There we defined inequality as a true statement containing the signs
'greater than' (>) or 'less than' (<).
We also saw that the vertex of inequality
symbol always lies towards small quantity.
We also saw how to write two or
more inequalities in one sentence, with example.
We also saw various other
signs, the inequality may contain with examples.
We also defined inequation
as an open sentence containing one of those signs.
We also understood set of
roots or solution set of an inequation.
We defined domain as the replacement
set of the variable of an inequation.
we also solved an example by trial and error.
Here we will solve more inequations by applying the properties of inequations.
We may call Linear Inequations as Linear Inequalities.
Properties of Linear Inequalities :
Property 1 of Linear Inequalities :
Adding (or Subtracting) the same number to both sides of an
Algebra Inequality does not change the order of the inequality sign
( i.e., '>', or '<'). i.e., if a < b then a + c < b + c and a - c < b - c
Similarly, if a > b then a + c > b + c and a - c > b - c
for any three numbers a, b, c.
Property 2 of Linear Inequalities :
Multiplying (or Dividing) both sides of an Algebra Inequality by the same positive number does not change the order of the inequality sign ( i.e., '>', or '<'). For any three numbers a, b, c where c > 0, (i) if a < b then ac < bc and a⁄c < b⁄c (ii) if a > b then ac > bc and a⁄c > b⁄c
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Multiplying (or Dividing) both sides of an
Algebra Inequality by the same negative number reverses the order
of the inequality sign ( i.e., '>' to '<' and '<' to '>').
For any three numbers a, b, c where c < 0,
(i) if a < b then ac > bc and a⁄c > b⁄c
(ii) if a > b then ac < bc and a⁄c < b⁄c
Property 4 of Linear Inequalities :
If three numbers are related in such a way that the first is less (greater)
than the second and the second is less (greater) than the third, then the
first is less (greater) than the third.
This is called transitive property.
Property 5 of Linear Inequalities :
If a and b are of the same sign and a < b (a > b), then
1⁄a > 1⁄b (1⁄a < 1⁄b).
If reciprocals are taken to quantities of the same sign on both sides
of an inequality, then the order of the inequality is changed.
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To solve linear inequation, collect terms containing unknown quantity
(variable) on left side and constants on the right side.
Then reduce the coefficient of the unknown quantity to unity.
While doing this, remember the properties of the algebra inequalities.
Check whether x = 6 is a solution of 9x + 1 > 7x + 5
Solution:
When x = 6, the L.H.S. of the inequation = 9x + 1 = 9(6) + 1 = 55
and the R.H.S. of the inequation = 7x + 5 = 7(6) + 1 = 43
We know 55 > 43. So x = 6 satisfies the given inequation.
∴ x = 6 is a solution of 9x + 1 > 7x + 5.
Solved Example 2 : Linear Inequalities
Check whether t = -3 is a solution of (7t + 5)⁄6 ≥ (t - 1)⁄2
Solution:
When t = -3, the L.H.S. of the inequation = (7t + 5)⁄6 = {7(-3) + 5}⁄6 = -8⁄3
and the R.H.S. of the inequation = (t - 1)⁄2 = (-3 - 1)⁄2 = -2
We know -8⁄3 < -2. So t = -3 does not satisfy the given inequation.
∴ t = -3 is not a solution of (7t + 5)⁄6 ≥ (t - 1)⁄2
Solved Example 3 : Linear Inequalities
If the domain of the variable is N (the natural number set),
solve the inequation (x + 2) ≤ (3x - 4)
Solution:
(x + 2) ≤ (3x - 4) ⇒ x - 3x ≤ -4 -2 ⇒ -2x ≤ -6
Dividing throughout by -3
(negative number division causes reversing of inequality sign), we get
⇒ x ≥ 3. Ans.
Since the domain is the set of Natural numbers, x = { 3, 4, 5, 6, ..............}. Ans.
Solved Example 4 : Linear Inequalities
If the domain of the variable is Q (the set of rational numbers),
solve the inequation (2x - 3)⁄4 > (3x + 4)⁄3
Solution:
(2x - 3)⁄4 > (3x + 4)⁄3
Multiplying both sides by 12, we get
3(2x - 3) > 4(3x + 4) ⇒ 6x - 9 > 12x + 16 ⇒ 6x - 12x > 16 + 9
⇒ - 6x > 25
Dividing throughout by -6
(negative number division causes reversing of inequality sign), we get
⇒ x < -25⁄6. Ans.
Solved Example 5 : Linear Inequalities
If the domain of the variable is Q (the set of rational numbers),
solve the inequation 3t⁄4 - 2⁄3 > t⁄2 + 1⁄3
Solution:
3t⁄4 - 2⁄3 > t⁄2 + 1⁄3
Multiplying both sides by 12, we get
9t - 8 > 6t + 4 ⇒ 9t - 6t > 4 + 8 ⇒ 3t >12 ⇒ t > 4 . Ans.
Exercise : Linear Inequalities
Check whether y = 2 is a solution of (2y + 3)⁄3 < (3y - 4)
Check whether z = -5 is a solution of (z⁄5 - z⁄2) < 3
Solve the inequation -7x + 78 ≥ 3x - 72 where x is a variable on Z (the set of integers).
Solve the inequation 13y⁄6 - 5⁄2 ≥ y⁄3 + 73⁄6 where y is a variable on N (the set of natural numbers).
Solve the inequation 2x + 3 < 10, where x is a variable in Q (the set of rational numbers).
For Answers see at the bottom of the page.
Answers to Exercise : Linear Inequalities
(1) No (2) Yes. (3) x ≤ 15 (4) y ≥ 8 (5) x < 7⁄2
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