Question

Help needed

Solve for p.

2p+5>2(p−3)

no solution

all real numbers

p>−9

p>−3

What is the solution to the compound inequality in interval notation?

4(x+1)>−4  or  2x−4≤−10

(−∞, −3] or (2, ∞)

(−3, −2]

(−∞, −2) or [3, ∞)

(−∞, −3] or (−2, ∞)

Answer 1

2p+5 > 2(p−3)
2p+5 > 2p -6
2p+5-5 > 2p -6-5
2p > 2p-11
2p-2p > 2p-11-2p
0 > -11
not possible, therefore NO SOLUTION

4(x+1) > −4  or  2x−4 ≤ −10
solve each:
4x+4 > -4
4x+4-4 > -4-4
4x > -8
x > -2
that solution is: (-2, ∞)
2x-4 ≤ −10
2x-4+4 ≤ −10+4
2x ≤ −6
x ≤ −3
that solution is: (-∞, -3)
So the combined solution is the last one:
(−∞, −3] or (−2, ∞)

Answer 2

First problem:

2p + 5 > 2(p − 3)

Distribute 2.

2p + 5 > 2p - 6

Subtract 2p from both sides.

5 > -6

Since 5 > -6 is a true statement, the inequality is true for all real numbers.

Answer: all real numbers.

Second problem:


4(x + 1) > −4  or  2x − 4 ≤ −10

Distribute 4.

4x + 4 > -4 or 2x - 4 ≤ −10

Subtract 4 from both sides in left inequality.
Add 4 to both sides in right inequality.

4x > -8 or 2x ≤ −6

Divide both sides of left inequality by 4.
Divide both sides of right inequality by 2.


x > -2 or x ≤ −3

x is greater than -2, so we start the interval at 2, but not including 2, so we use a left parenthesis symbol. The interval goes to positive infinity.
x is less than or equal to -3, so -3 is included in the interval, so we use a right square bracket to include the -3. We start this interval at negative infinity.

(-∞, -3] or (-2, ∞)

Notice that with infinity and negative infinity we always use a regular parenthesis, not a square bracket.