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In this section, we shall find the solution of system of linear inequalities in one variable. A system of linear inequations can have(a) two inequalities connected by

(b) two inequalities connected by

(c) more than two inequalities connected by

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.

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

which is nothing but x > 3.

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

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

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.

From system of inequalities to Home

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