2. A System of Equations is a pair of equations with two variables What is a System of Equations? A System of Equations is a pair of equations with two variables A system of equations has only one solution (or sometimes no solution) The solution is where the lines intersect Example 1: Example 2: y = 2x – 4 x + y = 3 y = 5x – 13 3x – 5y = 17 Slope-Int Form Standard Form Slope-Int Form Standard Form

You can find the solution (solve) to a system of equations by graphing 3. Finding Solution to Systems of Equations By Graphing Step 1 Graph both equations Step 2 The solution is the point (x, y) where the lines intersect

3. y = x – 3 (Equation 1) y = -x + 5 (Equation 2) y = 0 – 3 = -3 Finding Solution to Systems of Equations By Graphing Example 1 y = x – 3 (Equation 1) y = -x + 5 (Equation 2) Equation 1: Equation 2: X Y X Y -3 5 2 -1 2 3 y = 0 – 3 = -3 y = 0 + 5 = 5 y = 2 – 3 = -1 y = -2 + 5 = 3

3. X Y X Y -3 5 2 -1 2 3 Make sure to EXTEND YOUR LINES Equation 1: Equation 2: 3. Finding Solution to Systems of Equations By Graphing Example 1 X Y X Y -3 5 2 -1 2 3 Make sure to EXTEND YOUR LINES The solution is: ( 4 , 1 )

3. y = 3x + 5 (Equation 1) y = -2x + 10 (Equation 2) y = 3(0)+5 = 5 Finding Solution to Systems of Equations By Graphing Example 2 y = 3x + 5 (Equation 1) y = -2x + 10 (Equation 2) Equation 1: Equation 2: X Y X Y 5 10 -1 2 2 6 y = 3(0)+5 = 5 y = -2(0)+10 = 10 y = 3(-1)+5 = 2 y = -2(2)+10 = 6

3. X Y X Y 5 10 -1 2 2 6 Make sure to EXTEND YOUR LINES Equation 1: Equation 2: 3. Finding Solution to Systems of Equations By Graphing Example 2 X Y X Y 5 10 -1 2 2 6 Make sure to EXTEND YOUR LINES The solution is: ( 1 , 8 )

4. Finding Solution to System of Equations Set equations equal to each other and solve for x This is called substitution since we are substituting in one equation for y 4. Finding Solution Systems of Equations By Substitution Both Equations Slope-Intercept I DO y = 2x + 3 y = x + 1 2x + 3 = x + 1

4. 2x + 3 = x + 1 - x -x . x + 3 = 1 - 3 -3 x = -2 Now, solve for x: I DO 4. Finding Solution Systems of Equations By Substitution Both Equations Slope-Intercept Now, solve for x: 2x + 3 = x + 1 - x -x . x + 3 = 1 - 3 -3 x = -2 Check: 2(-2) + 3 = (-2) + 1 -4 + 3 = -2 + 1 -1 = -1

4. Remember, a solution is (x, y), so you need BOTH COORDINATES Check your solution by plugging both (x, y) into the OTHER equation I DO 4. Finding Solution Systems of Equations By Substitution Both Equations Slope-Intercept x = -2 Check: y = x + 1 y = 2(x) + 3 y = 2(-2) + 3 = -1 Solution: (-2, -1) -1 = -2 + 1 -1 = -1

4. 3x + 2 = -2x + 7 +2x +2x . 5x + 2 = + 7 - 2 -2 5x = + 5 ÷ 5 ÷ 5 WE DO Set them equal and solve for x 4. Finding Solution Systems of Equations By Substitution Both Equations Slope-Intercept 3x + 2 = -2x + 7 +2x +2x . Solve: y = 3x + 2 y = -2x + 7 5x + 2 = + 7 - 2 -2 5x = + 5 ÷ 5 ÷ 5 Plug-in x and solve for y: y = 3(x) + 2 y = 3(1) + 2 = 5 Solution: (1, 5) x = 1 Check: 3(1) + 2 = -2(1) + 7 3 + 2 = -2 + 7 5 = 5 Check: 5 = -2(1) + 7

4. X Y 2 1 5 X Y 7 1 5 The solution is: ( 1 , 5 ) Equation 1: y = 3x+2 Finding Solution Systems of Equations By Substitution Both Equations Slope-Intercept X Y 2 1 5 X Y 7 1 5 The solution is: ( 1 , 5 )

4. #1: y = 3x – 4 y = 2x + 2 #2: y = 5x + 8 y = -3x + 16 YOU DO 4. Finding Solution Systems of Equations By Substitution Both Equations Slope-Intercept #1: y = 3x – 4 y = 2x + 2 #2: y = 5x + 8 y = -3x + 16

5. We can use substitution to solve this type of system Finding Solution Systems of Equations By Substitution 1 Equation Slope Int. 1 Equation Standard Form We can use substitution to solve this type of system Substitute or plug in for y in the standard form equation y = 2x + 3 3x + 2y = 13 Slope-Int Form Standard Form

1. “Substitute” or Plug in the first equation into the second equation I DO y = 2x + 3 3x + 2y = 13 5. Finding Solution Systems of Equations By Substitution 1 Equation Slope Int. 1 Equation Standard Form Slope-Int Form Standard Form 1. “Substitute” or Plug in the first equation into the second equation 3x + 2(2x + 3) = 13 2. Solve for X 3x + 4x + 6 = 13 7x + 6 = 13 -6 = -6 7x = 7 ÷ 7 ÷ 7 y = 2(x) + 3 y = 2(1) + 3 = 5 Solution: (1, 5) x = 1 3. Plug in and solve for y

1. “Substitute” or Plug in the first equation into the second equation WE DO y = 2x + 3 7x – 1y = 7 5. Finding Solution Systems of Equations By Substitution 1 Equation Slope Int. 1 Equation Standard Form Slope-Int Form Standard Form 1. “Substitute” or Plug in the first equation into the second equation 7x – 1(2x + 3) = 7 2. Solve for X 7x – 2x – 3 = 7 5x – 3 = 7 +3 = +3 5x = 10 ÷ 5 ÷ 5 y = 2(x) + 3 y = 2(2) + 3 = 7 Solution: (2, 7) x = 2 3. Plug in and solve for y

5. #1: y = x – 6 x + y = -2 #2: 9x – 2y = 3 y = 3x – 6 YOU DO 5. Finding Solution Systems of Equations By Substitution 1 Equation Slope Int. 1 Equation Standard Form #1: y = x – 6 x + y = -2 #2: 9x – 2y = 3 y = 3x – 6

6. Some systems of equations have both equations in standard form Solving Systems of Equations by Elimination Both Standard Form Equations Some systems of equations have both equations in standard form We can use elimination to solve this type of system Stack the equations on top of each other Bring everything down by adding One variable will be eliminated Solve for the remaining variable Solutions have TWO COORDINATES! Plug-in to find final solution

The x variable is eliminated 6. Solving Systems of Equations by Elimination Both Standard Form Equations 2x + 3y = 12 -2x + 5y = 4 Standard Form Standard Form Stack up and bring down. The x variable is eliminated 0 + 8y = 16 2. Solve for Y ÷ 8 ÷ 8 y = 2 2x + 3y = 12 2x + 3(2) = 12 2x + 6 = 12 2x = 6 x = 3 Solution: (3, 2) 3. Plug in and solve for x

The y variable is eliminated 6. Solving Systems of Equations by Elimination Both Standard Form Equations 2x – 3y = 8 4x + 3y = 16 Standard Form Standard Form Stack up and bring down. The y variable is eliminated 6x + 0 = 24 2. Solve for X ÷ 6 ÷ 6 x = 4 2x – 3y = 8 2(4) – 3y = 8 8 – 3y = 8 -3y = 0 y = 0 Solution: (4, 0) 3. Plug in and solve for y

6. #1: -4x – 2y = -12 4x + 8y = -24 #2: x – y = 11 2x + y = 19 Solving Systems of Equations by Elimination Both Standard Form Equations #1: -4x – 2y = -12 4x + 8y = -24 #2: x – y = 11 2x + y = 19

7. 2x + 3y = 11 - x + y = 2 x + 4y = 13 WE HAVE A PROBLEM! Solving Systems of Equations by Elimination with Distribution Both Standard Form Equations 2x + 3y = 11 - x + y = 2 Standard Form Standard Form x + 4y = 13 WE HAVE A PROBLEM! No variable was eliminated, so we can’t find the solution! Remember, one equation with two variables has INFINITE solutions

7. 2x + 3y = 11 2(- x + y = 2 ) -2x + 2y = 4 2x + 3y = 11 -2x + 2y = 4 Solving Systems of Equations by Elimination with Distribution Both Standard Form Equations 2x + 3y = 11 2(- x + y = 2 ) Let’s try to eliminate x (we chose this) -2x + 2y = 4 What is the inverse of 2x? We need -2x. We need to multiply the second equation by 2 using DISTRIBUTION 2x + 3y = 11 -2x + 2y = 4 You MUST multiply ALL terms in the second equation

The x variable is eliminated 7. Solving Systems of Equations by Elimination with Distribution Both Standard Form Equations 2x + 3y = 11 -2x + 2y = 4 Stack up and bring down. The x variable is eliminated 0 + 5y = 15 2. Solve for Y ÷ 5 ÷ 5 y = 3 2x + 3y = 11 2x + 3(3) = 11 2x + 9 = 11 - 9 -9 2x = 2 x = 1 Solution: (1, 3) 3. Plug in and solve for x

7. 3x + 3y = 9 x + 2y = -3 4x + 5y = 6 WE HAVE A PROBLEM! Solving Systems of Equations by Elimination with Distribution Both Standard Form Equations 3x + 3y = 9 x + 2y = -3 Standard Form Standard Form 4x + 5y = 6 WE HAVE A PROBLEM! No variable was eliminated, so we can’t find the solution!

7. 4x + 4y = 12 2x + y = -3 6x + 5y = 9 WE HAVE A PROBLEM! Solving Systems of Equations by Elimination with Distribution Both Standard Form Equations 4x + 4y = 12 2x + y = -3 Standard Form Standard Form 6x + 5y = 9 WE HAVE A PROBLEM! No variable was eliminated, so we can’t find the solution!

7. 4x + 4y = 12 -4(2x + y = -3) 4x + 4y = 12 -8x – 4y = 12 Solving Systems of Equations by Elimination with Distribution Both Standard Form Equations 4x + 4y = 12 -4(2x + y = -3) Let’s try to eliminate y What is the inverse of 4y? We need -4y. We need to multiply the second equation by -4 using DISTRIBUTION 4x + 4y = 12 -8x – 4y = 12

The y variable is eliminated 7. Solving Systems of Equations by Elimination with Distribution Both Standard Form Equations 4x + 4y = 12 -8x – 4y = 12 Stack up and bring down. The y variable is eliminated -4x + 0 = 24 2. Solve for X ÷ -4 ÷ -4 x = -6 4x + 4y = 12 3(-6) + 4y = 12 -18 + 4y = 12 +18 +18 4y = 30 y = 7.5 Solution: (-6, 7.5) 3. Plug in and solve for Y

7. 3x + 3y = 9 -3( x + 2y = -3) 3x + 3y = 9 -3x – 6y = 9 Solving Systems of Equations by Elimination with Distribution Both Standard Form Equations 3x + 3y = 9 -3( x + 2y = -3) Let’s try to eliminate x What is the inverse of 3x? We need -3x. We need to multiply the second equation by -3 using DISTRIBUTION 3x + 3y = 9 -3x – 6y = 9

The x variable is eliminated 7. Solving Systems of Equations by Elimination with Distribution Both Standard Form Equations 3x + 3y = 9 -3x – 6y = 9 Stack up and bring down. The x variable is eliminated 0 – 3y = 18 2. Solve for Y ÷ -3 ÷ -3 y = -6 3x + 3y = 9 3x + 3(-6) = 9 3x – 18 = 9 +18 +18 3x = 27 x = 9 Solution: (9, -6) 3. Plug in and solve for x

4. #1: -4x + 9y = 9 x – 3y = -6 #2: 5x + y = 9 10x – 7y = -18 For You to Do #1: -4x + 9y = 9 x – 3y = -6 #2: 5x + y = 9 10x – 7y = -18