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Lesson 10.4
Objective: To solve equations with variables on
both sides of the equal sign
Some equations have variables on both sides of the
equal sign. To solve such equation, You may collect
the variable terms on one side of the equal sign.
Example:
12 + 4x = 10x
– 4x – 4x
12 = 6x
6
6
2=x
Subtract 4x from both sides.
Divide both sides by 6.
Solution
Check: 12 + 4(2) = 10(2)
20 = 20
Example:
50 + 9x = 4x
– 9x – 9x
Subtract 9x from both sides.
50 = – 5x
–5
–5
Divide both sides by – 5.
– 10 = x
Solution
Check:
50 + 9(-10) = 4(-10)
50 + (-90) = -40
– 40 = – 40
Example:
1
10 x  20  3  2 x
2
5x – 10 = 3 + 2x
– 2x
– 2x
3x – 10 = 3
+10 +10
3x
3
= 13
3
1
13
x=
or 4
3
3
1
Distribute the
2
Subtract 2x from both sides.
Add 10 to both sides
Divide both sides by 3
Write an equation to find the value of x so that each pair of
polygons have the same perimeter. then find each perimeter.
x+6
3x
3x
7
3x
7
3x
x+6
3x
Pentagon
=
Rectangle
3x+3x+ 3x+ 3x+ 3x = 7 + x + 6 + x + 6 + 7
15x = 2x + 26
- 2x -2x
Perimeter
13x = 26
15(2) =2(2) + 26
X=2
30 = 30
Verbal Models/Problem Solving
Example:
Hank’s video store charges $8.00 to rent a video game
for five days and does not charge an annual membership
fee. Bunker’s video only charges $3.00 for a five day
rental but has a $50.00 membership fee per year.
Find the number of rentals that would cost the same from each store.
Hanks
Fee
8
Number
Rented
•
x
=
Bunkers
Fee
=
3
Number
Rented
•
x
Membership
+ Fee
+
50
Subtract 3x from both sides
8x = 3x + 50
5x = 50
Divide both sides by 5
X = 10
If you rent 10 games per year, the
cost would be the same.