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Chapter one Linear Equations

Chapter one Linear Equations. Linear Equations. Definition of a Linear Equation A linear equation in one variable x is an equation that can be written in the form ax + b = 0 , where a and b are real numbers and a is not equal to 0.

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Chapter one Linear Equations

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  1. Chapter oneLinear Equations

  2. Linear Equations Definition of a Linear Equation A linear equation in one variable x is an equation that can be written in the form ax + b = 0, where a and b are real numbers and a is not equal to 0. An example of a linear equation in x is 4x + 2 = 6. Linear equations in x are first degree equations in the variable x.

  3. LINEAR EQUATION IN TWO VARIABLES

  4. LINEAR EQUATION IN TWO VARIABLES Definition of an equation: an equation is any expression containing an “equals” sign. Consider the following equation involving the two quantities, x and y: y = -3 + 2x The equation can easily be used to work out the value of y for any given value of x. All you need to do is insert your given value for x, and then evaluate the expression on the right hand side of the equation – this will be the corresponding value of y.

  5. LINEAR EQUATION IN TWO VARIABLES Repeating this for other values of x we get:

  6. LINEAR EQUATION IN TWO VARIABLES Presenting the concept of the equation of a straight line To plot y against x on a graph as illustrated in Figure by plotting the points given by the pairs of co-ordinates (e.g. (-2, -7) and (4,5)) in the table above: 1 2 3 4 5 6 7 -8 -7 -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 7 8 -7 -6 -5 -4 -3 -2 -1

  7. System of equations A pair of linear equations in two variables is said to form a system of simultaneous linear equations. For Example, 2x – 3y + 4 = 0 x + 7y – 1 = 0 Form a system of two linear equations in variables x and y.

  8. general form of a linear equation in two variables • The general form of a linear equation in two variables x and y is: ax + by + c = 0 , a not = 0, b not=0, where a, b and c being real numbers. • A solution of such an equation is a pair of values, one for x and the other for y, which makes two sides of the equation equal. • Every linear equation in two variables has infinitely many solutions which can be represented on a certain line.

  9. GRAPHICAL SOLUTIONS OF A LINEAR EQUATION • Let us consider the following system of two simultaneous linear equations in two variable. • 2x – y = -1 • 3x + 2y = 9 Here we assign any value to one of the two variables and then determine the value of the other variable from the given equation.

  10. GRAPHICAL SOLUTIONS OF A LINEAR EQUATION X 0 2 Y 1 5 For the equation 2x –y = -1 ---(1) 2x +1 = y Y = 2x + 1 3x + 2y = 9 --- (2) 2y = 9 – 3x 9- 3x Y = ------- 2 X 3 -1 Y 0 6

  11. GRAPHICAL SOLUTIONS OF A LINEAR EQUATION Y (-1,6) (2,5) (0,3) (0,1) X X’ X= 1 Y=3 Y’

  12. ALGEBRAIC METHODS OF SOLVING SIMULTANEOUS LINEAR EQUATIONS The most commonly used algebraic methods of solving simultaneous linear equations in two variables are • Method of elimination by substitution • Method of elimination by equating the coefficient • Method of Cross- multiplication

  13. ELIMINATION BY SUBSTITUTION Let us take an example x + 2y = -1 ------------------ (i) 2x – 3y = 12 -----------------(ii)

  14. SUBSTITUTION METHOD x + 2y = -1 x = -2y -1 ------- (iii) Substituting the value of x in equation (ii), we get: 2x – 3y = 12 2 ( -2y – 1) – 3y = 12 - 4y – 2 – 3y = 12 - 7y = 14 , y = -2 ,

  15. SUBSTITUTION METHOD Putting the value of y in eq (iii), we get x = - 2y -1 x = - 2 x (-2) – 1 = 4 – 1 = 3 Hence the solution of the equation is ( 3, - 2 )

  16. ELIMINATION METHOD • Example: we want to solve, 3x + 2y = 11 2x + 3y = 4

  17. ELIMINATION METHOD Let 3x + 2y = 11 --------- (i) 2x + 3y = 4 ---------(ii) Multiply 3 in equation (i) and 2 in equation (ii) and subtracting eq iv from iii, we get 9x + 6y = 33 ------ (iii) 4x + 6y = 8 ------- (iv) 5x = 25 => x = 5

  18. ELIMINATION METHOD • putting the value of y in equation (ii) we get, 2x + 3y = 4 2 x 5 + 3y = 4 10 + 3y = 4 3y = 4 – 10 3y = - 6 y = - 2 Hence, x = 5 and y = -2

  19. Graphing Linear Equations in Two Variables

  20. The Rectangular Coordinate System y-axis 2nd quadrant 1st quadrant x-axis 3rd quadrant 4th quadrant EXAMPLE Plot the points (3,2) and (-2,-4). SOLUTION (3,2) (-2,-4)

  21. The Graph of an Equation The graph of an equation in two variables is the set of points whose coordinates satisfy the equation. An ordered pair of real numbers (x,y) is said to satisfy the equation when substitution of the x and y coordinates into the equation makes it a true statement. For example, in the equation y = 2x + 6, the ordered pair (1,8) is a solution. When we substitute this point the sentence reads 8 = 8, which is true. The ordered pair (2,3) is not a solution. When we substitute this point, the sentence reads 3 = 10, which is not true.

  22. Solution of an Equation in Two Variables Example: Given the equation 2x + 3y = 18, determine if the ordered pair (3, 4) is a solution to the equation. We substitute 3 for x and 4 for y. 2(3) + 3 (4) ? 18 6 + 12 ? 18 18 = 18 True. Therefore, the ordered pair (3, 4) is a solution to the equation 2x + 3y = 18.

  23. Finding Solutions of an Equation Find five solutions to the equation y = 3x + 1. Start by choosing some x values and then computing the corresponding y values. If x = -2, y = 3(-2) + 1 = -5. Ordered pair (-2, -5) If x = -1, y = 3(-1) + 1 = -2. Ordered pair ( -1, -2) If x =0, y = 3(0) + 1 = 1. Ordered pair (0, 1) If x =1, y = 3(1) + 1 =4. Ordered pair (1, 4) If x =2, y = 3(2) + 1 =7. Ordered pair (2, 7)

  24. Graph of the Equation Plot the five ordered pairs to obtain the graph of y = 3x + 1 (2,7) (1,4) (0,1) (-1,-2) (-2,-5)

  25. Linear equations in two variables y This is the graph of the equation 2x + 3y = 12. (0,4) (6,0) x 2 -2 Equations of the form ax + by = c are called linear equations in two variables. The point (0,4) is the y-intercept. The point (6,0) is the x-intercept.

  26. Slope of a line

  27. Slope of a line y 1 1 m = 2 4 x 2 -2 m = - The slope of a line is a number, m, which measures its steepness. m is undefined m = 2 m = 0

  28. Slope of a line y2 – y1 , (x1 ≠ x2). m = x2 – x1 (x2, y2) y y2–y1 change in y (x1, y1) x2–x1 change in x x The slope of the line passing through the two points (x1, y1) and (x2, y2) is given by the formula The slope is the change in y divided by the change in x as we move along the line from (x1, y1) to (x2, y2).

  29. Slopeformula y2 – y1 5 – 3 2 1 m = = = = x2 – x1 2 4 – 2 y x Example: Find the slope of the line passing through the points (2,3) and (4,5). Use the slope formula with x1= 2, y1 = 3, x2 = 4, and y2 = 5. (4, 5) 2 (2, 3) 2

  30. Slope-intercept form y x change in y 2 m = = 1 change in x (0,-4) (1, -2) Example: Graph the line y = 2x– 4. • The equation y = 2x– 4 is in the slope-intercept form. So, m = 2 and b = -4. 2. Plot the y-intercept, (0,-4). 3. The slope is 2. 2 4. Start at the point (0,4). Count 1 unit to the right and 2 units up to locate a second point on the line. 1 The point (1,-2) is also on the line. 5. Draw the line through (0,4) and (1,-2).

  31. Slope-intercept form 1 1 1 2 2 2 Example:The graph of the equation y – 3 = -(x – 4) is a line of slope m = - passing through the point (4,3). y m = - 8 (4, 3) 4 x 4 8 A linear equation written in the form y–y1 = m(x – x1) is in point-slope form. The graph of this equation is a line with slope m passing through the point (x1, y1).

  32. Slope-intercept form Point-slope form y – y1 = m(x – x1) y – y1 = 3(x – x1) Let m = 3. y – 5 = 3(x – (-2)) Let (x1, y1)= (-2,5). y – 5 = 3(x + 2) Simplify. y = 3x + 11 Slope-intercept form Example: Write the slope-intercept form for the equation of the line through the point (-2,5) with a slope of 3. Use the point-slope form, y – y1 = m(x – x1), with m = 3 and (x1, y1)= (-2,5).

  33. Slope-intercept form m = Calculate the slope. 1 1 3 3 Point-slope form y – y1 = m(x – x1) 5 – 3 2 1 = - = - Use m = -and the point (4,3). -2 – 4 6 3 y – 3 = - (x – 4) 13 Slope-intercept form 3 y = - x + Example: Write the slope-intercept form for the equation of the line through the points (4,3) and (-2,5).

  34. Slope-intercept form y (0, 4) y = 2x + 4 x y = 2x– 3 (0, -3) Two lines are parallel if they have the same slope. If the lines have slopes m1 and m2, then the lines are parallel whenever m1 = m2. Example: The lines y = 2x – 3 and y = 2x + 4 have slopes m1 = 2 and m2 = 2. The lines are parallel.

  35. Slope-intercept form y 1 1 1 1 m2= - 3 3 3 m1 x Example: The lines y = 3x – 1 and y = - x + 4 have slopes m1 = 3 and m2 = - . y = - x + 4 Two lines are perpendicular if their slopes are negative reciprocals of each other. If two lines have slopes m1 and m2, then the lines are perpendicular whenever or m1m2 = -1. y = 3x– 1 (0, 4) (0, -1) The lines are perpendicular. Reciprocals = مقلوب

  36. Example C y x A certain factory has daily fixed overhead expenses of $2000, while each item produced Costs $100. Find an equation that relates the daily cost C to the number X of items Produced each day. Solution: The fixed overhead expense of $2000 represents the fixed cost, the cost incurred no matter how many items are produced. Since each item produced costs $100, the variable cost of producing X items is 100X. Thus the total daily cost C of production is C = 100x + 2000 The graph of this equation is given by the line Shown. Notice that the fixed cost $2000 is Represented by the y-intercept, while the $100 Cost of producing each item is the slope. Also Notice that a different scale is used on each axis. 2200 2100 2000 Dollars 1 2 3 4 5 No. of items

  37. Solved problems 1- Check whether the graphs of the equations given below represent parallel lines. Explain your answer. line a:y = –5x + 3 line b:y + 5x = –2 Solution line a: y = –5x + 3 line b: y = –5x – 2 Identify the slope of each equation. For line a: y = –5x + 3, Slope = –5 For line b: y = –5x – 2 Slope = –5 Line aand line b have equal slope, –5. Parallel lines have equal slope. So, line a and line b are parallel lines.

  38. Solved problems 2- Check whether the graphs of the equations given below represent parallel lines. Explain your answer. line a:y = x + 7 line b:x – y = –2 Solution line a: y = x+ 7 line b: y = x + 2 Identify the slope of each equation. For line a: y = x + 7, Slope = 1 For line b: y = x + 2 Slope = 1 Line aand line b have equal slope that is 1. Parallel lines have equal slope. So, line a and line b are parallel lines.

  39. Solved problems 3- Find the slope of the line that passes through the points (-1 , 0) and (3 , 8). Solution : The slope m is given by m = (y2 - y1) / (x2 - x1) = (8 - 0) / (3 - -1) = 2 4- Find the slope of the line that passes through the points (2 , 0) and (2 , 4). Solution : The slope m is given by m = (y2 - y1) / (x2 - x1) = (4 - 0) / (2 - 2) = 4 / 0 Division by zero is not allowed in math. Therefore the slope of the line defined by the points (2 , 0) and (2 , 4) is undefined. The line through the points (2 , 0) and (2 , 4) is perpendicular to the x axis.

  40. Solved problems 5- Find the slope of the line that passes through the points (7 , 4) and (-9 , 4). Solution : The slope m is given by m = (y2 - y1) / (x2 - x1) = (4 - 4) / (-9 - 7) = 0 / -16 = 0 The line defined by the points (7 , 4) and (-9 , 4) is parallel to the x axis and its slope is equal to zero. 6- Are the three points A(2 , 3) , B(5 , 6) and C(0 , -2) collinear? Solution :We first find the slope defined by the points A(2 , 3) and B(5 , 6). m(AB) = (y2 - y1) / (x2 - x1) = (6 - 3) / (5 - 2) = 3 / 3 = 1 We next find the slope defined by the points B(5 , 6) and C(0 , -2). m(BC) = (y2 - y1) / (x2 - x1) = (-2 - 6) / (0 - 5) = -8 / -5 = 8 / 5 The two slopes are not equal therefore the points are not collinear.

  41. Solved problems 7- What is the slope of the line perpendicular to the line whose equation is given by - 2y = - 8 x + 9? Solution : The equation of the given line is - 2y = - 8 x + 9 Rewrite in slope intercept form. y = 4 x - 9/2 The slope of the given line is 4. Two perpendicular lines have slopes m1 and m2 related by: m1*m2 = -1 If we set m1 = 4 then m2, the slope of the line perpendicular to the given line, is equal to -1/4.

  42. Solved problems 8- Is the triangle whose vertices are the points A(0 , -1) , B(2 , 1) and C(-4 , 3) a right triangle? Solution : We first find the slope of the line defined by points A and B m(AB) = (y2 - y1) / (x2 - x1) = (1 - -1) / (2 - 0) = 1 We next find the slope of the line defined by points A and C m(AC) = (3 - -1) / (-4 - 0) = 4 / -4 = -1 The product of the slopes m(AB) and m(AC) is equal to -1 and this means that the lines defined by A,B and A,C are perpendicular and therefore the triangle whose vertices are the points A, B and C is a right triangle.

  43. Solved problems 9- What is the slope of the line -7y + 8x = 9 Solution : The given equation -7y + 8x = 9 Rewrite the equation in slope intercept form. -7y = -8x + 9 y = (8/7) x - 9/7 The slope of the given line is 8/7

  44. Solved problems 10- What is the slope of the line y = 9? Solution : The given equation y = 9 The given equation is a line parallel to the x axis therefore its slope is equal to 0. 11- What is the slope of the line x = -5? Solution : The given equation x = -5 The given equation is perpendicular to the x axis therefore its slope is undefined.

  45. Solved problems 12- Which of the following equations passes through the points (2,4) and (-3,-6)? Choose: y = (1/2)x - 2 y = 2xy = 2x + 4 y = (-1/2)x + 2 13- What is the slope of a line perpendicular to 2y = -6x - 10? Choose: -3 3 -1/3 1/3

  46. Solved problems 14- Given 3y - 4x = 5 and 4y + 6 = 3x.Are these lines parallel, perpendicular or neither? Choose : parallel perpendicular neither 15- What is the equation of a line that passes through the point (4,-5) and is parallel to 3x + 2y = 12? Choose : y = -3x + 6 y = (3/2)x + 1 y = 3x + 1 y = (-3/2)x + 1

  47. Solved problems 16- What is the equation of a line that passes through the point (-1,-2) and is perpendicular to -5x = 6y + 18? Choose : y = (6/5)x - (4/5)y = (-6/5)x + (6/5) y = (6/5)x + (4/5) y = (-6/5)x - (4/5) 17- Given 4y - 2x = 10 and -6y - 6 = -3x.Are these lines parallel, perpendicular or neither? Choose : parallel perpendicular neither

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