1. In a System of Equations you need to solve for both x and y values
In a System of Equations you need to solve for both x and y values. In the easiest of the types of problems, the equations will both be in slope intercept form. In these problems you will have two separate expressions that are each equal to y. You can then set the expressions equal to each other, and solve for x. Then plug the value of x in to solve for y. you need to make sure you are using the properties to solve equations by keeping them balanced.

2. Solve the system of equations using substitution.
y = 3x – 8
y = 6x + 4

3. 3x - 8 = 6x + 4
y = 3x – 8
y = 6x + 4

4. 3x - 8 = 6x + 4
y = 3x – 8
y = 6x + 4
The first thing I can do is to remove the x’s from one side of the equation. Remember, whatever you do to one side, you must do to the other.

5. 3x - 8 = 6x + 4
y = 3x – 8
y = 6x + 4
The first thing I can do is to remove the x’s from one side of the equation. Remember, whatever you do to one side, you must do to the other.

6. 3x - 8 = 6x + 4
y = 3x – 8
y = 6x + 4
Now I get the variable term isolated on one side of the equal sign. Since the tiles are different colors, I cannot just remove them. I must make neutral pairs to make them disappear. Remember, whatever you do to one side, you must do to the other.

7. 3x - 8 = 6x + 4
y = 3x – 8
y = 6x + 4
Now I get the variable term isolated on one side of the equal sign. Since the tiles are different colors, I cannot just remove them. I must make neutral pairs to make them disappear. Remember, whatever you do to one side, you must do to the other.

8. 3x - 8 = 6x + 4
y = 3x – 8
y = 6x + 4
Now I find out the value of a single x.
3x’s are worth x will be worth -4.

9. 3x - 8 = 6x + 4
y = 3x – 8
y = 6x + 4
Now I find out the value of a single x.
3x’s are worth x will be worth -4.

10. By setting the two equations equal to each other, you were able to find the value of x. Now you must plug that value in to find the value of y. You can plug the value of x into either or your original equations in order to solve for y.
y = 6x + 4
y = 6(-4) + 4
y =
y = -20
The solution to
the system is (-4, -20)

11. Solve the system of equations using substitution.
y = 5x – 11
y = -2x + 3

12. 5x - 11 = -2x + 3
y = 5x – 11
y = -2x + 3
The first thing I can do is to eliminate the x’s from one side of the equation. I’ll do it by making neutral pairs for the x’s.

13. 5x - 11 = -2x + 3
y = 5x – 11
y = -2x + 3
The first thing I can do is to eliminate the x’s from one side of the equation. I’ll do it by making neutral pairs for the x’s.

14. 5x - 11 = -2x + 3
y = 5x – 11
y = -2x + 3

15. 5x - 11 = -2x + 3
y = 5x – 11
y = -2x + 3
Now I get the variable term isolated on one side of the equal sign. Since the tiles are different colors, I cannot just remove them. I must make neutral pairs to make them disappear.

16. 5x - 11 = -2x + 3
y = 5x – 11
y = -2x + 3

17. 5x - 11 = -2x + 3
y = 5x – 11
y = -2x + 3
Now I find out the value of a single x.
7x’s are worth x will be worth 2.

18. 5x - 11 = -2x + 3
y = 5x – 11
y = -2x + 3
Now I find out the value of a single x.
7x’s are worth x will be worth 2.

19. By setting the two equations equal to each other, you were able to find the value of x. Now you must plug that value in to find the value of y. You can plug the value of x into either or your original equations in order to solve for y.
y = 5x – 11
y = 5(2) – 11
y = 10 – 11
y = -1
The solution to the
system is (2, -1)

20. Solve the system of equations using substitution.
y = 9x – 12
y = 8x - 3

21. 9x - 12 = 8x - 3 1x - 12 = -3 1x = 9 y = 9x – 12 y = 8x - 3 -8x -8x
+12
+12
1x = 9
y = 9x – 12
y = 9(9) – 12
y =
y = 69
The solution to the
system is (9,69)
It is too hard to try and read the coordinates for the point of intersection. You cannot always rely on a graph to get the solution

22. or 9x - 12 = 8x - 3 9x = 8x + 9 1x = 9 +12 +12 -8x -8x y = 8x – 3
The solution to the
system is (9,69)
It is too hard to try and read the coordinates for the point of intersection. You cannot always rely on a graph to get the solution

23. Solve the system of equations using substitution.
y = c -12
y = 3c + 12

24. c - 12 = 3c + 12 -12 = 2c + 12 -24 = 2c -12 = c -c - c -12 - 12 ÷ 2
y = c - 12
y = -12 – 12
y = -24
The solution to the
system is (-12,-24)
y = 3c + 12
y = 3(-12)+ 12
y =
y = -24