Solve for x.
1. 6(2x – 3) – 2(6x + 1) = 10x
2. 15 + 2x – 4 = 9x + 11 – 7x
Solve for h.
3. V =
1
𝜋𝑟 2 ℎ
3
1.
6(2x – 3) – 2(6x + 1) = 10x
12x – 18 – 12x – 2 = 10x
- 20 = 10x
10 10
x = –2
2.
15 + 2x – 4 = 9x + 11 – 7x
11 + 2x = 11 + 2x
-11
-11
2x = 2x
-2x - 2x
0=0
ALWAYS TRUE!!!
Infinite Solution!!
3.
1
V = 𝜋𝑟 2 ℎ Solve
3
1
2
3(V) = ( 𝜋𝑟 ℎ)(3)
3
2
3V = 𝜋𝑟 ℎ
𝜋𝑟 2
𝜋𝑟 2
3𝑉
𝜋𝑟 2
=ℎ
ℎ=
3𝑉
𝜋𝑟 2
for h.
Algebra II
 To
 To
solve & graph inequalities
write & solve compound
inequalities
Open circle on the number line
 Does NOT include the #
>
<
Greater than
Less than
Closed circle on the number line
 INCLUDES the #
≥
Greater than or equal to
≤
Less than or equal to
x
>7
x is greater than 7
x
≥7
x is greater than or
equal to 7
x
<7
x is less than 7
x
≤7
x is less than or equal
to 7
Less Than <
 Is
fewer than
 Is less than
Less Than or
Equal to ≤
 Is
at most
 Is no more
than
 Is at maximum
Greater Than >
 Is
more than
 Is larger than
Greater Than or
Equal to ≥
 Is
at least
 Is no less than
 Is at minimum
5
fewer than a number is at least
12
x – 5 ≥ 12
 The
quotient of a number and 3 is
no more than 15
𝑥
3
≤ 15
ALWAYS FLIP THE
INEQUALITY SIGN
WHEN YOU
DIVIDE BY – #
1. -3(2x – 5) + 1 ≥ 4
-6x + 15 + 1 ≥ 4
-6x + 16 ≥ 4
- 16 - 16
-6x ≥ - 12
-6 - 6 FLIP!!!
x≤2
2. -2(x + 9) + 5 < 3
-2x – 18 + 5 < 3
-2x – 13 < 3
+13 +13
-2x < 16
-2 - 2 FLIP!!!
x >-8
3. -2(3x + 1) > -6x + 7
-6x – 2 > - 6x + 7
+2
+2
-6x > -6x + 9
+6x +6x
0>9
NEVER TRUE!
No Solution
4. 5(2x – 3) – 7x < 3x + 8
10x – 15 – 7x < 3x + 8
3x – 15 < 3x + 8
-3x
-3x
- 15 < 8
ALWAYS TRUE! Infinite Solutions
5. 4(2x – 3) > 8(x + 1)
8x – 12 > 8x + 8
-8x
-8x
- 12 > 8
NEVER TRUE!
No Solution



Compound Inequality – join two
inequalities with the word and or or
“And” compound inequality – Find all
values of the variable that make
BOTH inequalities true. Need overlap
“Or” compound inequality – Find all
values of the variables that make at
least one of the inequalities true
6. 7 < 2x + 1 and 3x ≤ 18
-1
-1 & 3
3
6 < 2x
& x≤6
2
2
3<x
x > 3 and x ≤ 6
3<x≤6
7. 5 ≤ 3x - 1 and 2x < 12
+1
+1 & 2
2
6 ≤ 3x
& x<6
3
3
2≤x
x ≥ 2 and x < 6
2≤x<6
8. 5< x + 1 < 13
-1
-1 - 1
4 < x < 12
Read: x + 1 is greater than 5
AND less than 13
9. 7 + x ≥ 6
-7
-7
x ≥ -1
or 8 + x < 3
-8
-8
or x < - 5
0
0
NOT -5> x ≥ -1 NOR -1 ≤ x <- 5
10. 7x + 3 > 11 or 4x – 1 < -13
-3 -3
+1 +1
7x > 8
or 4x < -12
7
7
4
4
x>
8
7
or x < -3
0
0
11. 16 < 5x + 1 or 3x + 9 < 6
-1
-1
-9 -9
15 < 5x or
3x < -3
5
5
3
3
3<x
or
x < -1
x>3
or
x < -1
0
0
 Your
friend solved the problem as
shown.
 What was his error?