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LESSON 10-6
Warm Up
Lesson 10-6 Warm-Up
ALGEBRA READINESS
LESSON 10-6
Warm Up
Lesson 10-6 Warm-Up
ALGEBRA READINESS
“Solving Equations With Variables
on Both Sides” (10-6)
Tip: To solve an equation with variables on both sides, use the Addition
How can you solve an
(2-2)
or Subtraction Properties of Equality to get the variables on one side of
equation with variables
on both sides?
the equal sign.
Example: 5m + 3 = 3m – 9
-3m
-3m
2m + 3 = + 0 - 9
-3
-3
-12 1
2 +0=
1 2m
1 2
2 1
1m
=
-6
ALGEBRA READINESS
Solving Equations with Variables on Both Sides
LESSON 10-6
Additional Examples
Solve 9 + 2p = – 3 – 4p.
9 + 2p = – 3 – 4p
+ 4p
+ 4p
9 + 6p = – 3 + 0
–9
–9
0 + 6p = – 12
1 6p
1 6
= – 12
2
6 1
1p = – 2
Addition Property of Equality
Combine like terms.
Subtraction Property of Equality
Simplify.
Division Property of Equality
Simplify.
Check 9 + 2p = – 3 – 4p
9 + 2(–2)
– 3 – 4(–2)
5=5
Substitute –2 for p.
The solution checks.
ALGEBRA READINESS
Solving Equations With Variables on Both Sides
LESSON 10-6
Additional Examples
Solve 5x – 3 = 2x + 12.
5x – 3 = 2x + 12
– 2x
– 2x
Subtract 2x from each side to get rid
of the x term from the right side.
3x – 3 = 0 + 12
+3
+3
Combine like terms.
Add 3 to each side to get rid of -3.
3x + 0 =
Simplify.
5
1
3x
3
15
=
1
15
3
Divide each side by 3 to isolate the x.
1
1x = 5
Simplify.
ALGEBRA READINESS
Solving Equations with Variables on Both Sides
LESSON 10-6
Additional Examples
Each week you set aside $18 for a stereo and put the remainder
of your weekly pay in a savings account. After 7 weeks, the amount you
place in the savings account is 4.2 times your total weekly pay. How
much do you earn each week?
Words: 7 times (weekly amount – 18) = 4.2 • weekly amount
Let x = the amount earned weekly.
Equation: 7 •
(x
–
18)
=
4.2
•
x
ALGEBRA READINESS
Solving Equations with Variables on Both Sides
LESSON 10-6
Additional Examples
(continued)
7(x – 18) = 4.2x
7x – 7(18) = 4.2x
Distributive Property
7x – 126 = 4.2x
Simplify.
– 7x
– 7.0x
0 –126 = –2.8x
1
–126
–2.8x
=
–2.8
–2.8
45 =
Subtraction Property of Equality
Combine like terms.
Division Property of Equality
1
1x
Simplify.
You earn $45 per week.
ALGEBRA READINESS
“Solving Equations With Variables
on Both Sides” (10-6)
If the variable on both sides of an equation cancel each other out when you use the
What happens when
(2-2)
Addition or Subtraction
Properties of Equality (Example: 2x = 2x), it will have an
the variables on both
sides of the equation
cancel each other out?
infinite number of solutions called an “identity” if both sides of the equation are
equal (Example: 10 = 10) or no solution if both sides of the equation are not equal
(Example: 10 = 7)
Example: 3x + 9 = 3 + 3(x + 2)
3x + 9 = 3 + 3x + 3(2) Distributive Property
3x + 9 = 3x + 9
Simplify
-3x
-3x
9=9
Identity (infiniite number of solutions)
Proof: The following function table shows that any solution for x works.
x
3x + 9 =
3 + 3(x + 2)
True
Statement?
105
3(105) +
9
= 324
=
3 + 3(105 + 2) = 3 + 3(107) =
3 + 321 =324
Yes
324 = 324
1,42
0
3(1,420)
+9=
4,269
=
3 + 3(1,420 + 2) = 3 +
3(1,422) = 3 + 4,266 = 4,269
Yes
4,269 =
4,269
3 + 3(23,246 + 2) = 3 +
3(23,248) = 3 + 69,744 =
69,747
Yes
69,747=69,7
47
23,2
46
3(23,246) =
+9=
69,747
ALGEBRA READINESS
“Solving Equations With Variables
on Both Sides” (10-6)
Example: 3x + 10 = 3 + 3(x + 2)
(2-2)
3x + 10 = 3 + 3x + 6 Distributive Property
3x + 10 = 3x + 9
-3x
-3x
10 ≠ 9
Simplify
no solutions
Proof: The following function table shows that no solution work for x.
x
3x + 10 =
3 + 3(x + 2)
= 3 + 3(12 + 2) = 3 + 3(14) = 3 +
42 =45
True
Statement?
12
3(12) +
10
= 46
No
45 ≠ 45
105
3(105) +
10
= 325
=
3 + 3(105 + 2) = 3 + 3(107) =
3 + 321 =324
Yes
325 ≠324
1,42
0
3(1,420)
+ 10 =
4,270
=
3 + 3(1,420 + 2) = 3 +
3(1,422) = 3 + 4,266 = 4,269
Yes
4,270 ≠
4,269
ALGEBRA READINESS
Solving Equations With Variables on Both Sides
LESSON 10-6
Additional Examples
Solve each equation.
a.
4 – 4y = –2(2y – 2)
4 – 4y = –2(2y) – (-2)(2)
4 – 4y =
-4y
+
4
+ 4y = +4y
Use the Distributive Property.
Use the Distributive Property.
Add 4y to each side.
4+0 = 0
Always true!
+ 4
The equation is true for every value of y, so the equation is an identity.
b.
–6z + 8 = z + 10 – 7z
–6z + 8 = 1z + 10 – 7z
Identity Property (1x = x)
–6z + 8 = –6z + 10
+ 6z
+ 6z
0 + 8 = 0 + 10
Combine like terms.
Add 6z to each side.
Not true for any value of z!
This equation has no solution.
ALGEBRA READINESS
Solving Equations with Variables on Both Sides
LESSON 10-6
Lesson Quiz
Solve each equation.
1. 4(3u – 1) = 20
2
2. 5t – 4 = t – 8
–1
3. 3(k – 8) = –k
6
ALGEBRA READINESS