Combined Inequalities?

Question

2x > -6 and x - 4 < 3

x + 3 > 2x + 1 and -4x < -8

-6 < x + 3 < 6

-3x > -6 or x - 5 < 2

x - 2 > 2x + 1 or 10 > -2x + 2

Could someone please help me solve these? They are the last questions on my test review.

Answer 1

Just solve for x, like you would with an equation. The one tricky thing is you have to flip the inequality if you multiply through by a negative number.

2x > -6 and x - 4 < 3
2x/2 > -6/2 and x - 4 + 4 < 3 + 4
x > -3 and x < 7

Therefore -3 < x < 7 ... or x is between -3 and 7.

-3x > -6 or x - 5 < 2
-3x/-3 < -6/-3 or x - 5 + 5 < 2 + 5 (NOTE THE FLIPPED INEQUALITY)
x < 2 or x < 7

In this case, since it's an OR (union of solution sets), the answer would be x < 7, as that is a superset of the values that satisfy x < 2. If it was an AND (intersection of solution sets), it would be x < 2, since everything less than 2 is also less than 7.

Do the same sort of thing for the rest. Good luck on the test.

Answer 2

In each inequality, solve for x by moving terms to the other side, or dividing by numbers (remember to reverse the direction of the inequality if you multiply or divide by a negative number!).

You'll get two intervals in which x can lie, one for each inequality. If the two inequalities are connected by an "and" and the intervals do not overlap, there is no solution. If they do overlap, then the solution can be expressed using a combined inequality of the form a < x < b.

First one:
x > -3, and x < 7
Therefore, -3 < x < 7.

Second one:
2 > x and x > 2 (note < becomes > after dividing by -4). You get x < 2 and x < 2, which has no solution.

Third one:
Add -3 to all sides and get -9 < x < 3.

Fourth one:
x < 2 (> becomes < after dividing by -3) or x < 7. The solution is x < 7, because if x < 2, then certainly x < 7.

Fifth one:
-3 > x or 8 > -2x, so x < -3 or x > -4. Every real number satisfies one of these inequalities, so any value of x is a solution. -infinity < x < infinity is one way to describe it. Note: if it had been "and", then the answer would be -4 < x < -3.

Answer 3

What you're doing when you are defining multiple inequalities is you are establishing a domain where "x" is valid. For instance, the first one:

2x > -6 and x - 4 < 3

in the left inequality solve for x, and you'll find x >-3
do the same with the right inequality: x < 7
combine the two to define on what domain x is acceptable:

-3
It works the same for the ones that ask "or". But for the middle one, the domain has already been compiled for you, all you have to do is perform the same operation on both inequalities:

-6 < x+3 < 6 subtract 3 from both sides and the middle

-6 - 3 < x < 6 - 3
-9
ta da!

Answer 4

2x>-6 so x>-3 and x-4<3 so x<7
-3
x+3>2x+1 so -x>-3 so x<3 and -4x<-8 so x>2
2
-6 -9 x>-9
x+3<6
x<3
3>x>-9

-3x>-6
x<2
x-5<2
x<7

x-2>2x+1
-x>3
x<-3
10>-2x+2
-2x+2<10
-2x<8
2x>-8
x>-4
-4