Solve 15 ≥ - 3x or 2/3 x ≥ - 2.

Solve each of the equations independently, then determine if the are continuous or discontinuous.

15≥-3x [start here]

-5≤x [divide both sides by (-3). * Dividing by a negative number means the direction of the sign changes!]

x≥-5 [just turned around for analysis]

Next equation:

2/3x≥-2 [start here]

x≥-2 (3/2) [multiply both sides of the equation by the reciprocal, 3/2)

x≥-3

So, (according to the first equation) all values of x must be greater than, or equal to - 5.

(According to the second equation) all values of x must be greater than, or equal to - 3.

So, when graphed on a number line, both equations graph in the same direction, so they are continuous.

The 'or' means that it has to be one OR the other, the numbers where both are true are not solutions

solve each independenlty, remember that dividing or multiplying both sides by a negaive flips the inequality sign direction

15≥-3x

divide both sides by - 3 and don't forget to flip sign

-5≤x

x≥-5

ok, solve 2nd one

2/3 x≥-2

multiply both sides by 3/2

x≥-3

so we got

x≥-5 and x≥-3

they intersect in x≥-3, so x≥-3 is not a solution reigon

the solution set is x≥-5 but stops at - 3, so it would be - 5≤x<-3