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3) 4y ≥ -16
2x ≤ 4
x-2y ≥ 6

4) 2x+3y ≥ -7
x-3y ≤ 28
x+6y ≤ -26

Solve the system of linear inequalities

see below

Step-by-step explanation:

3) 4y ≥ -16

2x ≤ 4

x-2y ≥ 6

First inequality

Divide by 4

4y/4 ≥ -16/4   y   ≥ -4

second inequality

Divide by 2

2x ≤ 4     2x/2  ≤ 4 /2   x  ≤ 2

Third inequality

Subtract x from each side

x-2y ≥ 6    x-2y -x ≥ -x +6   -2y  ≥ -x +6

Then divide by -2 remembering to flip the inequality

-2y/-2 ≤ (-x+6)/-2     y  ≤1/2x -3

Graph is below

4) 2x+3y ≥ -7

Putting each in line form

3y ≥ -7  -2x

y≥ -2x/3 -7/3

Second

x-3y ≤ 28

-3y≥ -x+28

Divide and flip since it is a negative

y  ≤ 1/3x -28/3

Third

x+6y ≤ -26

6y ≤ -x-26

y ≤ -1/6x -26/6

3. The solution is the region below the line y = x/2 - 1, above y = -4 and towards left of x = 2. (All bold)

4. The solution is the region above the lines y = -2x/3 -7/3 and y = x/3 - 28/3 and below the line y = -x/6 - 13/3. (All bold)

Step-by-step explanation:

3) 4y ≥ -16

y ≥ -4

2x ≤ 4

x ≤ 2

x-2y ≥ 6

2y ≤ x - 2

y ≤ x/2 - 1

The solution is the region below the line y = x/2 - 1, above y = -4 and towards left of x = 2. (All bold)

4) 2x+3y ≥ -7

3y ≥ -2x -7

y ≥ -2x/3 -7/3

x-3y ≤ 28

3y ≥ x - 28

y ≥ x/3 - 28/3

x+6y ≤ -26

6y ≤ -x - 26

y ≤ -x/6 - 13/3

The solution is the region above the lines y = -2x/3 -7/3 and y = x/3 - 28/3 and below the line y = -x/6 - 13/3. (All bold)

answer: (c) m∠qpo + (2x + 16)° = 180°

step-by-step explanation:

the opposite angles of a quadrilateral are supplementary.

so, ∠o + ∠q = 180°   and   ∠p + ∠r = 180°

since ∠r = 2x + 16°, we can use substitution as follows:

∠p + ∠r = 180°

∠p + (2x + 16)° = 180°

answer: i believe the answer is e

step-by-step explanation: