Download Solving Equations With Variables on Both Sides

Solving Equations
With Variables on Both
Sides
Objective: Solve equations with the variable on each
side and solve equations involving grouping symbols.
Variables on Each Side
•
To solve an equation that has variables on each side, use the
Addition or Subtraction Property of Equality to write an
equivalent equation with the variable terms on one side.
•
Collect the variable terms on the side of the equation with
the larger variable coefficient (number in front of variable).
•
Collect the constant terms (numbers) on the other side of the
equation.
•
Solve as a multi-step equation in one variable.
Example 1
•
Solve 8 + 5c = 7c – 2. Check your solution.
8 + 5c = 7c – 2
-5c -5c
8 = 2c – 2
+2
+2
10 = 2c
2
2
5=c
c=5
8 + 5c = 7c – 2
8 + 5(5) = 7(5) – 2
8 + 25 = 35 – 2
33 = 33 
Grouping Symbols
•
If equations contain grouping symbols such as parenthesis or
brackets, use the Distributive Property first to remove the
grouping symbols.
Example 2
•
Solve the following equations. Check your solution.
a.
1/ (18)
3
1/
3
(18 + 12q) = 6(2q – 7)
+ 1/3(12q) = 6(2q) – 6(7)
6 + 4q = 12q – 42
-4q -4q
6 = 8q – 42
+42
+42
48 = 8q
8 8
6=q
q=6
1/ (18
3
+ 12q) = 6(2q – 7)
1/ (18 + 12 • 6) = 6(2 • 6 – 7)
3
1/ (18 + 72) = 6(12 – 7)
3
1/ (90) = 6(5)
3
30 = 30 
Example 2
•
Solve the following equations. Check your solution.
b. 5x
.  2  3x
12
4
12 5x  2 3x 12 3

14 1
1 12
5x + 2 = 9x
-5x
-5x
2 = 4x
4
4
½=x
x=½
5x  2 3x

12
4
5( 1 2 )  2 3( 1 2 )

12
4
2.5  2 1.5

12
4
4.5 1.5

12
4
0.375 = 0.375 
Special Solutions
•
Some equations may have no solution.
•
That is, there is no value of the variable that will result in a
true equation.
•
Some equations are true for all values of the variables.
•
These are called identities.
Example 3
•
Solve each equation.
a.
8(5c – 2) = 10(32 + 4c)
8(5c) – 8(2) = 10(32) + 10(4c)
40c – 16 = 320 + 40c
-40c
-40c
-16 = 320 x
No Solution
Example 3
•
Solve each equation.
b. 4(t + 20) = 1/5 (20t + 400)
4(t) + 4(20) = 1/5 (20t) + 1/5 (400)
4t + 80 = 4t + 80
-4t
-4t
80 = 80 
Identity
(True for all values of t)
Steps for Solving Equations
1. Simplify the expressions on each side. Use the Distributive
Property as needed.
2. Use the Addition and/or Subtraction Properties of Equality
to get the variables on one side and the numbers without
variables on the other side. Simplify.
3. Use the Multiplication or Division Property of Equality to
solve.
Example 4
•
Find the value of h so that the figures have the same area.
A.
B.
C.
D.
1
3
4
5
h
(h – 2)
6
10
½ (6)h = ½ (10)(h – 2)
3h = 5(h – 2)
3h = 5(h) – 5(2)
3h = 5h – 10
-3h -3h
0 = 2h – 10
+10
+10
10 = 2h
2
2
Check Your Progress
•
Choose the best answer for the following.
•
Solve 9f – 6 = 3f + 7.
A. 1/3
B. 1/12
C. 2 1/6
D. 2
9f – 6 = 3f + 7
-3f
-3f
6f – 6 = 7
+6 +6
6f = 13
6
6
Check Your Progress
•
Choose the best answer for the following.
•
Solve 6(3r – 4) = 3/8 (46r + 8).
A.
B.
C.
D.
38
28
10
36
6(3r) – 6(4) = 3/8 (46r) + 3/8 (8)
18r – 24 = 69/4 r + 3
-18r
-18r
-24 = - ¾ r + 3
-3
-3
- 4/3 • -27 = - ¾ r • - 4/3
Check Your Progress
•
Choose the best answer for the following.
•
Solve 2(4a + 8) = 3(8a/3 – 10).
A. 1/3
B. 2
C. True for all values of a.
D. No solution.
2(4a) + 2(8) = 3(8a/3) – 3(10)
8a + 16 = 8a – 30
-8a
-8a
16 = -30
Check Your Progress
•
Choose the best answer for the following.
•
Solve 1/7 (21c – 56) = 3(c – 8/3).
A. 1/3
1/ (21c) – 1/ (56) = 3(c) – 3(8/ )
7
7
3
B. 0
3c – 8 = 3c – 8
C. True for all values of c.
-3c
-3c
D. No solution.
-8 = -8
Check Your Progress
•
Choose the best answer for the following.
•
Find the value of x so that the figures have the same area.
A.
B.
C.
D.
1
2
3
4
x
6
6x = ½ (4)(x + 4)
6x = 2(x + 4)
6x = 2(x) + 2(4)
6x = 2x + 8
-2x -2x
4x = 8
4 4
4
x+4