y 2 10 x y 3 x 10
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D AY 82 E LIMINATION U SING THE ADDITION PROPERTY OF EQUALITY 1. Let x represent the number of correct answers. Let y represent the incorrect answers. Write a system of equations. Janets scoring Matthews scoring system system 2

  1. D AY 82 – E LIMINATION

  2. U SING THE ADDITION PROPERTY OF EQUALITY 1. Let x represent the number of correct answers. Let y represent the incorrect answers. Write a system of equations. Janet’s scoring Matthew’s scoring system system 2 spaces for 3 spaces for a correct answer a correct answer 1 space forward for 1 space backward for + an incorrect answer + an incorrect answer 10 spaces forward 10 spaces forward ? + ? =10 ? + ? =10

  3. U SING THE ADDITION PROPERTY OF EQUALITY 1. Let x represent the number of correct answers. Let y represent the incorrect answers. Write a system of equations. Janet’s scoring Matthew’s scoring system system 2 spaces for 3 spaces for a correct answer a correct answer 1 space forward for 1 space backward for + an incorrect answer + an incorrect answer 10 spaces forward 10 spaces forward 2x + y = 10 3x + (-y) = 10

  4. 2. Rewrite the equation representing Matthew's scoring system using a minus sign. Model this system of equations using the x-tiles and y-tiles.  y  2 10 x

  5.  y  3 x 10

  6. 3. Combine the two models. Remove neutral pairs, and write the new equation.      ( 2 x y ) ( 3 x y ) 10 10

  7. 3. Combine the two models. Remove neutral pairs, and write the new equation. 5x = 20      ( 2 x y ) ( 3 x y ) 10 10

  8. 4. Which variable was eliminated? Why? Solve the resulting equation for the remaining variable.

  9. 4. Which variable was eliminated? Why? Solve the resulting equation for the remaining variable.  y = variable; neutral pairs; x 4

  10. 5. Use substitution in either of the original equations to find the value of the variable that was eliminated by removing neutral pairs. The solutions to the system is (?,?).

  11. 5. Use substitution in either of the original equations to find the value of the variable that was eliminated by removing neutral pairs. The solutions to the system is (?,?). ( 4 , 2 )

  12. 6. Henry answered _?_ questions correctly and _?_ questions incorrectly.

  13. 6. Henry answered _?_ questions correctly and _?_ questions incorrectly. 4 correctly , 2 incorrectly

  14. U SING THE SUBTRACTION P ROPERTY OF E QUALITY Use the x-titles and y-titles to model the system 1.    4 x y 10     x y 4  y  4 x 10

  15.  y  4 x

  16. 2. Combine the two models, remove any neutral pairs. Is a variable eliminated? Why or why not?

  17. 2. Combine the two models, remove any neutral pairs. Is a variable eliminated? Why or why not?  y  5 x 2 14 no variable are eliminated because neither combines with an equal number of opposites.

  18. 3. Use tiles to model the system of equations again.

  19. 4. You can eliminate the y-variable in this system by using the Subtraction Property of Equality. Recall that to subtract, you add the opposite. First model the opposite of the equation x + y = 4.

  20.  y  x 4

  21.     1 ( x y ) 1 ( 4 )     x y 4

  22. 5. Now subtract by adding the opposite. Combine the model of the opposite of x + y = 4 with the model of 4x y = 10, and remove neutral pairs. What is the resulting equation? What variable was eliminated?        ( 4 ) ( ) 10 ( 4 ) x y x y

  23. 6.Use the model to solve the resulting equation for x. 7. Substitute the solution for x into one of the original equations, and solve for y.

  24. 6.Use the model to solve the resulting equation for x.  x 2 7. Substitute the solution for x into one of the original equations, and solve  for y. y 2

  25. 8.Check your solution in both of the original equations. 9.Describe how to solve a system of equation using the Subtraction Property of Equality.

  26. 8.Check your solution in both of the original equations.     2 2 4 ( true ) and 8 2 10 ( true ) 9.Describe how to solve a system of equation using the Subtraction Property of Equality.

  27. S OLVE THE SYSTEM USING THE ELIMINATION METHOD .     2 x 5 y 4     4 x 5 y 22

  28. The coefficients of the y-terms are opposites. By adding the equations, you can eliminate the y-term. This leaves an equation with only one variable, which you can solve    for x. 2 5 4 x y   4 x 5 y 22 Add the equations, Solve for x  6 x 18  3 x

  29. To find y, substitute 3 for x in either of the original equations.    2 ( 3 ) 5 y 4 Substitute for x in Equation 1    6 5 y 4 Simplify.    5 y 10 Add 6 to both sides.   2 Solve for y y

  30. Check the solution, (3,-2), in both original equations. If the coefficients of the variable you want to eliminate are identical, subtract the equations. If the equations are not in standard form Ax + By = C, begin by writing them in standard form.

  31. S OLVE BY THE ADDITION METHOD :    3 10 1 x y     2 5 2 x y Since the coefficients of the y terms are additive inverses, adding the corresponding sided of the equations will eliminate the y terms.

  32.   3 10 1 x y  2 x y 5  2 5 x 15    x 3 3 Substituti on x 3 in Equations 1 or in Equation 2

  33.  y 1 3   3x y 10 1   3(3) y 10  y 1 4   x 3  Thus, Is an equivaltent system   y 1

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