Solve the following inequalities graphically in two-dimensional plane: \[-3x+2y\ge -6\] - Noon Academy

Solve the following inequalities graphically in two-dimensional plane:

$-3x+2y\ge -6$

Class 11 | Exercise 6.2 | Linear Inequalities | Maths | NCERT

Solve the following inequalities graphically in two-dimensional plane:

$-3x+2y\ge -6$

Solution:

The given inequality is

$-3x+2y\ge -6$

Let us draw a solid line

$-3x+2y=-6$

in the graph (∵

$-3x+2y=-6$

is included in the given question)

Let us consider

$-3x+2y\ge -6$

A point

$(0,0)$

is selected

⇒

$-3\times (0)+2\times (0)\ge -6$

⇒

$0\ge -6$

(this is true)

Therefore, the solution region of

$-3x+2y\ge -6$

is above the line

$-3x+2y\ge -6$

. (That is origin

is included in the region)

The graph is represented as:

i

More Material

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A line passes through the point (2, 1, -3) and is parallel to the vector Find the equations of the line in vector and Cartesian forms.

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Evaluate:

Evaluate:

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If $f(x) = |x| - 3,find\mathop {\lim }\limits_{x \to 3} f(x)$

Let $\begin{array}{l} f(x) = \left\{ \begin{array}{l} \frac{x}{{|x|}},x \ne 0\\ 0,x = 0 \end{array} \right.\\ \end{array}$ . Show that $\begin{array}{l} \mathop {\lim }\limits_{x \to 0} f(x) \end{array}$ does not exist.

Let $\begin{array}{l} f(x) = \left\{ \begin{array}{l} \frac{|x-3|}{{(x-3)'}},x \ne 3\\ 0,x = 3 \end{array} \right.\\ \end{array}$ . Show that $\begin{array}{l} \mathop {\lim }\limits_{x \to 3} f(x) \end{array}$ does not exist.

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