If the inequality is ‘>’, the inequality is a line (excluding the particular point) extending towards?

Respuesta :

Answer:

We can use the number line to solve inequalities containing <, ≤, >, and ≥. To solve an inequality using the number line, change the inequality sign to an equal sign, and solve the equation. Then graph the point on the number line (graph it as an open circle if the original inequality was "<" or ">"). The number line should now be divided into 2 regions -- one to the left of the point and one to the right of the point

Next, pick a point in each region and "test" it -- see if it satisfies the inequality when plugged in for the variable. If it satisfies the inequality, draw a dark line from the point into that region, with an arrow at the end. This is the solution set to the equation: if one point in the region satisfies the inequality, the entire region will satisfy the inequality.

Example: -3(x - 2)≤12

Solve -3(x - 2) = 12:

x - 2 = - 4

x = - 2

Graph x = - 2, using a filled circle because the original inequality was ≤:

Graph of x = - 2

Plug values into the equation -3(x - 2)≤12:

Pick a point on the left of

-2

(

-3

, for example):

-3(- 3 - 2)≤12

?

15≤12

? No.

Pick a point on the right of

-2

(

0

, for example):

-3(0 - 2)≤12

?

6≤12

? Yes..

We can use the number line to solve inequalities containing <, ≤, >, and ≥. To solve an inequality using the number line, change the inequality sign to an equal sign, and solve the equation. Then graph the point on the number line (graph it as an open circle if the original inequality was "<" or ">"). The number line should now be divided into 2 regions -- one to the left of the point and one to the right of the point

Next, pick a point in each region and "test" it -- see if it satisfies the inequality when plugged in for the variable. If it satisfies the inequality, draw a dark line from the point into that region, with an arrow at the end. This is the solution set to the equation: if one point in the region satisfies the inequality, the entire region will satisfy the inequality.

Example: -3(x - 2)≤12

Solve -3(x - 2) = 12:

x - 2 = - 4

x = - 2

Graph x = - 2, using a filled circle because the original inequality was ≤:

Graph of x = - 2

Plug values into the equation -3(x - 2)≤12:

Pick a point on the left of

-2

(

-3

, for example):

-3(- 3 - 2)≤12

?

15≤12

? No.

Pick a point on the right of

-2

(

0

, for example):

-3(0 - 2)≤12

?

6≤12

? Yes.

Thus, x≥ - 2.

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