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PROPERTIES OF
   STRAIGHT LINES
A. WRITING EQUATION OF A LINE :
  1. GIVEN TWO POINTS.
  2. GIVEN THE SLOPE AND A POINT
B. PARALLEL AND PERPENDICULAR
   LINES.
Writing Equations of Lines
Equations of lines come in several different forms.
Two of those are:

2. Slope-intercept form    y = mx + b
       where m is the slope and b is the y-intercept

 2. General form
                     Ax + By + C = 0
  Answers will be written in either of these two
                   forms only
Writing Equations of Lines
A. Given a Point and a Slope
           Find the equation of the line that goes
     through the point (4, 5) and has a slope of 2.
Solution: m = 2 x1 = 4 y1 = 5


y − y1 = m( x − x1 )   Substitute the above given to point-
                       slope form equation a of line.

 y − 5 = 2( x − 4)      Simplify
Slope-intercept form     General Form
  y − 5 = 2( x − 4)      y − 5 = 2( x − 4)

   y = 2x − 8 + 5         y − 5 = 2x − 8

    y = 2x − 3           −2 x + y + 3 = 0

                          2x − y − 3 = 0


                 FINAL ANSWER
Find the equation of the line that
             goes through the point (-3, 2) and
             has a slope of -4/5.

Solution: m = -4/5     x1 = -3   y1 = 2

y − y1 = m( x − x1 )   Substitute the above given to point-
                       slope form equation a of line.
         4
y − 2 = − ( x − −3)
         5
Slope-intercept form       General Form
           4                     4   2
 y − 2 = − [ x − (−3)]      y = − x−
           5                     5   5
          4                 5 y = −4 x − 2
y − 2 = − ( x + 3)
          5
          4 12              4x + 5 y + 2 = 0
y−2= − x−
          5     5
        4   2
   y = − x−
        5   5
                         FINAL ANSWER
Writing Equations of Lines
B. Given Two Points
               Find the equation of the line that passes
               through the points (-2, 3) and (1, -6).
Solution:   x1 = -2 y1 = 3   x2 = 1   y2 = -6
    y2 − y1        Substitute the above given to slope
 m=                formula to find the slope.
    x2 − x1
     −6 − 3              −9
  m=                  m=              m = −3
     1+ 2                3
Slope-intercept form       General Form
 y − y1 = m( x − x1 )        y = −3x − 3
y − 3 = −3[ x − (−2)]
y − 3 = −3( x + 2)          3x + y + 3 = 0
 y − 3 = −3x − 6
  y = −3x − 3


                        FINAL ANSWER
Definitions
Perpendicular Two lines that makes a 90° angle.
    Lines     The slopes of perpendicular lines
              are negative reciprocal of each
              other .

Parallel Lines   Lines that never meet .
                 The slopes of parallel lines are
                 the same.
Writing Equations of Lines
• Given a point and equation of a line parallel
  to it.
             A Find the equation of the line that passes
              through (1, -5) and is parallel to 4x – 2y =3.

 Solution:   x1 = 1   y1 = -5
  4x - 2y =3.    Rewrite the equation to slope- intercept
                 form to get the slope.
- 2y =-4x +3.
     -4     3                  3
 y= x+ .                y =2x - .           m=2
     -2    -2                  2
Slope-intercept form       General Form
 y − y1 = m( x − x1 )       y = 2x − 7
y − (−5) = 2( x − 1)
 y + 5 = 2x − 2             2x − y − 7 = 0
 y = 2x − 2 − 5

   y = 2x − 7


                        FINAL ANSWER
Writing Equations of Lines
 • Given a point and equation of a line
   perpendicular to it.
          B   Find the equation of the line that passes
     through (1, -5) and is perpendicular to 4x – 2y =3.

 Solution: x1 = 1   y1 = -5
  4x - 2y =3.   Rewrite the equation to slope- intercept
                form to get the slope.
- 2y =-4x +3.
     -4     3                3
 y= x+ .              y =2x - .            m=2
     -2    -2                2
                                           m = -½
Slope-intercept form       General Form
  y − y1 = m( x − x1 )            1   9
                             y = − x−
                                  2   2
            1
y − (−5) = − ( x − 1)
            2                 2y = −x − 9
            1     1          x + 2y + 9 = 0
   y+5= − x+
            2     2

         1   9
    y = − x−
         2   2           FINAL ANSWER
LINEAR EQUATION WITH TWO
        VARIABLES
SOLVING SYSTEM OF EQUATION BY:
             2.Graph
          3.Substitution
          4.Elimination
         5.Cramer’s Rule
Solving Systems of Linear Equations
  For two-variable systems, there are then three
  possible types of solutions:

                                       Properties
A. Independent system:
   one solution and       1.   two distinct non-parallel lines
 one intersection point
                          2.   cross at exactly one point
                          3.   "independent" system
                          4.   one solution at (x,y )point.
Solving Systems of Linear Equations

 B. Inconsistent system:            Properties
     no solution and
  no intersection point.   1. two distinct parallel lines
                           2. never cross
                           3. No point of intersection
                           4. "inconsistent" system
                           5. no solution.
Solving Systems of Linear Equations
                                    Properties
 C. Dependent system:
 infinitely many solution
                            1. only one line.
                            2. same line drawn twice.
                            3. "intersect" at every point
                            4. "dependent" system,
                            5. Infinitely many solutions.
Methods of Solving
      Systems
of Linear Equations
A.        Systems of Linear Equations:
              Solving by Graphing

               Solve the following system by graphing.
               2x – 3y = –2
               4x + y = 24
      Solve for y for each equation
     Equation 1                       Equation 2
     2x – 3y = –2
     2x + 2 = 3y                      4x + y = 24

     y = (2/3)x + (2/3)               y = –4x + 24
Get the ( x, y) values for both equation to facilitate easy
           graphing. The table below shows it

  x           y = (2/3)x + (2/3)           y = –4x + 24

  –4       –8/3 + 2/3 = –6/3 = –2          16 + 24 = 40

  –1            –2/3 + 2/3 = 0             4 + 24 = 28

  2          4/3 + 2/3 = 6/3 = 2           –8 + 24 = 16

  5         10/3 + 2/3 = 12/3 = 4          –20 + 24 = 4

  8         16/3 + 2/3 = 18/3 = 6         –32 + 24 = –8
Using the table of values we can now graph
       and look for the intersection:
    y=
    (2/3
     )x +
         (2/3




                           24
                        x+      solution: (x, y) = (5, 4)
           )




                   –4
                y=
B.          Systems of Linear Equations:
               Solving by Substitution

                Solve the following system by substitution.
                2x – 3y = –2
                4x + y = 24
Solution:
 4x + y = 24             solve the second equation for y:
 y = –4x + 24
2x – 3(–4x + 24) = –2    substitute it for "y" in
2x + 12x – 72 = –2       the first equation
14x = 70
x=5                      solve for x
Equation 1                              Equation 2
    x=5                                    x=5
4x + y = 24     plug this x-value back   2x – 3y = –2
                 into either equation,
4( 5 ) + y = 24     and solve for y      2( 5 ) – 3y = –2
20 + y = 24                              10 – 3y = –2
y = 24 - 20                              - 3y = –2 - 10
                                         - 3y = - 12
 y=4
                                            y=4


  Then the solution is ( x, y ) = (5, 4).
C.         Systems of Linear Equations:
              Solving by Elimination
               Solve the following system using elimination.
               2x + y = 9
               3x – y = 16
 Solution:

 2x + y = 9         add down, the y's will cancel out
 3x – y = 16
 5x = 25
     x=5            divide through to solve for x
Equation 1                               Equation 2
    x=5                                     x=5
  2x + y = 9      using either of the     3x – y = 16
                 original equations, to
2( 5 ) + y = 9       find the value       3( 5 ) – y = 16
10 + y = 9                 of y           15 – y = 16
y = 9 - 10                                - y = 16 - 15
                                          -y= 1
 y = -1
                                             y = -1


   Then the solution is ( x, y ) = (5, -1).
D.          Systems of Linear Equations:
              Solving by Cramer’s Rule
             Solve the following system using cramer’s rule.
             2x – 3y = –2
             4x + y = 24

Solution:             Nx        Ny
                 x=       ,y=
                      D          D
D - determinant of the coefficient of the variables
Nx - determinant taken from D replacing the coefficient of x
Ny - and y by their corresponding constant terms leaving all
     other terms unchanged
2   −3
D=              D = (2) −( −12) =14
     4    1


     −2 −3       N x = (−2) − (−72) = 70
Nx =                           N x 70
     24 1                   x=    =    =5
                               D    14

                N y = (48) − (−8) = 56
         2 −2
Ny =                   Ny 56
         4 24       y=  =    =4
                       D 14
  FINAL                ( 5, 4 )
  ANSWER
ANSWER THE FOLLOWING PROBLEMS
1. (a) Explain why the simultaneous equations 8x –
   4y = 20 and y =2x – 3 have no solution . What can
   you say about the straight lines representing these
   two equations?
               They are parallel

•    The diagram shows the graph of 2y = x - 2.
     The values of a and b are respectively.

     2 and -1
•   The graphs of x - 2y - 3 = 0 and 6 + 4y - 2x = 0
             are identical lines

•   Find the graph of y = -2x - 1?




5. The diagram shows the graph of y = ax + b. Find the
    values of a and b.

        a = 2, b = 2
Find the solution for each system of equation
using any method:

 1. 5x – 2y = 0              3.    y=x+3
     4x + y = 13                  5y + 6x = 15
       Solution :                    Solution :
    ( x , y) = ( 2, 5 )           ( x , y) = ( 0, 3 )

 2. 5y = 6x – 3              4. 1    1
                                  x + y =1
     2y = x – 4                 3    3
                                1
         Solution :               y =1 − x
     ( x , y) = ( -2, -3 )      2
                                            Solution :
                                        ( x , y) = ( -1, 4 )

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Solving Systems of Linear Equations Graphically

  • 1. PROPERTIES OF STRAIGHT LINES A. WRITING EQUATION OF A LINE : 1. GIVEN TWO POINTS. 2. GIVEN THE SLOPE AND A POINT B. PARALLEL AND PERPENDICULAR LINES.
  • 2. Writing Equations of Lines Equations of lines come in several different forms. Two of those are: 2. Slope-intercept form y = mx + b where m is the slope and b is the y-intercept 2. General form Ax + By + C = 0 Answers will be written in either of these two forms only
  • 3. Writing Equations of Lines A. Given a Point and a Slope Find the equation of the line that goes through the point (4, 5) and has a slope of 2. Solution: m = 2 x1 = 4 y1 = 5 y − y1 = m( x − x1 ) Substitute the above given to point- slope form equation a of line. y − 5 = 2( x − 4) Simplify
  • 4. Slope-intercept form General Form y − 5 = 2( x − 4) y − 5 = 2( x − 4) y = 2x − 8 + 5 y − 5 = 2x − 8 y = 2x − 3 −2 x + y + 3 = 0 2x − y − 3 = 0 FINAL ANSWER
  • 5. Find the equation of the line that goes through the point (-3, 2) and has a slope of -4/5. Solution: m = -4/5 x1 = -3 y1 = 2 y − y1 = m( x − x1 ) Substitute the above given to point- slope form equation a of line. 4 y − 2 = − ( x − −3) 5
  • 6. Slope-intercept form General Form 4 4 2 y − 2 = − [ x − (−3)] y = − x− 5 5 5 4 5 y = −4 x − 2 y − 2 = − ( x + 3) 5 4 12 4x + 5 y + 2 = 0 y−2= − x− 5 5 4 2 y = − x− 5 5 FINAL ANSWER
  • 7. Writing Equations of Lines B. Given Two Points Find the equation of the line that passes through the points (-2, 3) and (1, -6). Solution: x1 = -2 y1 = 3 x2 = 1 y2 = -6 y2 − y1 Substitute the above given to slope m= formula to find the slope. x2 − x1 −6 − 3 −9 m= m= m = −3 1+ 2 3
  • 8. Slope-intercept form General Form y − y1 = m( x − x1 ) y = −3x − 3 y − 3 = −3[ x − (−2)] y − 3 = −3( x + 2) 3x + y + 3 = 0 y − 3 = −3x − 6 y = −3x − 3 FINAL ANSWER
  • 9. Definitions Perpendicular Two lines that makes a 90° angle. Lines The slopes of perpendicular lines are negative reciprocal of each other . Parallel Lines Lines that never meet . The slopes of parallel lines are the same.
  • 10. Writing Equations of Lines • Given a point and equation of a line parallel to it. A Find the equation of the line that passes through (1, -5) and is parallel to 4x – 2y =3. Solution: x1 = 1 y1 = -5 4x - 2y =3. Rewrite the equation to slope- intercept form to get the slope. - 2y =-4x +3. -4 3 3 y= x+ . y =2x - . m=2 -2 -2 2
  • 11. Slope-intercept form General Form y − y1 = m( x − x1 ) y = 2x − 7 y − (−5) = 2( x − 1) y + 5 = 2x − 2 2x − y − 7 = 0 y = 2x − 2 − 5 y = 2x − 7 FINAL ANSWER
  • 12. Writing Equations of Lines • Given a point and equation of a line perpendicular to it. B Find the equation of the line that passes through (1, -5) and is perpendicular to 4x – 2y =3. Solution: x1 = 1 y1 = -5 4x - 2y =3. Rewrite the equation to slope- intercept form to get the slope. - 2y =-4x +3. -4 3 3 y= x+ . y =2x - . m=2 -2 -2 2 m = -½
  • 13. Slope-intercept form General Form y − y1 = m( x − x1 ) 1 9 y = − x− 2 2 1 y − (−5) = − ( x − 1) 2 2y = −x − 9 1 1 x + 2y + 9 = 0 y+5= − x+ 2 2 1 9 y = − x− 2 2 FINAL ANSWER
  • 14. LINEAR EQUATION WITH TWO VARIABLES SOLVING SYSTEM OF EQUATION BY: 2.Graph 3.Substitution 4.Elimination 5.Cramer’s Rule
  • 15. Solving Systems of Linear Equations For two-variable systems, there are then three possible types of solutions: Properties A. Independent system: one solution and 1. two distinct non-parallel lines one intersection point 2. cross at exactly one point 3. "independent" system 4. one solution at (x,y )point.
  • 16. Solving Systems of Linear Equations B. Inconsistent system: Properties no solution and no intersection point. 1. two distinct parallel lines 2. never cross 3. No point of intersection 4. "inconsistent" system 5. no solution.
  • 17. Solving Systems of Linear Equations Properties C. Dependent system: infinitely many solution 1. only one line. 2. same line drawn twice. 3. "intersect" at every point 4. "dependent" system, 5. Infinitely many solutions.
  • 18. Methods of Solving Systems of Linear Equations
  • 19. A. Systems of Linear Equations: Solving by Graphing Solve the following system by graphing. 2x – 3y = –2 4x + y = 24 Solve for y for each equation Equation 1 Equation 2 2x – 3y = –2 2x + 2 = 3y 4x + y = 24 y = (2/3)x + (2/3) y = –4x + 24
  • 20. Get the ( x, y) values for both equation to facilitate easy graphing. The table below shows it x y = (2/3)x + (2/3) y = –4x + 24 –4 –8/3 + 2/3 = –6/3 = –2 16 + 24 = 40 –1 –2/3 + 2/3 = 0 4 + 24 = 28 2 4/3 + 2/3 = 6/3 = 2 –8 + 24 = 16 5 10/3 + 2/3 = 12/3 = 4 –20 + 24 = 4 8 16/3 + 2/3 = 18/3 = 6 –32 + 24 = –8
  • 21. Using the table of values we can now graph and look for the intersection: y= (2/3 )x + (2/3 24 x+ solution: (x, y) = (5, 4) ) –4 y=
  • 22. B. Systems of Linear Equations: Solving by Substitution Solve the following system by substitution. 2x – 3y = –2 4x + y = 24 Solution: 4x + y = 24 solve the second equation for y: y = –4x + 24 2x – 3(–4x + 24) = –2 substitute it for "y" in 2x + 12x – 72 = –2 the first equation 14x = 70 x=5 solve for x
  • 23. Equation 1 Equation 2 x=5 x=5 4x + y = 24 plug this x-value back 2x – 3y = –2 into either equation, 4( 5 ) + y = 24 and solve for y 2( 5 ) – 3y = –2 20 + y = 24 10 – 3y = –2 y = 24 - 20 - 3y = –2 - 10 - 3y = - 12 y=4 y=4 Then the solution is ( x, y ) = (5, 4).
  • 24. C. Systems of Linear Equations: Solving by Elimination Solve the following system using elimination. 2x + y = 9 3x – y = 16 Solution: 2x + y = 9 add down, the y's will cancel out 3x – y = 16 5x = 25 x=5 divide through to solve for x
  • 25. Equation 1 Equation 2 x=5 x=5 2x + y = 9 using either of the 3x – y = 16 original equations, to 2( 5 ) + y = 9 find the value 3( 5 ) – y = 16 10 + y = 9 of y 15 – y = 16 y = 9 - 10 - y = 16 - 15 -y= 1 y = -1 y = -1 Then the solution is ( x, y ) = (5, -1).
  • 26. D. Systems of Linear Equations: Solving by Cramer’s Rule Solve the following system using cramer’s rule. 2x – 3y = –2 4x + y = 24 Solution: Nx Ny x= ,y= D D D - determinant of the coefficient of the variables Nx - determinant taken from D replacing the coefficient of x Ny - and y by their corresponding constant terms leaving all other terms unchanged
  • 27. 2 −3 D= D = (2) −( −12) =14 4 1 −2 −3 N x = (−2) − (−72) = 70 Nx = N x 70 24 1 x= = =5 D 14 N y = (48) − (−8) = 56 2 −2 Ny = Ny 56 4 24 y= = =4 D 14 FINAL ( 5, 4 ) ANSWER
  • 28. ANSWER THE FOLLOWING PROBLEMS 1. (a) Explain why the simultaneous equations 8x – 4y = 20 and y =2x – 3 have no solution . What can you say about the straight lines representing these two equations? They are parallel • The diagram shows the graph of 2y = x - 2. The values of a and b are respectively. 2 and -1
  • 29. The graphs of x - 2y - 3 = 0 and 6 + 4y - 2x = 0 are identical lines • Find the graph of y = -2x - 1? 5. The diagram shows the graph of y = ax + b. Find the values of a and b. a = 2, b = 2
  • 30. Find the solution for each system of equation using any method: 1. 5x – 2y = 0 3. y=x+3 4x + y = 13 5y + 6x = 15 Solution : Solution : ( x , y) = ( 2, 5 ) ( x , y) = ( 0, 3 ) 2. 5y = 6x – 3 4. 1 1 x + y =1 2y = x – 4 3 3 1 Solution : y =1 − x ( x , y) = ( -2, -3 ) 2 Solution : ( x , y) = ( -1, 4 )