Transcript Tutorial 6a - C on T ech Math : : An application

Simple Equations in 1 Variable:
Tutorial 6a
A Solution Set

Consider the different meanings of the word
solution.

The solution to the mystery escaped him.


The town’s solution to its landfill problem is to
encourage recycling.


The word solution here refers to an explanation.
Solution here refers to a method of solving a problem.
A chemist mixes two solutions to obtain a 15% acid
solution.

Solution here refers to a homogeneous molecular mixture
Solution Set
In Mathematics we also have different kinds
of solutions and, therefore, different kinds
of solution sets.
 Study the table below:

Equation/Inequality
3x + 5 = 14
|x|  5
x+3=x-7
(x + 6)(x – 3)=0
Solution Set
{3}
{-5  x  5}
No Solution
{-6, 3}
These examples
illustrate that a
solution set may have
one member, more
than one member, or
no members.
Solving:
Addition & Subtraction Equations
One way to solve an equation is to get the
variable alone on one side of the equal sign.
 You can do this by using inverse operations,
which are operations that undo one another.
 Addition and subtraction are inverse
operations.
 You can use subtraction to undo addition and
addition to undo subtraction .

Solving:
Addition & Subtraction Equations
Example #1: Solve the equation x + 4 = 7



Think to yourself: What is being
done to the variable (x)?
A 4 is being added to the variable
(x). Subtraction undoes addition
therefore you should subtract a 4
on the left to get x alone on one
side.
However, whatever you do to one
side of an equation you must also
do to the other side.
x+4=7
-4 -4
x =3
Always check your answers!
x + 4 = 7; Does x = 3?
3 + 4 = 7 is true therefore x = 3!
Solving:
Addition & Subtraction Equations
Example #2: Solve the equation x - 12 = 20



Think to yourself: What is being
done to the variable (x)?
A 12 is being subtracted from the
variable (x). Addition undoes
subtraction, therefore you should
add a 12 on the left to get x alone
on one side.
However, whatever you do to one
side of an equation you must also
do to the other side.
x - 12 = 20
+12 +12
x = 32
Always check your answers!
x - 12 = 20; Does x = 32?
32 - 12 = 20 is true therefore x = 32!
Problem Solving:
A veterinary assistant holds a dog and steps on a scale.
The scale reads 193.7 lb. Alone, the assistant weighs
135 lb. To find the weight of the dog, solve the equation
w + 135 = 193.7
w + 135 = 193.7
 Think to yourself: What is being done
to the variable (w)?
-135 -135
 A 135 is being added to the variable (w).
w = 58.7

Subtraction undoes addition therefore
you should subtract a 135 on the left to
get w alone on one side.
However, whatever you do to one side
of an equation you must also do to the
other side.
The dog weighs 58.7 lb.
Always check your answers!
w + 135 = 193.7; Does w = 58.7?
58.7 + 135 = 193.7 is true!
Solving:
Multiplication & Division Equations
Multiplication and division are inverse
operations.
 You can use division to undo multiplication
and multiplication to undo division.

Solving:
Multiplication & Division Equations
Example #1: Solve the equation 5x = 35



Think to yourself: What is being
done to the variable (x)?
A 5 is being multiplied to the
variable (x). Division undoes
multiplication, therefore you
should divide a 5 to the left side
to get x alone on that side.
However, whatever you do to one
side of an equation you must also
do to the other side.
15x = 357
1
5
5
1
x =7
Always check your answers!
5x = 35; Does x = 7?
5•7 = 35 is true therefore x = 7 !
Solving:
Multiplication & Division Equations
r
4
6
Example #2: Solve the equation

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
Think to yourself: What is being
done to the variable (r)?
A 6 is being divided into the
variable (r). Multiplication
undoes division, therefore you
should multiply a 6 to the left
side to get r alone on that side.
However, whatever you do to one
side of an equation you must also
do to the other side.
1 r
6•
1
6
 4• 6
r = 24
Always check your answers!
r
 4 ; Does r  24?
6
24
 4 is true therefore r  24!
6
Solving:
Multiplication & Division Equations



5
r  90
Example #2: Solve the equation
6
1
1
Think to yourself: What is being
18
6 5
6
 r  90
done to the variable (r)?
51
5
1 5 61
A /6 is being multiplied to the
variable (r). Multiplying by the
reciprocal will eliminate the
fraction, therefore you should
multiply a 6/5 to the left side to
get r alone on that side.
However, whatever you do to one
side of an equation you must also
do to the other side.
r = 108
Always check your answers!
5
r  90 ; Does r  108?
6
5
108  90 is true therefore r  108
6