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Noor Language Schools Prep(3) Algebra Unit(1) Equations

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**Solving two equations of first degree in two variables**

[1] Graphically [2]Algebraically

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

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

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

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**Ex(1) find the s.s graphically**

3x +4y = , x + 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 = L2: 2x + y -4=0 9 5 1 X -4 -1 2 y 3 2 1 X -2 y

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**Step(2) graph the two lines**

3 2 1 -1 -2 -3 -4 4 X y y\ 5 -5 6 7 8 9 L1: 3x +4y = 11 9 5 1 X -4 -1 2 Y The s.s = {(1,2)} L2: 2x + y -4=0 3 2 1 X -2 y

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**The number of solutions is zero**

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**The S.S. in R2= {(X , Y) : y = 2x – 4, (x , y) R2 }**

Example (3): Find the S.S. of the following two equations graphically: L1: y = 2x – 4 L2: 4x = 2y + 8. Solution: L1: y = 2x –5 X 1 2 Y -4 -2 3 2 1 -1 -2 -3 -4 4 y y\ 5 -5 L2: 4x = 2y + 8. X 2 3 1 Y -2 X\ X The S.S. in R2= {(X , Y) : y = 2x – 4, (x , y) R2 } L1 and L2 are coincident

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

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**A) Substitution method**

Ex(1) find the solution set for the following pair of equations a lgebriac 2x – y = (1) , x + 3y + 1 = (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= sub in y = 2x -5 Then y = 2(2) -5 = -1 Then the s.s = {(2,-1)}

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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 y = 15 x 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 = 14 X = 2 Sub in eq (1) 2(2) – y = 5 Y = -1 S.S = {(2 , -1 )}

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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 = L2: 4x = -6y +10 Answer L1 : 2x + 3y = L2: 4x +6y =10 2:4 = 1 : , 3 : 6 = 1 : , 5 : 10 = 1 : 2 then L1 , L2 are coincident then the number of solutions is infinite

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**5 : 7 does not equal 1 : 2 then L1 , L2 are parallel**

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 = L2: 4x = -6y +7 Answer L1 : 2x + 3y = L2: 4x +6y =7 2:4 = 1 : , 3 : 6 = 1 : 2 5 : 7 does not equal 1 : 2 then L1 , L2 are parallel then the number of solutions is zero

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**H.W 2) sheet( 1) from the booklet**

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

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