Download presentation

Published byMelany Bloomfield Modified over 3 years ago

1
Noor Language Schools Prep(3) Algebra 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 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 = ⍉

5
**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

6
**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

7
**The number of solutions is zero**

8
**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

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

10
**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)}

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 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 )}

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 = 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

13
**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

14
**H.W 2) sheet( 1) from the booklet**

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

Similar presentations

OK

Chapter 7.3. Objective NCSCOS 4.03 Students will know how to solve a system of equations using addition.

Chapter 7.3. Objective NCSCOS 4.03 Students will know how to solve a system of equations using addition.

© 2018 SlidePlayer.com Inc.

All rights reserved.

To make this website work, we log user data and share it with processors. To use this website, you must agree to our Privacy Policy, including cookie policy.

Ads by Google