Question

Please help me with both questions!!!!​​

Answer 1

Answer:

Step-by-step explanation:

16.

points are (0,-3),(2,1),(4,-3)

eq. of line through (0,-3) and (2,1) is

y+3=\frac{1+3}{2-0}(x-0)

or y+3=2x

2x-y=3

as the line is dotted so either < or >.

consider 2x-y>3

put x=0,y=0

0>3

which is not possible .

(0,0) does not satisfy the inequality

Hence shaded region is the required region.

now eq of line through (2,1)and (4,-3) is

y-1=\frac{-3-1}{4-2}(x-2)

y-1=-2(x-2)

y-1=-2x+4

2x+y=5

it is also a  dotted line so either < or >

consider 2x+y<5

put x=0,y=0

0<5

which is true.

so (0,0) satisfy this inequality.

so the two inequalities are

2x-y>3

and 2x+y<5

17.

consider the points (0,4),(2,3),(4,4)

eq. of line through (0,4) and (2,3) is

y-4=\frac{3-4}{2-0}(x-0)

y-4=-1/2(x)

2y-8=-x

or x+2y=8

as the line is solid

so either≤ or ≥

consider x+2y≥8

put x=0,y=0

0≥8

which is impossible.

(0,0) does not satisfy the graph.

which is true as graph lies above the line.

again eq. of line through (2,3) and (4,4) is

y-3=\frac{4-3}{4-2}(x-2)

y-3=1/2(x-2)

2y-6=x-2

x-2y=-4

consider x-2y≤-4

put x=0, y=0

0≤-4

which is impossible.

so (0,0) does not satisfy the graph.

so inequality is true as shaded portion is above and left of the line.

so two inequalities are

x+2y≥ 8

and x-2y≤-4