notes show how to solve the system algebraically using ELIMINATION with addition and subtraction. • Elimination is easiest when the equations are in standard form.

    3. The Linear Combination Method aka The Addition Method, aka The Elimination Method. • Add (or subtract) a multiple of one equation to (or from) the other equation, in such a way that either the x-terms or the y-terms cancel out. • Then solve for x (or y, whichever's left) and substitute back to get the other coordinate. • Solve this system of equations using the addition or subtraction method. • When solving systems of linear equations using elimination, you will sometimes need to multiply one or both equations by a factor in order to get the same coefficients for a variable. • Best time to use this method is when you have two coefficients that the same

    4. Steps in Elimination • Line up the two equations using standard form (Ax + By = C). 2. GOAL: The coefficients of the same variable in both equations should have the same value but opposite signs. 3. If this doesn’t exist, multiply one or both of the equations by a number that will make the same variable coefficients opposite values. • Add the two equations (like terms). • The variable with opposite coefficients should be eliminated. • Solve for the remaining variable. • Substitute that solution into either of the two equations to solve for the other variable.

    5. Color Coded Steps in solving a system of equations by elimination. Standard Form: Ax + By = C Step 1: Put the equations in Standard Form. Step 2: Determine which variable to eliminate. Look for variables that have the same coefficient. Step 3: Add or subtract 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.

    6. 1) Solve the system using elimination. Add into your spirals x + y = 5 3x – y = 7 Step 1: Put the equations in Standard Form. They already are! Step 2: Determine which variable to eliminate. The y’s have the same coefficient. Add to eliminate y. x + y = 5 (+) 3x – y = 7 4x = 12 x = 3 Step 3: Add or subtract the equations.

    7. 1) Solve the system using elimination. x + y = 5 3x – y = 7 x + y = 5 (3) + y = 5 y = 2 Step 4: Plug back in to find the other variable. (3, 2) (3) + (2) = 5 3(3) - (2) = 7 Step 5: Check your solution. The solution is (3, 2). What do you think the answer would be if you solved using substitution?

    8. 2) Solve the system using elimination. Add into your spirals 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. Subtract to eliminate x. 4x + y = 7 (-) 4x – 2y = -2 3y = 9 y = 3 Step 3: Add or subtract the equations. Remember to “keep-change-change”

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

    10. 3) Solve the system using elimination. Add into your spirals y = 7 – 2x 4x + y = 5 Step 1: Put the equations in Standard Form. 2x + y = 7 4x + y = 5 Step 2: Determine which variable to eliminate. The y’s have the same coefficient. Subtract to eliminate y. 2x + y = 7 (-) 4x + y = 5 -2x = 2 x = -1 Step 3: Add or subtract the equations.

    11. 2) Solve the system using elimination. y = 7 – 2x 4x + y = 5 y = 7 – 2x y = 7 – 2(-1) y = 9 Step 4: Plug back in to find the other variable. (-1, 9) (9) = 7 – 2(-1) 4(-1) + (9) = 5 Step 5: Check your solution.

    12. Example 1 (In note packet pg 10) x - 2y = 14 x + 3y = 9 Both equations are already in Standard Form. Step 1: Put the equations in Standard Form. We will eliminate the variable “x” Since they both have same coefficient. Step 2: Determine which variable to eliminate. Subtraction is easiest, just remember this Is like multiplying the entire equation by -1. x - 2y = 14-( x + 3y = 9) -5y = 5 y = -1 Step 3: Add or subtract the equations.

    13. Example 1 continued… x - 2y = 14 x + 3y = 9 Since y = -1, we will substitute in x - 2y = 14 x- 2 (-1) = 14 Therefore x+2=14 my solution x=12 is ( 12, -1) Step 4: Plug back in to find the other variable. Substitute ( 12, -1) into: x - 2y = 14 12 – 2(-1) = 14 12+2 = 14 14 = 14 Step 5: Check your solution.

    14. Example 2 (In note packet pg 11) 4x + 3y = -15x + 4y = 1 Both equations are already in Standard Form. Step 1: Put the equations in Standard Form. So for this one we need to multiply each equation to create same Coefficients. So we will multiply EQ1 by 4 and EQ2 by -3, then add to eliminate y. Step 2: Determine which variable to eliminate. 4(4x + 3y = -1)-3(5x + 4y = 1) 16x + 12y = -4)-15x- 12y = -3) x = -7 Step 3: Add or subtract the equations.

    15. Example 2 continued… 4x + 3y = -15x + 4y = 1 4x + 3y = -1 4(-7) + 3y = -1 Therefore -28 +3y = -1 my solution is 3y = 27 y = 9 (-7, 9) Step 4: Plug back in to find the other variable. 4x + 3y = -1 5x + 4y = 14(-7) +3(9) = -1 (-7) + 4(9) = 1  -28 + 27 = -1 -35 + 36 = 1  -1 = -1  (check!) 1 = 1  Step 5: Check your solution. The third example we will do together in class.

    16. EXTRA EXAMPLES Example 1: Equation ‘a’: 2x - 4y = 13 Equation ‘b’: 4x - 5y = 8 Multiply equation ‘a’ by –2 to eliminate the x’s: Equation ‘a’: -2(2x - 4y = 13) Equation ‘b’: 4x - 5y = 8

    17. EXTRA EXAMPLES Extra Example 1, continued: Equation ‘a’: -2(2x - 4y = 13) ------> -4x + 8y = -26 Equation ‘b’: 4x - 5y = 8 ------> 4x - 5y = 8 Add the equations (the x’s are eliminated): -4x + 8y = -26 4x - 5y = 8 3y = -18 y = -6

    18. EXTRA EXAMPLES Example 1, continued: Equation ‘a’: -2(2x - 4y = 13) ------> -4x + 8y = -26 Equation ‘b’: 4x - 5y = 8 ------> 4x - 5y = 8 Substitute y = -6 into either equation: 4x - 5(-6) = 8 4x + 30 = 8 4x = -22 x = x = Solution: ( , -6)

    19. EXTRA EXAMPLES Extra Example 2: Equation ‘a’: -9x + 6y = 0 Equation ‘b’: -12x + 8y = 0 Multiply equation ‘a’ by –4 and equation ‘b’ by 3 to eliminate the x’s: Equation ‘a’: - 4(-9x + 6y = 0) Equation ‘b’: 3(-12x + 8y = 0)

    20. EXTRA EXAMPLES Example 2 continued… Equation ‘a’: - 4(-9x + 6y = 0) Equation ‘b’: 3(-12x + 8y = 0) 36x - 24y = 0 -36x + 24y = 0 0 = 0 What does this answer mean? Is it true?

    21. EXTRA EXAMPLES 36x - 24y= 0 -36x + 24y = 0 0 = 0 Example 2, continued: When both variables are eliminated, • if the statement is TRUE (like 0 = 0), then they are the same lines and there are infinite solutions. • if the statement is FALSE (like 0 = 1), then they are parallel lines and there is no solution. 36x - 24y = 0 -36x + 24y = 0 0 = 0 Since 0 = 0 is TRUE, there are infinite solutions.

    22. Summary • When solving systems of linear equations using elimination, you will sometimes need to multiply one or both equations by a factor in order to get the same coefficients for a variable. This process is very similar to getting a common denominator for fractions. So to summarize, this is the most complex method of solving systems of of equations. Although you just need to be sure to follow the steps to ensure success. Remember:

    23. ASSIGNMENT • Do your summary from the guided questions online. • Do practice problems in note packet on page 12 • Do practice quiz on page 13. • Section 6.3 pages 347-34 • Do Problems 1-10; 33 & 34, and 43-45 9