x
When a one step equation is solved, essentially the constants are sorted to
one side of the equation and the variable to the other. x 
3 x
20

8
8

5 x

3 x 3x

2x
Think of this as “moving” the constant 8 to the other side of the equation.
“Move” terms involving variables the same way.

2 Choose the term containing the variable with the smaller coefficient.
This doesn’t matter now, but will make things easier in other

2 equations.
2 2 Now there is only a two step equation to solve.
10

2x
10
2

2 x
2
5
 x
To solve equations with variables on both sides of the equal sign, first distribute. 2 ( 3 x

8 )

5

7 x
Then combine like terms.
Move all the variables to the same side.
6
13 x x


16
21


5
9
 x
7
 x
6


8

9 x
Solve the two step equation.

9 x
4 x

21

6
Subtract or add, and then divide. 
21

21
Every step isn’t needed in every equation. 4 x

27
8

( 2


2
 x x )

8 x

8 x x

27
4

6 3
4
Practice: a) 5x - 8 = 7x + 3 b) 3d - 1 = 4d + 9 c) 7x + 4 = 5x - 8 d) 6 - 5c = 2 + 9c e)
2 x
3

7
4 x

8

2 x

7 f) 0.23x-7x+4=4x-0.6+4x-8 g) 4

2 x

8
 
2

3 x

2s - 8 = 3s + 9
3s - 9 = 2s +10
4s - 10 = s + 11
-11 + s = 4s + 12

4 2
3
 x

1 7
8

7 x

9
12x-(5-3x)=7-9(2-3x)
3 2
3 x

7

2 5
8 x

3 1
2

3
4 x

8

3
4

8 x
50X - 80 = 70X + 30
4L - 12 = 8 +5L
2K + 3 = 6K - 4
4x - 10 = 10x + 4
3
 s
5
12

16


7
 s
4

15

3 x


2 x

4

3 ( 2 x

8 )

4 x

5 x

8 ( 2 x

1 )

h) 5 t

3

2

7 t


12

3 t


2
 t
 i) 15
 
8

3 x


4 x

8

3
 x
  j)
 x  5

 9 x x k) l) m)
3 x  ( x x
23 x  ( x  )  x 
3 x   x   4  x   2
123

8

2 x

5


15 ( 3 x

5 )

23
5 x

2


5
 x

3

4
 x
 
3 1
3
 7
8
 3 x  2 x  7
8 n) o) p) q) r) s) t)
5

3 x

2


34 x

7

24 x

8
4

2 x

17 x

4

3 x

5

10 x   x  10
 x  2 


 x  x x x  x  x x
 d

7
  d
 
4 d
     d
 
4 d
8 t
  t

7

5 t 9 21
 
2 t

7
 
10 t
    x

8
   
3 x

9

  x
 
13 x

4 x
  
5 x

8

 z 8
  z 9
  z



3 b
  b

9
 

3 b

10
5

9 8 x
 
  x 5

2


7 x

5
9

  n

4

5


3 n
  n
  n

8
 
3 x

8
  

2(4 x

5)

(2
 x
2(3 - (4 y + 8)) - (9 -2(3 y + 5)) = (2 y - 8) - 8 - (5y ) - 9 y + 10
 f
  f

8
    f

9
 

5
 f

5
 
 f


7
  f

8
 

12


15

The proportions on this page require the algebra methods learned to solve.
Example:
2
9 x
8(2
16
 7 x x
7 x x



5

 x


8
5) 7(
40 7
40 14 x
7
2 x x

 2)
14
Cross multiply.
Then distribute.
Subtract to get the variables to the same side of the equation.
Solve the two step equation.
 40  40
9 x  54 x  6
Practice: a)
2 a


3
9

3 a
5
 8 2 x
3
 5

3 x
5
 8 m b)

2
3 a a


9
2

8
5 m m
 10

4
5  2 3
 10
7

2 9
5  f f

10
15 m
5
 4 c)
9 8 s  9
 1 5 s  10
3
4
 t t 

3
8
24
8

10 k k
 8 d)
6
8

5 y
8
 10 3
16
 w
 w
8
 5 u
45
 8

5 u
3
 10

Remember to solve proportion problems, cross multiply, and then solve.
Examples: x

8
7

3
4
2
5 x

5 x

100
1
4
 x

7


24 100

2 x

5

5 x

1

4 x

28

24 200 x

25 x

5

28

28
4 x

52
175 x

5 x

1
35 x
4 x

4
13

52
4
Practice: a) x

4
5
 b) x
6
2
3

8 c)
2 x
9
 4 1
3 d)
3 x
5


5 x

6
2

3 x
5
5
8

2 x
3 x

5

8

( x x


7
3 3 8
)
3
2
( x x

 )

3
3 x
 x
3

 x
5
2

2

4
3
  x

5

9 x

6

4 x

20

4 x

4 x
5 x

6

6

20

6
5 x

26 x

5 1
5
2 x
8

7

5
3 x
2 x

6

7
3 x

5
8 x 
2
6

3  x  5 
5
6 
3 x
 x


2