It is the process of obtaining all the possible solution of an inequation.
* SOLUTION SET:
The set of all possible solutions of an inequation is known as its solution set.
=> SOLVING LINEAR INEQUATION IN ONE VARIABLE:
• RULE-1: Same number may be added to (or subtracted from) both sides of an inequation without changing the sign of inequality.
•RULE-2: Both sides of an inequation can be multiplied(or divided) by the same positive real number without changing the sign of inequality. However, the sign of inequality is reversed when both sides of an inequation are multiplied or divided by a negative number.
• RULE-3: Any term of an inequation may be taken to the other side with its sign changed without affecting the sign of inequality.
A linear inequation in one variable is of the form
ax+b<0 or, ax+b≤0 or, ac+b> 0 or, ax+ b ≥ 0.
We follow the following STEP to solve a linear inequation in one variable.
• STEP-1: Obtain the linear inequation.
• STEP -2 : Collect all terms involving the variable on one side of the inequation and the constant terms on the other side.
• STEP -3: Simplified both sides of inequality in their simplest form to reduce the inequation in the form
ax <b, or ax≤b, or ax>b, or ax≥b
• STEP -4: Solve the inequation obtained in step - 3 by dividing both sides of the inequation by the coefficient of the variable.
• STEP -5: Write the solution set obtained in step -4 in the form of an interval on the real line.
EXERCISE - 1
😀 ------------------ 😀
* Solve the following inequation:
1) 2x - 4≤ 0. (-∞, 2]
2) - 3x +12< 0. (4,∞)
3) 4x - 12 ≥ 0. [3,∞)
4) 12x < 50, when
a) x ∈ R. (-∞,25/6)
b) x ∈ Z. {....,-3,-2,-1,0,1,2,3,4}
c) x ∈ N. {1,2,3,4}
5) - 4x > 30, when
a) x ∈ R. (-∞,-15/2)
b) x ∈ Z. {....,-9,-8}
c) x ∈ N. null set
6) 7x +9> 30. (3,∞)
7) 4x -2 < 8, when
a) x ∈ R. (-∞,5/2)
b) x ∈ Z. {....-2,-1,0,1,2}
c) x ∈ N. {1,2}
8) 5x -3< 3x+1 when
a) x is a real number. (-∞, 2)
b) x is integer number. {..,-4,-3, -2, -1, 0,1}
c) x is a natural number. {1}
8) 3x - 7 > x+1. (4,∞)
9) x+ 5 > 4x -10. (-∞, 5)
10) 3x+ 9 ≥ - x+19. (5/2,∞)
11) 3x+17≤ 2(1-x). (-∞, 3]
12) 2(2x+3)-10≤6(x-2). [4,∞)
13) 2(3-x)≥ x/5 + 4. (-∞, 10/11]
14) -(x-3)+4< 5- 2x. (-∞,-2)
15) (2x-3)/4 +9≥ 3+ 4x/3. (-∞,63/10]
16) (3x-2)/5 ≤ (4x-3)/2. [11/14,∞)
17) (5x-2)/3 -(7x-3)/5 > x/4. (4,∞)
18) 1/2(3x/5 +4) ≥ (x- 6)/3. (-∞,120]
19) 3(x- 2)/5 ≥ 5(2 - x)/3. [2,∞)
20) x/5< (3x-2)/4 - (5x-3)/5. (-∞,2/9)
21) 2(x-1)/5 ≤ 3(2+x)/7. [-44,∞)
22) 5x/2 + 3x/4 ≥ 39/4. [3,∞)
23) (x-1)/3 +4< (x -5)/5 - 2. (-∞,50)
24) (2x+3)/4 - 3 < (x-4))3 - 2. (-∞, - 13/2)
25) (5-2x)/3 < x/6 - 2. (8,∞)
26) (4+2x)/3 ≥x/2 - 3. (-26,∞)
27) (2x+3)/5 - 2 < 3(x-3)/5. (-1,∞)
28) (x-2)≤(5x+8)/3. (-7,∞)
29) 1/(x -2)< 0. (-∞,2)
30) (x+1)/(x +2)≥ 1. (-∞,-2)
------------------------------------------------------
Type ::2
EQUATION OF THE FORM
* (ax+b)/(cx+d)> k, OR
* (ax+b)/(cx+d)≥ k, OR
* (ax+b)/(cx+d)< k, OR
* (ax+b)/(cx+d) ≤ k.
STEP -1: Obtain the inequation.
STEP-2:Transpose all terms on LHS
STEP-3: Simplify LHS of the inequation obtained in STEP-2 to obtain an inequation of the form
(px+q)/(rx+s)> 0 OR
(px+q)/(rx+s)≥0 OR
(px+q)/(rx+s)< 0, OR
(px+q)/(rx+s) ≤ 0
STEP-4: Make coefficient x positive in numerator and denominator if they are not.
STEP-5: Equate numerator and denominator separately to zero and obtain the values of x. These values of x are generally called critical points.
STEP-6: Plot the critical points obtained in STEP-5 on real line. These points will divide the real line in three regions.
STEP-7: In the right most region the expression on LHS of the inequation obtained in STEP-4 will be positive and in other regions it will be alternatively negative and positive. So, mark positive signs in the right most region and then mark alternatively negative and positive signs in other regions.
STEP-8: Select appropriate region on the basis of the sign of the inequation obtained in STEP-4 , Write these region in the form of interval to obtain the desired solution sets of the given inequation.
EXERCISE-2
-----------------
Solve the inequation of followings:
1) (2x+4)/(x-1) ≥ 5. (1,3]
2) (x+3)/(x-2) ≤2. (-∞,2)U(7,∞)
3) (2x-3)/(3x-7) ≤2. (-∞,3/2) U (7/3,∞)
4) 3/(x-2) < 1 (-∞,2)U(5,∞)
5) 1/(x-1) ≤2. (-∞,2)U(3/2,∞)
6) (4x+3)/(2x-5) ≤2. (-∞,5/2) U (33/8,∞)
7) (5x-6)/(x+6) <1. (-6,3)
8) (5x+8)/(4-x) <2. (-∞,0) U (4,∞)
9) (x-1)/(x+3) >2. (-7,-3)
10) (7x-5)/(8x+3)>4 (-17/25, -3/8)
11) x/(x-5) > 1/2. (-∞,-5)U(5,∞)
12) (x-3)/(x+4) >0, x ∈R {x∈ R: x < -4}U {x ∈ R: x> 3}
13) (x+5)/(x-2) >0, x ∈R. {x∈ R: x ≤ -5}U {x ∈ R: x> 2}
14) (2x+5)/(x+3) >1, x ∈R. {x∈ R: x < -3} U {x∈ R: x > -2}
15) (x+7)/(x+4) >1, x ∈R. {x∈ R: x > - 4}
16) (x+4)/(x+6) >1, x ∈R. {x∈ R: x < -6}
17) 3/(x -2) >2, x ∈R. {x∈ R: 2< x < 7/2}
18) (x-3)/(x+1) < 0, x ∈R. {x∈ R: -1 < x < 3}
-----------------------------------------------------
TYPE -3:
-------------
* If a is a positive real number, then
1) |x| < a <=> -a<x< a i.e. x∈ (-a, a).
2) |x|≤a <=> -a≤x≤a i.e. x x∈ [-a, a].
3) | x|> a <=> x<-a or x> a.
4) | x|≥ a <=> x≤-a or x≥ a.
* Let r be a positive real number and a be a fixed real number. Then,
1) |x - a| < r <=> a-r <x< a+r i.e. x∈ (a - r, a+ r).
2) |x -a|≤r <=> a-r ≤x≤a+r i.e.x∈ [a - r, a +r].
3) | x - a|> r <=> x<a -r or x> a +r
4) | x -a|≥ r <=> x≤-a- r or x≥ a+r.
EXERCISE -3
------------------
Solve the inequation of followings:
1) | x | < 5, x ∈R. {x∈ R: -5< x < 5}
2) |x |≥ 5, x ∈R. {x∈ R: x < -5} U {x∈ R: x ≥ 5}
3) |3x -2|≤ 1/2. {x∈ R: x < -5}
4) | 4x - 5 | ≤ 1/3, x∈[1/2,5/7}
5) | x - 2|≥5 (-∞,-3)U[7,∞)
6) |5 - 2x | ≤ 3, x ∈R. {x∈ R: 1≤x< 4}
7) | 3x - 7| > 4, x ∈R. {x∈ R: x < 1} U {x∈ R: x > 11/3}
8) |2(3-x)/5| < 9/5, x ∈R. {x∈ R: -3/2 < x < 15/2}
9) |x +1/3| > 8/3. (-∞,-3)U(7/3,∞)
10) |4 - x| + 1<3. (2,6)
11)|(3x-4)/2|≤ 5/12. (19/18,29/18)
12) |x -1|≤ 5, |x|≥ 2. (-∞,-2]U[2,∞)
13) 1≤| x-2|≤ 3. [-1,1]U[3,5]
14) |x- 1|≤5, | x |≥2. (-∞,-2]U[2,∞)
15) |x-2|/(x-2) > 0. (2,∞)
16) (|x| -1)/(|x| -2) ≥0, x∈R, x≠ ±2. [-1,1]U(-∞,-2]U(2,∞)
17) -1/(|x| -2) ≥ 1, where x∈R, x≠ ±2. [-2,-1]U[1,2)
18) |2/(x-4)|> 1, x ≠4. (2,4)U(4,6)
19) (|x+3| + x)/(x+2) > 1. (-5,-2) U(-1,∞)
20) 1/(|x| -3)< 1/2. (-∞,-5) U(-3, 3) U(5,∞)
21) (|x +2| - x)/x < 2. (-∞,2]U(1,∞)
22) |x -1| + |x-2|≥ 4. (-∞,-1/2]U [7/2,∞)
23) |x-1|+ |x-2| + |x -3|≥6. (-∞,0] U [4,∞).
-------------------------------------------------------
TYPE:: 4
STEP-1: Convert the given inequation, say ax + by≤ c, into the equation ax+ by = c which represents a straight line in xy-plane put.
STEP-2: Put y= 0 in the equation obtained in STEP- 1 to get the point where the line meets with x-axis. Similarly, put x= 0 to obtain a point where the line meets y-axis.
STEP-3: Join the points obtained in STEP-2 to obtain the graph of the line obtained from the given inequation. In case of strict inequalities like ax+ by< c or ax + by > c, draw the dotted line, otherwise mark it thick line.
STEP-4: Choose a point, if possible (0,0), not lying on this line: substitute its coordinates in the inequation. If the inequation is satisfied, then shade the portion of the plane which contains the chosen point; otherwise shade the portion which does not contain the chosen point.
STEP-5: The shaded regions obtained in STEP-4 represents the desired solution set.
*** NOTE::
In case of inequalities ax+ by≤ c and ax + by ≥ c points on the line are also a part of the shaded region while in the case of inequalities ax + by < c and ax + by > c points on the line ax + by = c are not shaded region.
EXERCISE-4
-----------------
** Solve graphically in a two- dimensional plane.
1) 5x - 8 ≥ 0.
2) 2y -3 < 0.
3) 2x - 3 ≥ 0.
4) 4x - 3 ≤ 0.
5) x - 1 < 0.
6) x - 2 > 0.
7) 2y - 3≥ 0.
8) 5y - 4 ≤ 0.
9) y - 1< 0.
10) 2y +5> 0.
11) 2x+3y ≤ 6.
12) 2x - y≥ 1
13) x ≥ 2
14) y ≤ -3
15) x+ 2y - y ≤ 0.
16) x+ 2y ≥ 6.
17) - 3x + 2y≤ 6.
18) 0≤ 2x - 5y +10.
19) 3x - 2y≤ x+y - 8.
20) |x|≤ 3.
21) |y - x|≤ 3.
22) |x - y|≥ 1.
--------------------------------------------------------
LINEAR INEQUATIONS(In Two Variables)
TYPE: 5 (A)
------------
STEP-1: Consider the equation ax+ by+ c= 0.
Draw the graph of this equation, which is a line.
In case of strict inequation > or < draw the line dotted, otherwise make it thick.
This line divides the plane into two equal parts.
STEP-2: Choose a point [if possible (0,0)], not lying on this line. If this point satisfies the given inequation then shade the part of the plane containing this point, otherwise shade the other part.
The shaded portion represents the solution set of the given inequation. The dotted line is not a part of the sulution, while thick line is a part of it.
EXAMPLE:
Draw the graph of the solution set of the inequation 2x -y≥ 1.
Solution: consider the equation 2x- y= 1.
The value of (x,y) satisfying 2x - y= 1 are:
x: 2 0
y: 3 -1
Plot the points A(2,3) and B (0,-1) on a graph paper.
Join A and B by a thick line.
This line divides the plane of the paper into two equal parts. Consider the point (0,0). It does not lie on 2x - y= 1.
Clearly, (0,0) doesn't satisfy 2x - y≥ 1.
So shade that part of the plane divided by line AB which does not contain (0,0).
The shaded part of the plane together with all points on line AB constitute the solution set of the inequation 2x -y ≥ 1.
EXERCISE-5
-----------------
A) Draw the graph of the solution of each of the followings:
1) x+ y ≥ 4.
2) x - y ≤ 3.
3) x+ 2y > 1.
4) 2x - 3y < 4.
5) x ≥ y - 2.
6) y - 2 ≤ 3x.
7) 3x+ 5y < 15.
8) 4x - 3y > 12.
9) 3x+ 2y > 6.
-------------------------------++++++++---------
TYPE:5(B)
--------------
STEP-1:
B) Solve each of the following systems of inequation graphically:
1) 3x +2y ≤ 18, x+ 2y ≤10, x ≥0, y ≥ 0.
2) 2x+3y≤6,x+y≥2,x≥0,y≥0.
3) 3x+4y≥12,4x+ 7y≤28, y≥1, x≥0, y≥0.
4) x- 2y≤2, x+y≥3, -2x+y≤4, x≥0,y≥0.
5) x+2y≤100, 2x+y≤120, x+y≤70, x≥ 0, y≥ 0.
6) x+ 2y ≤2000, x+ y≤1500, y≤ 600, x≥ 0, y≥ 0.
No comments:
Post a Comment