2
,equations in 1 two equations
equivalent if they
are
have
said to
the
be
same
Valuable solution
↳
set
equation equivalent equations
•
is a statement saying that 3×+7 =
13
two expressions are equal 3x =
6
✗
=
2
examples check the solution
-
2x + 1 =
x -
7 3 (2) t 7 =
13
'
1
=
2 312) =
6
2- 2 4 -
Zz 2 =
2
.
V2 =
✗
✗ t 1 Y linear equation
y
•
in one variable is an equation
if an equation holds true for that can be written as
every permissible / real value
( in R ) ,
then the equation is ax 't b = 0 ,
a. b ER ,
a 1=0
called an
identity
2 an equation leading to a linear
↳ !z = is true for any 2=12
z y -
zz equation is one which upon ,
at
undefined algebraic manipulation ( transpose)
reduces the ax t b 0
form → =
if an
equation is never true for ↳
✗
any permissible / real value -5=2×+7 x 12=0 or
→ - -
( in R ) , then the equation is CELTALE ) ✗ t 12=0
called a contradiction
↳
how to solve linear equation
1 -
2x =3
-
2x is never true
of ✗ C- R 1 additive property ( transpose )
2
multiplicative property ( multiply
if an equation holds true for some both sides by a nonzero R )
value Is ,
then the equation is a ↳ or divide
conditional equation 3
simplify ( similar ) terms
↳ -8 size those with
2x t I =
x -
7 hold if ✗ =
* combine on one
variables and on the other those
a solution or root an equation those that don't
of
( in one variable ) is a value
of the variable that makes the example : 5×-5 = 2x t 7
equation true 5×-2 ✗ =
7+5
set:{
311 =
12
solution
a solution set of an equation 3
is the set
of all solutions of ✗
=
4
the equation
3×12 -
✗ ) t 7×1-3 = 5×-3×2 t 1
[ the solution sets that we con
-
-3×1 t 13×1-3 =
5×-3×2 t 1
set:{-
Sider are subsets of R ] 13×-511 =
1-3
solution
" 8x = -
z -1
real number ✗ =
8 4
?⃝
?⃝
,quadratic equation * works whether or not ax2 t
-
in ✗ is an equation which has bxtc is factorable
the general form axtbxtc =
0 ,
=/ 0 ↳ 6×2 11 10 0
where b. C ER and
-
=
a. a
-
* once a =
0 → linear equation a 6 b =
11 C = 10
-
= -
an
equation leading to a quadratic 11 In (-1112-41671-10)
✗ =
equation is one which upon , 2 (6)
algebraic manipulation ,
reduces
to the form × =
11 I - 121 -1 240
12
↳ 2
10 2
3×-10 0
✗ t 3✗ = → ✗ t =
11+-1361
'
2-
✗ =
✗ t = 0 → ✗ ✗ t 1 = 0 12
✗ ,
,
4
( transpose ✗ and multiply by ✗ -
1) ✗ =
11 t 19 ✗ = 11
-
19
12 12
how to solve quadratic equation
30 5 -8
-2-3
✗ = = ✗ =
=
2
1
factoring 12 12
so1uti0n-2/z
2
quadratic formula
[ via
factoring ] vet : { 512 ,
-
if ab = 0 .
'
then a / b = 0
2
↳ ✗ t 3✗ =
10
the discriminant
'
☐ of the quad
2
✗ t 3×-10 =
0
( ✗ + 5) ( x
-
2) =
0 ratio polynomial axztbxtc
a b comes from the quadratic formula
is
f
✗ + 5 = O x -
2 =
0 and equal to b2 -
4ac
✗ = -
g ✗ =
2
if distinct real solutions
set:{-5①
solution D > 0 →
two
if D= 0 →
One real solution
if ☐ so → two imaginary solutions
1- hare
conjugate of each other
-
↳ 6×2 -
11×-10 = 0
(3×+2712×-5)=0 how to folvefiatronal exp
a b
2×-5=0
1 LCD
/
3×1-2=0
311=-2 2x =
5
2 multiply both sides
3 2
3 simplify
✗ =
-2/3 ✗ = 5/2
set:{-2/z,5/z
✗ t 4 2×-3 3×-8
solution -
=
✗ , z X -
5 ✗ 2--3×-10
ts
↳ ( ✗ 1- 2) ( x -
5)
3) =/ Jcxtz
2x -
3×-8
) )
[ via
quadratic formula ] 1×+2 -
✗ -5
-
let a. with
b. C E R a =/ 0 (x -
5) ✗ +2 ✗ 2-3×-10 ( x
-
)
s
if 9×2 t b. ✗ t C =
0 ,
then
( ✗ 1- 4) ( x -
5) -
( 2x -
3) ( ✗ 1- 2) = 3×-8
-
b ± -1 b2 -
4ac
✗ =
2A
, rational expressions
2
✗ -
✗ -
20 -
2×2 -
✗ -16
2
-
✗ -
2×-14=3 X
-
8
✓ ✗
✓
f
2 2
-
✗
-
5×-6 = 0 ✗ X
✗
2
t 5×1-6 =
0 g
( ✗ 1- 3) ( ✗ t 2) =
0
2
a b 15 -2 ✗
/
=
✗
-
✗ 1- 3=0 ✗ 1- 2 = 0 ✗
2
t 2x -
15
=
0
✗ =-3 ✗ = -
2 ( ✗ t 5) ( x
-
3) =
0
3 0
5 Of
=
5 ✗
-
↳ extraneous ✗ t =
-
will make the denominator ✗ = - ✗ =3
°%e¥°?{-g,z
=
01 undefined
set:{3①
solution
- -
↳ 2x t 3 1
}
✗ t in
,
= - quadratic formula ,
4 2
the ± means at
solution
( ) :( ;-)
Most 2 real
2×+3 ✗ t 1 - ,
12 ,
-
yz
4 z
3×2 -
✗ 1- 5 =
✗
3 (2×1-3)
-
6( 1) ✗ 1- =
-
4 3×2 -
x -
✗ t 5 =
0
(6×+9) -
(6×+6) = -4 3×2 -
2×1-5 =
0
6×1-9 -
6×-6 =
-4 a b C
0 = -4 t 6 -
9
0 = -7 ✗ = 2 It 4 -
) (5)
↳
contradiction 2. (3)
set:{}0r solution
✗ = 2 If -56
6
I
not a
real no .
2×+9 =
30 -
6× + ✗
+ Gx ✗ =
30 9 ↳
imaginary
-
2x
-
7- ✗ =
21
set:{①
y
7 solution
rincon
✗ =3 equation
set:{ solution
4×2 t
(2×1-7) (2×1-1)=0
4X t 1 =
0
2×1-1 = 0
✗ t ✗ -
2 = 2×-3 t 1 2x = -1
2x -
2 =
2×-2 2
↳
identity ✗ =
-1/2
solution set : IR
?⃝