Linear Equation _ Linear Inequality

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					Admas University College Hargeisa Main Campus


Linear Equation & Linear Inequality
Definition: An Equation is an algebraic expression that involve equality only
Definition: Linear Equation is an equation of the form
              . ax + b=0 where a, b are fixed number and a≠0
Solution of ax + b =0 is {-b/a}
Ex.Solve

. 3 x  1  2 x  4   1
 1            1
 7            5
Solutin
3x  1 2 x  4                     Ex.
                 1
   7         5                     Solve :
53 x  1  72 x  4                  1     1     1
                         1        1.          
          35                          xx  1 x x  1
15 x  5  14 x  28  35
                                   2.. 2 x  4  3x  1  1
                                       1          1
x  23  35                            5          7
x  12
S .Set  {12}
3. The sum of four consecutive even integer is 44. Find the four integers.
4. If three times a number is decreased by 6, the result is 16. Find the number.

                                       Simultaneous Equations
       The general form of a pair of simultaneous linear equations in two variables is
                  a1 x  b1 y  c1  0
                                         where a, b, c fixed number a and b ≠0
                  a 2 x  b2 y  c2  0
       A system of simultaneous linear equation in two variables may have
                        A unique solution
                        An infinite number of solutions
                        No solution
                                                                 2 x  y  5
                                                                x  y  4 
              2x  y  5                                        
                                                                            
                                                                             
Eg. Solve                    Solution by elimination method   3 x  9  S.set={x=3, y=1}
              x y 4                                            x3 
                                                                            
                                                                  y 1 
                                                                            
Exercise: Solve
                                3
1.3x  y  1                      x  4y  2
                           2. 2
... x  y  5
                                2 x  4 y  12

3. Yonas’s father is 36 years old. Six years ago he was three times as old as Yonas was
then. How old is yonas now?



                                       Page 1 of 8             Prepared By: Mohammed.M
               Admas University College Hargeisa Main Campus



               Linear Inequality
              A linear inequality is an inequality that can be written in the form:
              Ax + b < 0 (or > 0) …. For real numbers a and b , with a≠0
              Ex. 4x +2 > 5x-1
               4x – 5x > -1 – 2
               -x > -3 multiplying both sides by -1
               x<3
              ;. Solution set ={x: x <3} = (- , 3)
              Ex. Solve  7 x  6  23  x   3
              Solution -7x +6 < -6-2x -3 -7x +2x < -9 -6  -5x < -15  x > 3
                                                                           x5
               Solution Set is ={x: x>3} = (3, ) Ex. Solve                     0 for x≠0
                                                                           x3
              Solution Method I Case approach
              Case 1 for x  (-,3)                                          Case 2 for x  (3,∞)
              x–3<0                                                       x-3 >0
                                   x5                                                     x5
              There for (x-3)(             0)                            There for (x-3)(       0)
                                   x3                                                     x3
              x+5>0                                                       x + 5 <0
               x > -5 and x < 3                                           x <-5 and x >3
              S.S1= x   : 5  x  3   5,3                       S.S2={ }=no common point = 
               Sol. set= S.S1  S.S2= x   : 5  x  3   5,3   = x   : 5  x  3   5,3
              Method II Sign chart
               x5
                        0       x + 5=0 at x=-5 x -3 = 0 at x=3
               x3
                     x  5 5
                                3 3 x 3
                   5 x  
                  x50
                  x  3      0  
                  x5
                           
                  x3
                                             x5
              There fore the sol. set of             0 is x   : 5  x  3   5,3
                                             x3
                              x2
              Ex. Solve            0
                              x 1
              Quadratic Equation
Definition: An equation which is reducible to the form ax 2  bx  c  0 where
a, b, and c are fixed numbers such that a≠0 is called a quadratic (second degree) equation.
              1. Solution By factorization Method
              Ex. Solve x 2  x  110  0 p, q  Z such that p + q=b=1 and p.q=a.c=1.(-110)=-110
               x 2  x  110  ( x  11)( x  10 ) =0  x +11=0 or x – 10=0  x=-11 0r x=10
              Sol. Set = {-11, 10}




                                                      Page 2 of 8            Prepared By: Mohammed.M
                Admas University College Hargeisa Main Campus


                           x 2  y 2   x  y  x  y 
                N.B                                 
                           x 3  y 3   x  y  x 2  xy  y 2   
                           x y
                            3        3
                                           x  y x
                                                xy  y  2    2
                                                                  
                Ex . Solve 4x   2-25=0

                        2 25         2  5 2      5  5               5  5
                Sol. 4 x    4 x       x   x    0  sol. Set=  ,   
                              4          2      2  2               2 2 
                                      
                Ex. Solve 4x2-64=0
                2. Sol. By “completing the square” method
                a  b 2  a 2  2ab  b 2
                a  b 2 a 2 2ab  b 2
                x 2  a, a  0
                x a
                Example 1.  x  1  9  x  1   9  3  x=-4 or x=2
                                   2


                       The truth set of the equation is {-4, 2}
                 2.2 x  4   8  8
                                 2


                  2 x  4   8  8
                                     2


                  2 x  4   16  4


                2x=0 or -8  x=0 or -4         The solution set is {-4, 0}
                3. Consider the equation we started with, i. e x 2  x  1  0
                x 2  x  1  0  x 2  x  1.
We must now change x 2  x to a perfect square so that we will be able to take the square root of
 both sides. This is done by adding the square of half the coefficient of the x-term and adding
 the result to both sides of the last equation.
                                                              2        2
                                                      1      1
                                x2  x  1  x2  x     1  
                                                      2      2
                                               2
                                  1     1 5
                             x    1 
                Therefore:        2     4 4
                                  1    5     5
                            x        
                                  2    4    2
                                     1  5 1 5
                            x          
                                     2 2        2
                                                  1 5 1 5 
               Getting the solution the set to be        ,         
                                                   2         2 
               4. 8x 2  18x  9  0
  We start by dividing the equation by 8 so as to make the coefficient of x2 term unity, hence


                                                             Page 3 of 8   Prepared By: Mohammed.M
Admas University College Hargeisa Main Campus


                               18   9
8 x 2  18 x  9  0  x 2       x 0
                                8   8
      18     9
 x2    x
       8     8
      18   81    9 81 9
 x2  x         
       8   64    8 64 64
            2
     9   9
 x   
     8   64
   9    9     3
 x     
   8    64    8
  9 3
 
  8 8
                          3 3
Hence the solution set is  , 
                          4 2
5. Derive the quadratic formula by using completing square method. i.e
ax 2  bx  c  0 a≠0
                b  b 2  4ac
Sol. Set is x 
                      2a
Proof      ax  bx  c  0
             2

   1                         b   c
 (ax 2  bx  c  0)  x 2  x   0
   a                         a   a
      b       c
 x2  x  
      a       a
                     2           2
     b   b    b   c
 x  x    
     2

     a   2a   2a  a
                2
      b    b2   c
 x       2 
      2a   4a   a
              b 2  4ac
                2
      b 
 x      
      2a       4a 2
         b     b 2  4ac
 x        
         2a       4a 2
     b    b 2  4ac  b  b 2  4ac
x              
     2a      2a           2a

                     b  b 2  4ac  b  b  4ac
                                           2

The sol. Set is =                 ,                    
                          2a              2a



                                          Page 4 of 8       Prepared By: Mohammed.M
               Admas University College Hargeisa Main Campus


               N.B The expression b2 – 4ac is called the discriminant of the given quadratic
               equation and using it we can deduce the following about the roots of the
               equation;
                  i. If b2 – 4ac>0, we have two possible values for x.
                                                                               b
                 ii. If b2 – 4ac=0, the equation has only one root. i.e x  
                                                                               2a
                iii. If b2 – 4ac<0, there cannot possible exist a real value for x to satisfy the
                         equation. Hence the solution set is 
               Ex.
                1. 2 x 2  3x  2  0
               In this equation a=2, b=-3 and c=-2
                                  b  b 2  4ac 3  9  16 3  5
               Therefore x                     =          
                                       2a            4        4
                             1
                x=2 or 
                             2
                                              1 
               Therefore the solution set is  ,2 
                                              2 
               2. 3x 2  2 x  5  0 Again a=3, b=2 and c=5 b  4ac  4  60  56  0
                                                                   2

                Consequently the equation has no real roots. Therefore the solution set is 
               3. Determine the values of k so that the quadratic equation x 2  4kx  20  0 has
               only one solution.
               Sol. Only one solution  b 2  4ac  0  (-4k)2-4(1)(20)=16k2-80=0
                16k2=80
               k2=5
               k=  5
Ex. Determine the values of k so that the quadratic equation x 2  kx  4  0 has only one solution
               N.B Let r1 and r2 are roots of the quadratic equation, ax 2  bx  c  0 then
                                    b
                    r1  r2  
                                    a
                                  c
                    r1  r2 
                                  a
                   Ex. Find the sum and the product of the roots of the following equations,
                   a.3x 2  4 x  1  0
                   b.  6 x 2  12 x  6  0
                   c.x 2  13x  36  0
                   d .x 2  3x  54  0
               Ex. If one root of x 2  7 x  k  0 is -2.Find the other root and the value of k.
                                 b                7
               Sol. r1  r2     2  r2    r2  7  2  5
                                 a                1



                                                     Page 5 of 8           Prepared By: Mohammed.M
             Admas University College Hargeisa Main Campus


                             c             k
             And r1  r2        2  5   k  10
                             a             1
            Ex. If one root of 2hx 2  24x  15h  7  0 is 1.Find the other root and the value of h.
            Quadratic Inequality
Definition; An Inequality that can be reduced to the form ax 2  bx  c  0 or
                  ax 2  bx  c  0 and so on., a≠0 is called a quadratic inequality .
Example solve 4 x 2  8 x  5
Solution
4 x 2  8x  5
 4 x 2  8x  5  0
 4 x 2  10 x  2 x  5  0
 2 x(2 x  5)  1(2 x  5)  0
 (2 x  5)(2 x  1)  0
Method I Case approach
Case 1                                                        Case 2
2x  5  0  2x  1  0                                        2x  5  0  2x  1  0
2 x  5  2 x  1                                             2 x  5  2 x  1
   5    1                                                         5    1
x x                                                        x x
   2    2                                                         2    2
     1                                                              5
x                                                            x
     2                                                              2
                                                                                5         5
S.S1= x   :x 
                     1
                         =  1 ,  
                            2                              S.S2= x   :x    =   , 
                     2                                                        2         2
                               5 1 
S.S= S.S1  S.S2=   ,    ,  
                               2 2 
Method II Sign Chart
(2 x  5)(2 x  1)  0
                     5         5       5          1     1           1
             x                          x               x
           2       2       2        2        2
                                                               2

2x  5             ----------0 +++++++++++++++++++++++++++++++++
2x  1            -------------------------------------0++++++++++
2 x  52 x  1 ++++++++0--------------------------0++++++++++
                                                                 5 1 
Therefore the solution set of (2 x  5)(2 x  1)  0 is =   ,    ,   and
                                                                 2 2 
                                                 5 1
The solution set of (2 x  5)(2 x  1)  0 is =  , 
                                                 2 2
Ex. Solve
1.x 2  7 x  12  0
2.x  2  3x  1
         2             2




                                                      Page 6 of 8           Prepared By: Mohammed.M
Admas University College Hargeisa Main Campus


       Admas University College Hargeisa Main Campus
                               Qunt 101 Worksheet II
   1. Solve
                                                         x2
         a.
               1
                 4 x  5  1 2 x  5  2         k.        0
               9             5                           x 1
                3x  2 y  7                             2 x
         b. {                                         l.        0
                5x  y  3                                x 1
         c. 2x-y=1, x + y=2                          m. x 2  3x  2  0
         d. 5x +2y=8, 9x-5y =23                      n. x 2  25  10 x
         e. 4x – y + 11=0, 24x-6y +2= 0               o. 3x 2  2 x  1  0
         f. 25  x   2 x  5                      p. x  34  x   0
         g. 3x  6  23  x   5                   q. 2 x  4x  6  0
         h. 2(x + 1)-2(1- x)≤ 0
         i. 2x  4  5x  3
               x2
         j.           0
                x 1
   2. Solve each of the following equation by factorization.
          a..x 2  15  26  0
            b..x 2  x  72  0
            c..4 x 2  10 x  6  0
   3. Solve the following quadratic equations by using Solution formula( or using completing square
                                    1
                                a. x 2  15 x  26  0
                                    2
                                b.4 x 2  10 x  6  0
       method)
                                  1
                               c. x 2  8 x  3  0
                                  2
                               d .5 x 2  17 x  18  0
   4. Find the value of k so that there is only one solution
              a.5 x 2  kx  5  0
                 b.x 2  kx  25  0
                 c.x 2  3kx  20  0
              d .5kx2  2kx  20  0
   5. For each of the following quadratic equations. Determine k so that the equation
      has (i) exactly one root, (ii) exactly two roots (iii) no roots
         a. 2 x 2  5 x  k  0
         b. 4 x 2  (k  6) x  9  0
         c. 2  k x 2  2kx  1  0




                                           Page 7 of 8               Prepared By: Mohammed.M
Admas University College Hargeisa Main Campus


   6. If one root of x 2  3x  k  0 is 1.Find the other root and the value of k.
   7. Find two consecutive even integers whose product is 624
   8. The sum of the digits of certain two digit number is 9. If the order of the digits is
       reversed. The number is increased by 45. What is the number?
   9. The area of a rectangle is 64 cm2. Its perimeter is 40 cm. Find the length and
       width.
   10. If three times a number is decreased by, the result is 16. Find the number
   11. A man is 25 years older than his son. Ten years ago, he was twice as old as his
       son was. What are their present ages?
   12. Solve the following equation
            a. x 2  y 2  29 ,. x  y  3
            b. x  y  30,.xy  216
            c. x  y  18,.xy  136
            d. 4 x 4  5x 2  1  0
            e. x 4  11x 2  28  0




                                       Page 8 of 8            Prepared By: Mohammed.M

				
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