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M1 L5 Remediation Notes
M1 L5 Remediation Notes
M1 L5 Remediation Notes
M1 L5 Remediation Notes
M1 L5 Remediation Notes
M1 L5 Remediation Notes
M1 L5 Remediation Notes
M1 L5 Remediation Notes
M1 L5 Remediation Notes
M1 L5 Remediation Notes
M1 L5 Remediation Notes
M1 L5 Remediation Notes
M1 L5 Remediation Notes
M1 L5 Remediation Notes
M1 L5 Remediation Notes
M1 L5 Remediation Notes
M1 L5 Remediation Notes
M1 L5 Remediation Notes
M1 L5 Remediation Notes
M1 L5 Remediation Notes
M1 L5 Remediation Notes
M1 L5 Remediation Notes
M1 L5 Remediation Notes
M1 L5 Remediation Notes
M1 L5 Remediation Notes
M1 L5 Remediation Notes
M1 L5 Remediation Notes
M1 L5 Remediation Notes
M1 L5 Remediation Notes
M1 L5 Remediation Notes
M1 L5 Remediation Notes
M1 L5 Remediation Notes
M1 L5 Remediation Notes
M1 L5 Remediation Notes
M1 L5 Remediation Notes
M1 L5 Remediation Notes
M1 L5 Remediation Notes
M1 L5 Remediation Notes
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M1 L5 Remediation Notes

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Math 3 CC

Math 3 CC

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  • 1. SOLVING SYSTEMS OF EQUATIONS AND INEQUALITIES Module 1 Lesson 5
  • 2. What is a system of equations?    A system of equations is when you have two or more equations using the same variables. The solution to the system is the point that satisfies ALL of the equations. This point will be an ordered pair. When graphing, you will encounter three possibilities.  Consistent Systems (one solution)  Inconsistent Systems (no solutions)  Dependent Systems (Infinite number of solutions)
  • 3. Consistent Systems    The lines will intersect. The point where the lines intersect is your solution. The solution of this graph is (1, 2) (1,2)
  • 4. Inconsistent Systems    These lines never intersect as the lines are parallel! Since the lines never cross, there is NO SOLUTION! Parallel lines have the same slope with different y-intercepts. 2 =2 1 y-intercept = 2 y-intercept = -1 Slope =
  • 5. Dependent Systems    These lines are the same! Since the lines are on top of each other, there are INFINITELY MANY SOLUTIONS! Coinciding lines have the same slope and y-intercepts. 2 =2 1 y-intercept = -1 Slope =
  • 6. What is the solution of the system graphed below? 1. 2. 3. 4. (2, -2) (-2, 2) No solution Infinitely many solutions
  • 7. 1) Find the solution to the following system by graphing: 2x + y = 4 x-y=2 Graph both equations. I will graph using x- and y-intercepts (plug in zeros), but you can also rewrite in to y =mx + b form. 2x + y = 4 (0, 4) and (2, 0) x–y=2 (0, -2) and (2, 0) Graph the ordered pairs.
  • 8. Graph the equations. 4 Where do the lines intersect? (2, 0) y= x-y=2 (0, -2) and (2, 0) + 2x 2x + y = 4 (0, 4) and (2, 0) x– y= 2
  • 9. Check your answer! To check your answer, plug the point back into both equations. 2x + y = 4 2(2) + (0) = 4 x-y=2 (2) – (0) = 2 Nice job…let’s try another!
  • 10. 2) Find the solution to the following system: y = 2x – 3 -2x + y = 1 Graph both equations. Put both equations in slope-intercept or standard form. I’ll do slopeintercept form on this one! y = 2x – 3 y = 2x + 1 Graph using slope and y-intercept
  • 11. Graph the equations. y = 2x – 3 m = 2 and b = -3 y = 2x + 1 m = 2 and b = 1 Where do the lines intersect? No solution! Notice that the slopes are the same with different y-intercepts. If you recognize this early, you don’t have to graph them!
  • 12. Check your answer! Not a lot to check…Just make sure you set up your equations correctly. I double-checked it and I did it right…
  • 13. What is the solution of this system? 3x – y = 8 2y = 6x -16 1. 2. 3. 4. (3, 1) (4, 4) No solution Infinitely many solutions
  • 14. Solving a system of equations by Graphing Let's summarize! There are 3 steps to solving a system using a graph. Step 1: Graph both equations. Graph using slope and y – intercept or x- and y-intercepts. Be sure to use a ruler and graph paper! Step 2: Do the graphs intersect? This is the solution! LABEL the solution! Step 3: Check your solution. Substitute the x and y values into both equations to verify the point is a solution to both equations.
  • 15. Solving Systems of Equations using Subs titutio n Steps: 1.Solve one equation for one variable (y = ; x = ; a =) 2.Substitute the expression from step one into the other equation. 3.Simplify and solve the equation. 4.Substitute back into either original equation to find the value of the other variable. 5. Check the solution in both equations of the system.
  • 16. Example #1: y = 4x 3x + y = -21 Step 1: Solve one equation for one variable. y = 4x (This equation is already solved for y.) Step 2: Substitute the expression from step one into the other equation. 3x + y = -21 3x + 4x = -21 Step 3: Simplify and solve the equation. 7x = -21 x = -3
  • 17. Example #1 cont: y = 4x 3x + y = -21 Step 4: Substitute back into either original equation to find the value of the other variable. 3x + y = -21 3(-3) + y = -21 -9 + y = -21 y = -12 Solution to the system is (-3, -12).
  • 18. y = 4x 3x + y = -21 Step 5: Check the solution in both equations. Solution to the system is (-3,-12). y = 4x 3x + y = -21 -12 = 4(-3) 3(-3) + (-12) = -21 -12 = -12 -9 + (-12) = -21 -21= -21
  • 19. Example #2: x + y = 10 5x – y = 2 Step 1: Solve one equation for one variable. x + y = 10 y = -x +10 Step 2: Substitute the expression from step one into the other equation. 5x - y = 2 5x -(-x +10) = 2
  • 20. Example #2 cont: x + y = 10 5x – y = 2 Step 3: Simplify and solve the equation. 5x -(-x + 10) = 2 5x + x -10 = 2 6x -10 = 2 6x = 12 x=2
  • 21. Example #2 cont: x + y = 10 5x – y = 2 Step 4: Substitute back into either original equation to find the value of the other variable. x + y = 10 2 + y = 10 y =8 Solution to the system is (2,8).
  • 22. Example #2 cont: x + y = 10 5x – y = 2 Step 5: Check the solution in both equations. Solution to the system is (2, 8). x + y =10 2 + 8 =10 10 =10 5x – y = 2 5(2) - (8) = 2 10 – 8 = 2 2=2
  • 23. Solving Systems of Equations using Elimination (a ls o c a lle d So lving by us ing A d itio n) d Steps: 1.Place both equations in Standard Form, Ax + By = C. 2.Determine which variable to eliminate with Addition or Subtraction. 3.Solve for the variable left. 4.Go back and use the found variable in step 3 to find second variable. 5. Check the solution in both equations of the system.
  • 24. EXAMPLE #1: STEP 1: form. STEP 2: STEP 3: 5x + 3y = 11 5x = 2y + 1 Write both equations in Ax + By = C 5x + 3y =1 5x - 2y =1 Multiply the 2nd equation by -1. 5x + 3y =11 -5x + 2y =1 Add like terms and solve. 5x + 3y =11 -5x + 2y = -1 5y =10 y =2
  • 25. 5x + 3y = 11 5x = 2y + 1 STEP 4: Solve for the other variable by substituting into either equation. 5x + 3y =11 5x + 3(2) =11 5x + 6 =11 5x = 5 x =1 The solution to the system is (1,2).
  • 26. 5x + 3y= 11 5x = 2y + 1 Step 5: Check the solution in both equations. The solution to the system is (1,2). 5x + 3y = 11 5(1) + 3(2) =11 5 + 6 =11 11=11 5x = 2y + 1 5(1) = 2(2) + 1 5=4+1 5=5
  • 27. Solving Systems of Equations using Elimination Steps: 1. Place both equations in Standard Form, Ax + By = C. 2. Determine which variable to eliminate with Addition or Subtraction. 3. Solve for the remaining variable. 4. Go back and use the variable found in step 3 to find the second variable. 5. Check the solution in both equations of the system.
  • 28. Example #2: x + y = 10 5x – y = 2 Step 1: The equations are already in standard form: x + y = 10 5x – y = 2 Step 2: Adding the equations will eliminate y. x + y = 10 x + y = 10 +(5x – y = 2) +5x – y = +2 Step 3: Solve for the variable. x + y = 10 +5x – y = +2 6x = 12 x =2
  • 29. x + y = 10 5x – y = 2 Step 4: Solve for the other variable by substituting into either equation. x + y = 10 2 + y = 10 y =8 Solution to the system is (2,8).
  • 30. x + y = 10 5x – y = 2 Step 5: Check the solution in both equations. Solution to the system is (2,8). x + y =10 2 + 8 =10 10=10 5x – y =2 5(2) - (8) =2 10 – 8 =2 2=2
  • 31. Using Elimination to Solve a Word Problem: Two angles are supplementary. The measure of one angle is 10 degrees more than three times the other. Find the measure of each angle.
  • 32. Using Elimination to Solve a Word Problem: Two angles are supplementary. The measure of one angle is 10 more than three times the other. Find the measure of each angle. x = degree measure of angle #1 y = degree measure of angle #2 Therefore x + y = 180
  • 33. Using Elimination to Solve a Word Problem: Two angles are supplementary. The measure of one angle is 10 more than three times the other. Find the measure of each angle. x + y = 180 x =10 + 3y
  • 34. Using Elimination to Solve a Word Problem, cont: x + y = 180 x =10 + 3y x + y = 180 x - 3y = 10 I will multiply the second equation by -1 then add like terms. x + y = 180 -x + 3y = -10 4y =170 y = 42.5
  • 35. Using Elimination to Solve a Word Problem: Substitute the 42.5 to find the other angle. x + 42.5 = 180 x = 180 - 42.5 x = 137.5 (137.5, 42.5)
  • 36. Using Elimination to Solve a Word problem: The sum of two numbers is 70 and their difference is 24. Find the two numbers. x = first number y = second number Therefore, x + y = 70
  • 37. Using Elimination to Solve a Word Problem: The sum of two numbers is 70 and their difference is 24. Find the two numbers. x + y = 70 x – y = 24
  • 38. Using Elimination to Solve a Word Problem: x + y =70 x - y = 24 2x = 94 x = 47 47 + y = 70 y = 70 – 47 y = 23 (47, 23)

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