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Unit(1) Equations. Solving two equations of first degree in two variables [1] Graphically [2]Algebraically.

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Presentation on theme: "Unit(1) Equations. Solving two equations of first degree in two variables [1] Graphically [2]Algebraically."— Presentation transcript:

1 Unit(1) Equations

2 Solving two equations of first degree in two variables [1] Graphically [2]Algebraically

3 [1]Graphically Solving two equations graphically means the point of intersection between the two straight lines if this exist

4 parallel coincident s.s = {(0,-2)} s.s = ⍉ s.s = {(x,y) :x,y Є R} intersecting

5 Ex(1) find the s.s graphically a) 3x +4y = 11, 2x + y -4 =0 Answer Step(1): Draw a table for each equation take any values of x then subsututing in the equation to find y L1: 3x +4y = 11 L2: 2x + y -4=0 951X -42y 321X -202y

6 Step(2) graph the two lines X y y\y\ X 2Y 321X -202y L1: 3x +4y = 11 L2: 2x + y -4=0 The s.s = {(1,2)}

7 The number of solutions is zero

8 Example (3): Find the S.S. of the following two equations graphically: L 1 : y = 2x – 4 L 2 : 4x = 2y + 8. Solution: X012 Y-4-20 L 1 : y = 2x –5 L 2 : 4x = 2y + 8. X231 Y02-2 X\X\ X L 1 and L 2 are coincident y y\y\ The S.S. in R 2 = {(X, Y) : y = 2x – 4, (x, y)  R 2 }

9 [2] Algebraic Solving two equations algebraic means searching about the values of the variables

10 Ex(1) find the solution set for the following pair of equations a lgebriac 2x – y = 5 (1), x + 3y + 1 = 0 (2) A) Substitution method 2x – y = 5,y = 2x – 5 Substituting in the second equation: X + 3 (2x – 5) + 1 = 0 X + 6x – = 0 7x – 14 = 0   7x = 14 x=2 sub in y = 2x -5 Then y = 2(2) -5 = -1 Then the s.s = {(2,-1)}

11 ++ B]Omitting method 2x – y = 5, x + 3y + 1 = 0 Solution: 1)Multiply the the first equation by 3 to make the coff of y of the first is the same coff of y of the second 6x - 3 y = 15 x + 3 y = -1 2)If the coff of y is the same sign subtract the equations if the coff of y have different sign add the equations add 7x + 0 = 14 X = 2 Sub in eq (1) 2(2) – y = 5 Y = -1 S.S = {(2, -1 )}

12 Notes: To determine the number of solutions of two equations L1: a1 X+b1 Y = C1 L2: a2X +b2Y = C2 1 ) If a1 : a2 = b1 : b2 = C1 : C2 Then L1 and L2 are coincident then the number of solutions is infinite Ex L1 : 2x + 3y = 5 L2: 4x = -6y +10 Answer L1 : 2x + 3y = 5 L2: 4x +6y =10 2:4 = 1 :2, 3 : 6 = 1 : 2, 5 : 10 = 1 : 2 then L1, L2 are coincident then the number of solutions is infinite

13 2) If a1 : a2 = b1 : b2 does not equal C1:C2 then L1 and L 2 are parallel so there is no point of intersection then the number of solution is zero Ex L1 : 2x + 3y = 5 L2: 4x = -6y +7 Answer L1 : 2x + 3y = 5 L2: 4x +6y =7 2:4 = 1 :2, 3 : 6 = 1 : 2 5 : 7 does not equal 1 : 2 then L1, L2 are parallel then the number of solutions is zero

14 H.W 1)school book p(8) first, second and third 2) sheet( 1) from the booklet


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