Solve. Write the solution set in interval notation. (Enter EMPTY or Ø for the empty set.) 3x + 7 < 10 or 6x - 14 > 4

Question

use set in interval notation

Answer 1

The question shown has an 'OR' between the two inequalities.  Therefore, i believe Kirill to be correct in showing the union of the two solutions:  (-∞;1)∪(3;∞).  'OR' means that x must solve one equation OR the other OR both to be part of the solution set.  So we place the union symbol between the two solution sets to include those values of x.

Answer 2

The goal is to isolate x in both equations by adding, subtracting, multiplying or dividing both sides of the inequality by an identical value.

 

3x + 7 < 10

Subtract 7 to both sides

3x < 3

Divide both sides by 3

x<1

 

 

6x - 14 > 4

Add 14 to both sides

6x>18

Divide both sides by 6

x> 3

 

So x must be both lower than 1 OR bigger than 3. Solution (-∞;1)∪ (3;+∞) (corrected from a previous version)

 

1 and 3 are not included as  it is a strict inequality.

Answer 3

3x+7<10 ⇔ 3x<3 ⇔ x<1

6x-14>4 ⇔ 6x>18 ⇔ x>3

 

If those two inequalities have to be satisfied simultaneously, then the solution is ∅. If any of those have to be satisfied, then the solution is (-∞;1)∪(3;∞)

Answer 4

• 3x + 7 < 10  equivalent to 3x < 10 - 7 that is 3x < 3 which is equivalent to x < 1 (by dividing both sides of the inequation by 3) The partial answer in interval notation is   x ∈ ( -∞ , 1 )
• 6x - 14 > 4  ====  6x > 4 + 14  ====  6x > 18  ==== x > 3  ====  x ∈ ( 3 , +∞ ) This is the other partial answer. The final answer contains x's that satisfy both partial answers that is no x !  x ∈ ∅

Answer 5

The question is asking for all real values of x for which the two inequalities are true.  

 

As an example, let's try a value of 4 for x.  Plugging that value in the first inequality we have 

 

3*4+7<10

 

Since 14 is greater than 10 (or not less than 10) our guess of 4 does not belong in the set of all values for which both inequalities are true

 

Let's try this analytically:

 

The first inequality is 3x+7<10 

 

Subtracting 7 from both sides we have a new inequality below

 

3x<3 or x<1

 

The second inequality is 6x-14>4

 

Transformations will yield another simplified inequality as shown below

 

6x>18 or x>3

 

The x values that hold for both inequalities will be all x values that are common to both x<1 and x>3

 

Since x cannot be both less than 1 and greater than 3, the solution set is empty.