2 2linear equations i

Linear Equations I
Frank Ma © 2011

Linear Equations I
We left the house with a certain amount of money, say $x.

Linear Equations I
We left the house with a certain amount of money, say $x. After we bought a hamburger for $4, we had $(x – 4) left.

Linear Equations I
We left the house with a certain amount of money, say $x. After we bought a hamburger for $4, we had $(x – 4) left. Suppose there was $14 remained, how much did we have in the beginning?

Linear Equations I
We left the house with a certain amount of money, say $x. After we bought a hamburger for $4, we had $(x – 4) left. Suppose there was $14 remained, how much did we have in the beginning? In symbols, we've the equation x – 4 =14.

Linear Equations I
Calculate backwards by adding 4 to14, we get 14 + 4 = 18.

Linear Equations I
In symbols x – 4 =14 add 4 to both sides

Linear Equations I
+4 +4

Linear Equations I
+4 +4
so x = 18

Linear Equations I
+4 +4
so x = 18
We bought 5 apples for $20. How much was each apple?

Linear Equations I
+4 +4
so x = 18
Again, let x be the price of an apple, then 5 apples cost $5x.

Linear Equations I
+4 +4
so x = 18
Again, let x be the price of an apple, then 5 apples cost $5x. Since the total cost was 20, by dividing 20/5 = 4 and we get that each apple was $4.

Linear Equations I
+4 +4
so x = 18
Again, let x be the price of an apple, then 5 apples cost $5x. Since the total cost was 20, by dividing 20/5 = 4 and we get that each apple was $4. In symbols,
5x = 20

Linear Equations I
+4 +4
so x = 18
5x = 20 both sides divided by 5

Linear Equations I
+4 +4
so x = 18
5x 20
so
=
5
5

Linear Equations I
+4 +4
so x = 18
5x 20
so x = 4
so
=
5
5

Linear Equations I
In the above examples, the symbolic versions may seem unnecessarily cumbersome.

Linear Equations I
In the above examples, the symbolic versions may seem unnecessarily cumbersome. But for complicated problems, the symbolic versions are indispensable.

Linear Equations I
An equationistwo expressions set equal to each other.

Linear Equations I
Equations have the form:
left expression = right expression

Linear Equations I
or
LHS = RHS

Linear Equations I
or
LHS = RHS
In the two examples above x – 4 = 14 and 5x = 20, are equations.

Linear Equations I
or
LHS = RHS
In the two examples above x – 4 = 14 and 5x = 20, are equations. In particular these are linear equations.

Linear Equations I
or
LHS = RHS
A linear equation is an equation where both sides are linear expressions.

Linear Equations I
or
LHS = RHS
A linear equation does not contain any higher power of x such as x2, x3; it does not have x's in the denominator.

Linear Equations I
or
LHS = RHS
x2 – 3x = 2x – 3 is not a linear equation because of the x2.

Linear Equations I
or
LHS = RHS
x2 – 3x = 2x – 3 is not a linear equation because of the x2.
Linear equations are the simpliest type of equations.

Linear Equations I
We want to solve equations.

Linear Equations I
We want to solve equations. That is, we want to find a value (or values) for the variable x such that it makes the two sides equal.

Linear Equations I
We want to solve equations. That is, we want to find a value (or values) for the variable x such that it makes the two sides equal. Such a value is called a solution of the equation.

Linear Equations I
Example A.
For 14 = 2x + 8, x = (3) is a solution

Linear Equations I
Example A.
For 14 = 2x + 8, x = (3) is a solution because
14 = 2(3) + 8 ?

Linear Equations I
Example A.
14 = 2(3) + 8 ?
14 = 6 + 8 ?
14 = 14 yes

Linear Equations I
Example A.
14 = 2(3) + 8 ?
14 = 6 + 8 ?
14 = 14 yes
Linear equations are the easiest equations to solve

Linear Equations I
Example A.
14 = 2(3) + 8 ?
14 = 6 + 8 ?
14 = 14 yes
Linear equations are the easiest equations to solve and the simplest of these are the ones where the a single operation is applied to x.

Linear Equations I
Example A.
14 = 2(3) + 8 ?
14 = 6 + 8 ?
14 = 14 yes
Linear equations are the easiest equations to solve and the simplest of these are the ones where the a single operation is applied to x. These are the one-step equations.

Linear Equations I
Example A.
14 = 2(3) + 8 ?
14 = 6 + 8 ?
14 = 14 yes
Linear equations are the easiest equations to solve and the simplest of these are the ones where the a single operation is applied to x. These are the one-step equations. For example,
x – 3 = 12,

Linear Equations I
Example A.
14 = 2(3) + 8 ?
14 = 6 + 8 ?
14 = 14 yes
x – 3 = 12,
12 = x + 3,

Linear Equations I
Example A.
14 = 2(3) + 8 ?
14 = 6 + 8 ?
14 = 14 yes
x – 3 = 12,
12 = x + 3,
3*x = 12,

Linear Equations I
Example A.
14 = 2(3) + 8 ?
14 = 6 + 8 ?
14 = 14 yes
x – 3 = 12,
12 = x + 3,
3*x = 12,
12 = are one-step equations.
x
3

Linear Equations I
Basic principle for solving one- step-equations:

Linear Equations I
To solve one-step-equations, isolate the x on one side byapplying theopposite operation to both sides of the equation.

Linear Equations I
Example B. Solve for x
a. x – 3 = 12
b. x + 3 = –12
c. 3x = 15

Linear Equations I
a. x – 3 = 12 Add 3 to both sides
+ 3 + 3
b. x + 3 = –12
c. 3x = 15

Linear Equations I
+ 3 + 3
x = 15
b. x + 3 = –12
c. 3x = 15

Linear Equations I
+ 3 + 3
x = 15
?
check: 15 – 3 = 12
b. x + 3 = –12
c. 3x = 15

Linear Equations I
+ 3 + 3
x = 15
?
check: 15 – 3 = 12
12 = 12 (yes)
b. x + 3 = –12
c. 3x = 15

Linear Equations I
+ 3 + 3
x = 15
?
check: 15 – 3 = 12
12 = 12 (yes)
b. x + 3 = –12 Subtract 3 from both sides
–3 –3
c. 3x = 15

Linear Equations I
+ 3 + 3
x = 15
?
check: 15 – 3 = 12
12 = 12 (yes)
–3 –3
x = –15
c. 3x = 15

Linear Equations I
+ 3 + 3
x = 15
?
check: 15 – 3 = 12
12 = 12 (yes)
–3 –3
x = –15check: –15 + 3 = –12
?
c. 3x = 15

Linear Equations I
+ 3 + 3
x = 15
?
check: 15 – 3 = 12
12 = 12 (yes)
–3 –3
x = –15check: –15 + 3 = –12
?
–12 = –12 (yes)
c. 3x = 15

Linear Equations I
+ 3 + 3
x = 15
?
check: 15 – 3 = 12
12 = 12 (yes)
–3 –3
x = –15check: –15 + 3 = –12
?
–12 = –12 (yes)
c. 3x = 15 Both sides divided by 3
3x
15
=
3
3

Linear Equations I
+ 3 + 3
x = 15
?
check: 15 – 3 = 12
12 = 12 (yes)
–3 –3
x = –15check: –15 + 3 = –12
?
–12 = –12 (yes)
3x
15
=
3
3
x = 5

Linear Equations I
+ 3 + 3
x = 15
?
check: 15 – 3 = 12
12 = 12 (yes)
–3 –3
x = –15check: –15 + 3 = –12
?
–12 = –12 (yes)
3x
15
=
3
3
?
x = 5
check: 3(5)= 15

Linear Equations I
+ 3 + 3
x = 15
?
check: 15 – 3 = 12
12 = 12 (yes)
–3 –3
x = –15check: –15 + 3 = –12
?
–12 = –12 (yes)
3x
15
=
3
3
?
x = 5
check: 3(5)= 15
15 = 15 (yes)

Linear Equations I
x
–12
=
d.
3

Linear Equations I
x
–12
Multiply both sides by 3
=
d.
3
x
)
(
(3)
–12
=
3

Linear Equations I
x
–12
=
d.
3
x
)
(
(3)
–12
=
3
x = –36

Linear Equations I
x
–12
=
d.
3
x
)
(
(3)
–12
=
3
– 36
–12
=
x = –36
Check:
3

Linear Equations I
x
–12
=
d.
3
x
)
(
(3)
–12
=
3
– 36
–12
=
x = –36
Check:
3
Fact: Given a linear equation if we +, –, * , /, to both sides by the same quantity, the new equation will have the same solution.

Linear Equations I
x
–12
=
d.
3
x
)
(
(3)
–12
=
3
– 36
–12
=
x = –36
Check:
3
Next we tackle equations that require two steps.

Linear Equations I
x
–12
=
d.
3
x
)
(
(3)
–12
=
3
– 36
–12
=
x = –36
Check:
3
Next we tackle equations that require two steps. These are the ones that we have to collect the x-terms or the constant terms first.

Linear Equations I
x
–12
=
d.
3
x
)
(
(3)
–12
=
3
– 36
–12
=
x = –36
Check:
3
Next we tackle equations that require two steps. These are the ones that we have to collect the x-terms or the constant terms first. We do this by addition or subtraction.

Linear Equations I
x
–12
=
d.
3
x
)
(
(3)
–12
=
3
– 36
–12
=
x = –36
Check:
3
Example C. Solve for x
a. x – 6 = 3x

Linear Equations I
x
–12
=
d.
3
x
)
(
(3)
–12
=
3
– 36
–12
=
x = –36
Check:
3
a. x – 6 = 3x Collect the x'sby subtracting x from both sides
–6 = 2x Divide by 2
–x
–x

Linear Equations I
x
–12
=
d.
3
x
)
(
(3)
–12
=
3
– 36
–12
=
x = –36
Check:
3
a. x – 6 = 3x Collect the x'sby subtracting x from both sides
Divide by 2
–x
–x
–6
2x
=
2
2
–3 = x

Linear Equations I
b. 3x – 6 = 3

Linear Equations I
b. 3x – 6 = 3 collect the numbers by adding 6
+6
+6

Linear Equations I
+6
+6
3x = 9

Linear Equations I
+6
+6
3x = 9 div. by 3
3x/3 = 9/3

Linear Equations I
+6
+6
3x = 9 div. by 3
3x/3 = 9/3
x = 3

Linear Equations I
+6
+6
3x = 9 div. by 3
3x/3 = 9/3
x = 3
The more general linear equations have the form
#x ± #=#x ± #,
where # can be any number.

Linear Equations I
+6
+6
3x = 9 div. by 3
3x/3 = 9/3
x = 3
#x ± #=#x ± #,
where # can be any number. We solve it by following steps:

Linear Equations I
+6
+6
3x = 9 div. by 3
3x/3 = 9/3
x = 3
#x ± #=#x ± #,
1. Add or subtract to move the x-term to one side of the
equation and get: #x ± #=# or #=#x ± #

Linear Equations I
+6
+6
3x = 9 div. by 3
3x/3 = 9/3
x = 3
#x ± #=#x ± #,
2. Add or subtract the # to separate the number-term from the
x-term to get: #x=# or#=#x.

Linear Equations I
+6
+6
3x = 9 div. by 3
3x/3 = 9/3
x = 3
#x ± #=#x ± #,
3. Divide or multiply to get x:
x= solution or solution = x

Linear Equations I
+6
+6
3x = 9 div. by 3
3x/3 = 9/3
x = 3
#x ± #=#x ± #,
The idea is to use + or –, to separate the x-term from the number term,

Linear Equations I
+6
+6
3x = 9 div. by 3
3x/3 = 9/3
x = 3
#x ± #=#x ± #,
The idea is to use + or –, to separate the x-term from the number term, then use * or / to find the value of x.

Linear Equations I
Example D.
Solve 3x – 4 = 5x + 2

Linear Equations I
Example D.
Solve 3x – 4 = 5x + 2
subtract 3x to remove
the x from one side.
–3x –3x

Linear Equations I
Example D.
Solve 3x – 4 = 5x + 2
–3x –3x
– 4 = 2x + 2

Linear Equations I
Example D.
Solve 3x – 4 = 5x + 2
subtract 2 to move the 2 to the other side.
–3x –3x
– 4 = 2x + 2
–2 –2

Linear Equations I
Example D.
Solve 3x – 4 = 5x + 2
–3x –3x
– 4 = 2x + 2
–2 –2
– 6 = 2x

Linear Equations I
Example D.
Solve 3x – 4 = 5x + 2
divide by 2 get x.
–3x –3x
– 4 = 2x + 2
–2 –2
– 6 = 2x
–6
2x
=
2
2

Linear Equations I
Example D.
Solve 3x – 4 = 5x + 2
divide by 2 get x.
–3x –3x
– 4 = 2x + 2
–2 –2
– 6 = 2x
–6
2x
=
2
2
–3 = x

Linear Equations I
Example D.
Solve 3x – 4 = 5x + 2
divide by 2 get x.
–3x –3x
– 4 = 2x + 2
–2 –2
– 6 = 2x
–6
2x
=
2
2
–3 = x
Finally, all linear equations can be reduced to the format
#x ± #=#x ± #
by simplifying each side by itself first.

Linear Equations I
Example D.
Solve 3x – 4 = 5x + 2
divide by 2 get x.
–3x –3x
– 4 = 2x + 2
–2 –2
– 6 = 2x
–6
2x
=
2
2
–3 = x
#x ± #=#x ± #
Example E. Solve the equation.
2x – 3(1 – 3x) = 3(x – 6) – 1

Linear Equations I
Example D.
Solve 3x – 4 = 5x + 2
divide by 2 get x.
–3x –3x
– 4 = 2x + 2
–2 –2
– 6 = 2x
–6
2x
=
2
2
–3 = x
#x ± #=#x ± #
Example E. Solve the equation.
2x – 3(1 – 3x) = 3(x – 6) – 1
Simplify each side by combining like-term first.

Linear Equations I
2x – 3(1 – 3x) = 3(x – 6) – 1

Linear Equations I
2x – 3(1 – 3x) = 3(x – 6) – 1
2x – 3 + 9x =

Linear Equations I
2x – 3(1 – 3x) = 3(x – 6) – 1
2x – 3 + 9x = 3x – 18 – 1

Linear Equations I
2x – 3(1 – 3x) = 3(x – 6) – 1
2x – 3 + 9x = 3x – 18 – 1
11x – 3 =

Linear Equations I
2x – 3(1 – 3x) = 3(x – 6) – 1
2x – 3 + 9x = 3x – 18 – 1
11x – 3 = 3x – 19

Linear Equations I
2x – 3(1 – 3x) = 3(x – 6) – 1
2x – 3 + 9x = 3x – 18 – 1
11x – 3 = 3x – 19
–3x–3x
gather the x’s, subtract 3x from both sides

Linear Equations I
2x – 3(1 – 3x) = 3(x – 6) – 1
2x – 3 + 9x = 3x – 18 – 1
11x – 3 = 3x – 19
–3x–3x
8x – 3 = –19

Linear Equations I
2x – 3(1 – 3x) = 3(x – 6) – 1
2x – 3 + 9x = 3x – 18 – 1
11x – 3 = 3x – 19
–3x–3x
8x – 3 = –19
+3 +3
gather the number-terms, +3 to both sides

Linear Equations I
2x – 3(1 – 3x) = 3(x – 6) – 1
2x – 3 + 9x = 3x – 18 – 1
11x – 3 = 3x – 19
–3x–3x
8x – 3 = –19
+3 +3
8x = –16

Linear Equations I
2x – 3(1 – 3x) = 3(x – 6) – 1
2x – 3 + 9x = 3x – 18 – 1
11x – 3 = 3x – 19
–3x–3x
8x – 3 = –19
+3 +3
8x = –16
divide by 8 to get x.

Linear Equations I
2x – 3(1 – 3x) = 3(x – 6) – 1
2x – 3 + 9x = 3x – 18 – 1
11x – 3 = 3x – 19
–3x–3x
8x – 3 = –19
+3 +3
8x = –16
8x
-16
=
8
8

Linear Equations I
2x – 3(1 – 3x) = 3(x – 6) – 1
2x – 3 + 9x = 3x – 18 – 1
11x – 3 = 3x – 19
–3x–3x
8x – 3 = –19
+3 +3
8x = –16
-2
8x
-16
=
8
8

Linear Equations I
2x – 3(1 – 3x) = 3(x – 6) – 1
2x – 3 + 9x = 3x – 18 – 1
11x – 3 = 3x – 19
–3x–3x
8x – 3 = –19
+3 +3
8x = –16
-2
8x
-16
=
8
8
x = -2

2 2linear equations i

by math123a on Jun 22, 2011

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2 2linear equations i — Presentation Transcript