Answer:
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