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Question

Question: Solve the given inequalities graphically: \(3x + 2y \leqslant 12,x \geqslant 1,y \geqslant 2\)...

Solve the given inequalities graphically:
3x+2y12,x1,y23x + 2y \leqslant 12,x \geqslant 1,y \geqslant 2

Explanation

Solution

First, we will solve the inequality given in the question and from the solution of inequality, we will plot those points on a graph. After plotting the point, we will connect those points by a straight line. For each line, we will check whether it’s inequality at origin and accordingly shade that portion on the graph.

Complete step-by-step answer:
The given inequalities are 3x+2y12,x1,y23x + 2y \leqslant 12,x \geqslant 1,y \geqslant 2.
The solution for 3x+2y123x + 2y \leqslant 12 is
Let’s, x = 0, then
2y=122y = 12
y=6\Rightarrow y=6
Now, let’s y=0, then
3x=123x=12
x=4\Rightarrow x=4

xx0044
yy6600

i.e., (0,6)(0,6) and (4,0)(4,0).
Plotting all these points on a graph,

Now we will check the position of origin with respect to the inequalities,
For 3x+2y123x + 2y \leqslant 12 at (0,0)(0,0) , 3(0)+2(0)12<03(0) + 2(0) \leqslant 12 < 0. Hence (0,0)(0,0) lies in the inequality 3x+2y123x + 2y \leqslant 12.
For x1x \geqslant 1 at (0,0)(0,0) , 01<00 - 1 < 0. Hence (0,0)(0,0) lies does not lie in the inequality x1x \geqslant 1.
For y2y \geqslant 2 at (0,0)(0,0) , 02<00 - 2 < 0. Hence (0,0)(0,0) lies does not lie in the inequality y2y \geqslant 2.
The inequality 3x+2y123x + 2y \leqslant 12is towards the origin and x1x \geqslant 1 and y2y \geqslant 2 are away from the origin.
After shading the inequality part in the graph, we get the graph as,

The shaded part in the above graph represents the solution of the given inequalities.

Note: Any solution to an inequality is the fee of that variable which makes inequality a true statement. For example, assume we've an inequality of x<5x < 5 . In one of these cases, all the values of x which are much less than 55 make this inequality a true inequality. While fixing inequalities we want to preserve a few policies in thought, identical numbers may be added or subtracted from both aspects of the inequality. Both facets of inequality may be extended (or divided) by the identical wonderful quantity. However, when both aspects are increased or divided by means of a bad quantity, then the sign of inequality is reversed. These operations do not affect inequality and may be used to simplify inequality for us.