**System of Linear Inequalities**

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(a) two inequalities connected by

**and**

(b) two inequalities connected by

**or**

(c) more than two inequalities connected by

**and**and so on.

Now we will discuss the above three cases.

(i) When the system of inequalities is connected by 'and' then the solution set is the common points of two solution sets. (intersection of two sets)

(ii) When the system of inequalities is connected by 'or' then the solution set belongs to either of the two solution sets. (union of two sets)

(iii) When the system of inequalities consists of three inequations and all are connected by 'and' then the solution set is intersection of three sets.

## Examples on system of linear inequalities

**Example 1 :**Find the solution set from system of linear inequalities 3 + 2x > 5 and -3 + 4x >9. Also represent the solution on number line.

**Solution**

3 + 2x > 5 Add -3 on both sides 3 - 3 + 2x > 5 - 3 2x > 2 (since 3 - 3 = 0) $\frac{2x}{2} > \frac{2}{2}$ ∴ x > 1 |
and -3 + 4x > 9 Add + 3 on both sides and -3 + 3 + 4x > 9 and 4x > 12 ( since - 3 + 3 =0) and $\frac{4x}{4} > \frac{12}{4}$ and x > 3 |

From the 2nd equation the solution set = { 4,5,6,7,8,...}

As there is an

**and**between the two equations so the solution set of linear inequalities will be the intersection of the two sets which is { 4,5,6,7,8,...}

which is nothing but x > 3.

Hence the solution set = {x | x > 3, x $\epsilon $ R }

**Example 2 :**Find the range of values of x which satisfy the inequalities

$\frac{-1}{5}\leq \frac{3x}{10}+1 < \frac{2}{5}$

**Solution :**$\frac{-1}{5}\leq \frac{3x}{10}+1 < \frac{2}{5}$

As there is just 1 means 1 over 1

$\frac{-1}{5}\leq \frac{3x}{10}+ \frac{1}{1} < \frac{2}{5}$

As there are fractions so the LCD of 5, 10 and 5 is 10, so multiply each fraction by 10 we get,

$\frac{-1\times 10}{5}\leq \frac{3x \times 10 }{10}+\frac{1\times 10}{1} < \frac{2\times 10}{5}$

-2 $\leq$ 3x + 10 < 4

∴ we have -2 $\leq$ 3x + 10 and 3x + 10 < 4

-2 $\leq$ 3x + 10 Add -10 on both sides -2 - 10 $\leq$ 3x -12 $\leq$ 3x $\frac{-12}{3} \leq \frac{3x}{3}$ ∴ -4 $\leq$ x |
and 3x +10 < 4 Add -10 on both sides and 3x + 10 -10 < 4 - 10 and 3x < -6 ( since 10 - 10 =0) and $\frac{3x}{3} < \frac{-6}{3}$ and x < -2 |

∴ -4 $\leq$ x < -2

The dark blue line is the solution set.

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