mathematics-class-11-unit-07-chapter-03-linear-inequalities-l-3-6-sz-npodeldm

##### <Success>Graphical solution of linear inequation in two variables</Success>
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<p>$(x, y)$</p>
<ol>
<li>$x+y=5$</li>
<li>$x+y>5$</li>
<li>$x+y<5$</li>
</ol>
<p>Consider the graph.
The graph of the line $x+y=5$ divides the points in the plane into three sets.</p>
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<h5 id="successtypes-of-half-planesuccess"><Success><strong>Types of Half Plane</strong></Success></h5>
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<p><strong>(i) Closed Half Plane:</strong></p>
<p>$a x+b y \geqslant c / a x+b y \leqslant c$</p>
<p><strong>(ii) Open Half Plane:</strong></p>
<p>$a x+b y>c / a x+b y<c$</p>
<p>$ 2 x+3 y \leqslant 1 \text { or } x+y \geqslant 3 $</p>
<p>$ x+y<2 \text { or } 3 x-y>1$</p>
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### <Success>Solution set and solution graph</Success>
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<p><strong>Solution set:</strong></p>
<p>The set of all ordered pair $(\alpha, \beta)$ of a real number which satisfy a given inequality is called the solution set of the given inequality.</p>
<p><strong>Solution region (or Graph):</strong></p>
<p>The region of the plane containing all the points whose co-ordinates satisfy a given inequality is called the solution region (or graph) of the inequality.</p>
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### <Success>Solution set and solution graph</Success>
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$3x-y<2$
<p>$(\alpha,\beta)$</p>
<p>$3 \alpha-\beta<2 \quad(\alpha, \beta) \in \text{solution set}$</p>
<p>$3 \alpha -\beta\nless 2 \quad(\alpha,\beta) \not\in\text{solution set}$</p>
<p>$\text{solution set}$ =$\{(\alpha,\beta)| (\alpha,\beta) \text{satisfy the inequality} ax+by \leqslant c ~ or ~ ax+by \geqslant c ~ or ~ ax+by < c ~ or ~ ax+by > c\}$</p>
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#### <Success>$\text{Solution region for} \ ax+by \leqslant c$</Success>
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<img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-11-Unit-07-chapter-03-linear-inequalities-l-3-6-sz-npodeldm-5a.jpg">
<img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-11-Unit-07-chapter-03-linear-inequalities-l-3-6-sz-npodeldm-5b.jpg"></p>
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### <Success>Example</Success>
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<p>Step I: $\quad 2 x-y \geqslant 1$</p>
<p>Associated equation is</p>
<p>$2 x-y=1$</p>
<p>Step II:</p>
<p>put $y=0 \Rightarrow x=1 / 2$</p>
<p>that is  $\quad(\frac{1}{2}, 0)$</p>
<p>put</p>
<p>$x=0 \Rightarrow-y=1 \Rightarrow y=-1$</p>
<p>that is  $(0,-1)$</p>
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<img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-11-Unit-07-chapter-03-linear-inequalities-l-3-6-sz-npodeldm-6a.jpg"></p>
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##### <Success>Example</Success>
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<p>Step III : Draw the graph of the line,</p>
<p>$2 x-y=1$</p>
<p>Step IV : $\quad 2 x-y \geqslant 1$</p>
<p>$(1,2)$</p>
<p>$2 \times 1-2=0 \geqslant 1$</p>
<p>$\therefore$ point $(1,2)$ does’nt lies in the solution  region of
$2 x-y \geqslant 1$</p>
<p><strong>Origin test:</strong></p>
<p>$O(0,0) \quad 2 x-y \geqslant 1$         <br>
$2 \times 0-0=0 \cancel{\geqslant} 1$</p>
<p>$\therefore O(0,0)$ does’nt lies in the solution region.</p>
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<img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-11-Unit-07-chapter-03-linear-inequalities-l-3-6-sz-npodeldm-7a.jpg"></p>
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### <Success>Solution</Success>
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<p>Arbitrary point test: .</p>
<p>$(-1,2)\quad  2 x+3 y \leqslant 6$</p>
<p>$2 \times(-1)+3 \times 2=  4<6$</p>
<p>$\therefore(-1,2)$ satisfy the inequation</p>
<p>$2 x+3 y \leqslant 6$</p>
<p>$\therefore$ Half plane II will be the solution Region as $(-1,2)$ belongs to this region.</p>
<p>$\therefore$ Shaded region will be solution region.</p>
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<img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-11-Unit-07-chapter-03-linear-inequalities-l-3-6-sz-npodeldm-10a.jpg"></p>
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### <Success>Example 2</Success>
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<p>Solve graphically: $2 y+x \geqslant 0$</p>
<p><strong>Solution:</strong> Associated equation  for the given inequation will be</p>
<p>$2 y+x=0$</p>
<p>put $y=0$
$ \Rightarrow x=0 \quad(0,0)$</p>
<p>put $x=1 $
$\Rightarrow y=-1 / 2 \quad(1,-1 / 2)$</p>
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<img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-11-Unit-07-chapter-03-linear-inequalities-l-3-6-sz-npodeldm-11a.jpg"></p>
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### <Success>Solution</Success>
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Arbitrary point test:
<p>$ (1,2) \quad 2 y+x \geqslant 0$</p>
<p>$2 \times 2+1=5>0$</p>
<p>$\therefore(1,2)$ satisfy the inequation $2 y+x \geqslant 0$</p>
<p>$\therefore(1,2) \in$ Solution Region.</p>
<p>$\therefore$ Half plane I will be the solution region.</p>
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<img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-11-Unit-07-chapter-03-linear-inequalities-l-3-6-sz-npodeldm-12a.jpg"></p>
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### <Success>Example 3</Success>
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<p>$|x| \leqslant 3$</p>
<p>$\Rightarrow\quad  -3 \leqslant x \leqslant 3$</p>
<p>$\Rightarrow\quad  x \geqslant-3$ & $x \leqslant 3$</p>
<p>Solve Graphically: $|y-x| \leqslant 3$</p>
<p><strong>Solution:</strong> Given, $|y-x| \leqslant 3$</p>
<p>$ \Rightarrow \quad-3 \leqslant y-x \leqslant 3 $</p>
<p>$ \Rightarrow y-x \geqslant-3,$<br>
$ \quad y-x \leqslant 3 $</p>
<p>$ \Rightarrow x-y \leqslant 3, \quad x-y \geqslant-3$</p>
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### <Success>Solution</Success>
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<p>$x-y \leqslant 3\cdots(i)$</p>
<p>$x-y \geqslant-3\cdots(ii)$</p>
<p>For (i):</p>
<p>$ \therefore \quad x-y \leqslant 3 $</p>
<p>$ \Rightarrow \quad \frac{x}{3}-\frac{y}{3} \leqslant  1 $</p>
<p>$ \Rightarrow \quad \frac{x}{3}+\frac{y}{-3}\leqslant 1$</p>
<p>Associated equation
$\frac{x}{3}+\frac{y}{-3}=1$</p>
<p>$\left[\frac{x}{a}+\frac{y}{b}=1\right]$</p>
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<img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-11-Unit-07-chapter-03-linear-inequalities-l-3-6-sz-npodeldm-14a.jpg"></p>
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### <Success>Solution</Success>
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<p>$ x-y \leqslant 3 $</p>
<p>$ (0,0) $</p>
<p>$ 0-0=0<3$</p>
<p>$\therefore$ origin $(0,0)$ belongs to solution region of  $x-y \leqslant  3$</p>
<p>Again for (ii),  $\quad x-y \geqslant-3 $</p>
<p>$\Rightarrow \quad \frac{x}{-3}+\frac{y}{3}=1$<br>
$\ \  \text { (Associated equation) }$</p>
<p>$ \therefore(-3,0)$ & $(0,3)$</p>
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<img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-11-Unit-07-chapter-03-linear-inequalities-l-3-6-sz-npodeldm-15a.jpg"></p>
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### <Success>Example 4</Success>
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<p>Solve graphically: $|x-y| \geqslant 1$</p>
<p><strong>Solution:</strong> $\quad|x-y| \geqslant 1$</p>
<p>$\Rightarrow \quad x-y \leqslant-1 \quad$ & $\quad x-y \geqslant 1$</p>
<p>$x-y=-1$</p>
<p>$\Rightarrow \frac{x}{-1}+\frac{y}{1}=1$</p>
<p>$x-y=1$</p>
<p>$\Rightarrow \frac{x}{1}+\frac{y}{-1}=1$</p>
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<img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-11-Unit-07-chapter-03-linear-inequalities-l-3-6-sz-npodeldm-17a.jpg"></p>
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### <Success>Solution</Success>
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origin test $\quad(0,0)$
<p>$x-y \leqslant-1$</p>
<p>$0-0=0 \leqslant-1  \text { (False) }$</p>
<p>$x-y \geqslant 1$</p>
<p>$0-0=0 \geqslant 1 \text { (False) }$</p>
<p>$\therefore$ Shaded portion will be the solution region for $|x-y| \geqslant 1$.</p>
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<img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-11-Unit-07-chapter-03-linear-inequalities-l-3-6-sz-npodeldm-18a.jpg"></p>
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<p>======</p>
<h3 id="successthank-yousuccess"><Success>Thank you</Success></h3>
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