1. 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 – x + y = 3
y = 5x – x – 5y = 17
Slope-Int Form
Standard Form
Slope-Int Form
Standard Form

2. You can find the solution (solve) to a system of equations by graphing3.
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. 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 = = 5
y = 2 – 3 = -1
y = = 3

4. 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 )

5. 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

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 )

7. 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

8. 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 =
x = -2
Check:
2(-2) + 3 = (-2) + 1 =
-1 = -1

9. 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 =
-1 = -1

10. 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 x
Solve:
y = 3x + 2
y = -2x + 7
5x + 2 =
5x =
÷ ÷ 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 =
5 = 5
Check: 5 = -2(1) + 7

11. 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 )

12. 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

13. 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

14. 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
y = 2(x) + 3
y = 2(1) + 3 = 5
Solution: (1, 5)
x = 1
3. Plug in and solve for y

15. 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
y = 2(x) + 3
y = 2(2) + 3 = 7
Solution: (2, 7)
x = 2
3. Plug in and solve for y

16. 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

17. 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

18. The x variable is eliminated6.
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
y = 16
2. Solve for Y
÷ ÷ 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

19. The y variable is eliminated6.
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 = 24
2. Solve for X
÷ ÷ 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

20. 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

21. 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

22. 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

23. The x variable is eliminated7.
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
y = 15
2. Solve for Y
÷ ÷ 5
y = 3
2x + 3y = 11
2x + 3(3) = 11
2x = 11
2x = 2
x = 1
Solution: (1, 3)
3. Plug in and solve for x

24. 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!

25. 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!

26. 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

27. The y variable is eliminated7.
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 = 24
2. Solve for X
÷ ÷ -4
x = -6
4x + 4y = 12
3(-6) + 4y = 12
y = 12
4y = 30
y = 7.5
Solution: (-6, 7.5)
3. Plug in and solve for Y

28. 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

29. The x variable is eliminated7.
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 – = 9
3x = 27
x = 9
Solution: (9, -6)
3. Plug in and solve for x

30. 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