Solve the following inequalities graphically in two-dimensional plane-x - y - 5

Question

Toppr Admin · Accepted Answer

X-y-lt-5 -xA0-is the shaded region shown in the graph-If we put O-0-0- in the LHS-we get LHS-0Now LHS-lt-5-xA0-Means O-0-0- lies on -quot-less than -quot- side of -xA0-the line-Hence required region is -quot-origin-quot- side of the-xA0-line-

Solve the following inequalities graphically in two-dimensional plane:
$x + y < 5$

$x + y < 5$ is the shaded region shown in the graph.
If we put $O (0, 0)$ in the LHS,we get LHS $= 0$
Now $L H S < 5$
Means $O (0, 0)$ lies on "less than " side of the line.
Hence required region is "origin" side of the line.

Solve the following inequalities graphically in two-dimensional plane:x+y<5

x+y<5 is the shaded region shown in the graph.If we put O(0,0) in the LHS,we get LHS=0Now LHS<5 Means O(0,0) lies on "less than " side of the line.Hence required region is "origin" side of the line.

Solve the following inequalities graphically in two-dimensional plane:
$x + y < 5$

$x + y < 5$ is the shaded region shown in the graph.
If we put $O (0, 0)$ in the LHS,we get LHS $= 0$
Now $L H S < 5$
Means $O (0, 0)$ lies on "less than " side of the line.
Hence required region is "origin" side of the line.