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 lines3 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