3x+1<7 or 7<2x-9?

Answer 1

Solve each inequality separately.
3x + 1< 7
Subtract 1 from each side
3x < 6
Divide both sides by 3.
x < 2

Now solve the other inequality.
7 < 2x - 9
Add 9 to each side.
16 < 2x
Divide both sides by 2.
8 < x
x > 8

Answer: x < 2 or x > 8
Interval notation: (-infinity, 2) U (8, infinity)

Answer 2

For all x over the set of real numbers:

3x + 1 < 7 or 2x - 9 > 7

3x < 6 or 2x > 16

x < 2 or x > 8

Therefore:

{x: x < 2 or x > 8}

Answer 3

3x+1<7 or 7<2x-9?

(a)
3x+1<7
3x <7 - 1
3x < 6
x < 6/3
x < 2

(b)
7 < 2x - 9
7 + 9 < 2x
16 < 2x
16/2 < x
8 < x

If x is the same variable in both questions, then they may be combined. Read it from the centre.
2 > x > 8

Answer 4

3x+1<7
3x<6
x<2

2x-9>7
2x>16
x>8

2 >x>8

Answer 5

3x + 1 < 7
3x < 6
x < 2

7 < 2x - 9
16 < 2x
x > 8

Answer 6

If x has the same value in both inequalities, then both can't be true at the same time. The first inequality is only true for x less than 1. The second inequality is only true for x greater than 9. Since x cannot be both less than 1 and greater than 9, both inequalities cannot be true at the same time.
I bet you're being asked to draw the inequality on a number line. If so, draw a circle at 1 and a circle at 9. Then draw arrows pointing outward from both circles.

Answer 7

only distribute the unfavorable sign by using the term (3x+a million), making it -3x -a million. From there you may forget approximately with regard to the parentheses and upload the like words like so: (2x-9)+(6x-7)-(3x+a million)= (2x-9)+(6x-7)+(-3x-a million)= 2x - 9 + 6x - 7 - 3x -a million= 2x + 6x -3x - 9 - 7 - a million= 5x - 17

Answer 8

if x is same in both then from first x <2 & the second x>8 so we don't have such an x

Answer 9

A. x<2
B. x>8

So x is either less than 2 or more than 8

Answer 10

if x=2 or is less than 2, the first statement is true. otherwise the 2nd one is true