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2.1 – Linear Equations in One Variable
Algebraic equation is a statement that two expressions have equal
value.
Solving algebraic equations involves finding values for a variable
that make the equation true.
Linear equation in one variable can be written in the form:
ax + b = c, a  0.
Equivalent equations are equations with the same solutions in
the form of:
variable = number, x = 3, or
number = variable, 3 = x.
2.1 – Linear Equations in One Variable
A linear equation in one variable can be written in the form:
ax + b = c
where a, b, and c are real numbers and a ≠ 0.
Addition Property of Equality
a = b and a + c = b + c
equivalent equations
Multiplication Property of Equality
a = b and ac = bc
equivalent equations
2.1 – Linear Equations in One Variable
Example
8+z=–8
Add –8 to each side
8 + (–8) + z = –8 + –8
Simplify both sides
z = –16
2.1 – Linear Equations in One Variable
Example
3z – 1 = 26
3z – 1 + 1 = 26 + 1
3z = 27
3 z 27

3
3
z=9
2.1 – Linear Equations in One Variable
Example
4p – 11 – p = 2 + 2p – 20
3p – 11 = 2p – 18
3p + (– 2p) – 11 = 2p + (– 2p) – 18
p – 11 = –18
p – 11 + 11 = –18 + 11
p = –7
2.1 – Linear Equations in One Variable
Example
5(3 + z) – (8z + 9) = – 4z
15 + 5z – 8z – 9 = – 4z
6 – 3z = – 4z
6 – 3z + 4z = – 4z + 4z
6+z=0
6 + (–6) + z = 0 + (–6)
z = –6
2.1 – Linear Equations in One Variable
Example
12x + 30 + 8x – 6 = 10
20x + 24 = 10
20x + 24 + (– 24) = 10 + (– 24)
20x = – 14
20 x  14

20
20
7
x
10
2.1 – Linear Equations in One Variable
Example
3( y  3)
 2y  6
5
5  3( y  3)
 52 y  6 
5
LCD: 5
3 y  9  10 y  30
3 y  (3 y )  9  10 y  (3 y )  30
9  (30)  7 y  30  (30)
 21 7 y

7
7
3  y
2.1 – Linear Equations in One Variable
Example
5x – 5 = 2(x + 1) + 3x – 7
5x – 5 = 2x + 2 + 3x – 7
5x – 5 = 5x – 5
Both sides of the equation are identical.
This equation will be true for every x that is substituted
into the equation. The solution is “all real numbers.”
The equation is called an identity.
2.1 – Linear Equations in One Variable
Example
3x – 7 = 3(x + 1)
3x – 7 = 3x + 3
3x + (– 3x) – 7 = 3x + (– 3x) + 3
–7 = 3
No value for the variable x can be substituted into this
equation that will make this a true statement.
There is “no solution.”
The equation is a contradiction.
2.2 - An Introduction to Problem Solving
General Strategy for Problem Solving
UNDERSTAND the problem.
Read and reread the problem.
.
Choose a variable to represent the unknown.
Construct a drawing.
Propose a solution and check.
TRANSLATE the problem into an equation.
SOLVE the equation.
INTERPRET the result:
Check proposed solution in problem.
State your conclusion.
2.2 - An Introduction to Problem Solving
Example
Twice a number plus three is the same as the number
minus six.
2x
+3
=
x–6
2x  3  x  6
2x  3  x  x  6  x
x  3  6
x  3  3  6  3
x  9
2.2 - An Introduction to Problem Solving
Example
The product of twice a number and three is the same as
the difference of five times the number and ¾. Find the
number.
2x ·3
=
5x
2x · 3 = 5x – ¾
6x = 5x – ¾
6x + (–5x) = 5x + (–5x) – ¾
x = –¾
2.2 - An Introduction to Problem Solving
Example
A car rental agency advertised renting a Buick Century
for $24.95 per day and $0.29 per mile. If you rent this
car for 2 days, how many whole miles can you drive on a
$100 budget?
x = the number of whole miles driven
0.29x = the cost for mileage driven
cost for mileage driven + daily rental cost = $100
0.29x + 2(24.95) = 100
0.29x + 49.90 = 100
2.2 - An Introduction to Problem Solving
0.29x + 49.90 = 100
0.29x + 49.90 – 49.90 = 100 – 49.90
0.29x = 50.10
0.29 x 50.10

0.29
0.29
x  172.75 miles
2.2 - An Introduction to Problem Solving
Example
A pennant in the shape of an isosceles triangle is to be
constructed an Athletic Club and sold at a fund-raiser. The
company manufacturing the pennant charges according to
perimeter, and the athletic club has determined that a
perimeter of 149 centimeters should make a nice profit. If
each equal side of the triangle is twice the length of the third
side, increased by 12 centimeters, find the lengths of the
sides of the triangular pennant.
Let x = the third side of the triangle
2x + 12
2x + 12
2x + 12 = the first side
2x + 12 = the second side
x
2.2 - An Introduction to Problem Solving
2x + 12
2x + 12
Perimeter is 149 cm.
x
2x + 12 + 2x + 12 + x = 149
2x  2x  x  12  12  149
5x  24  149
5x  125
x  25 cm
2(25) + 12
= 62 cm
2.2 - An Introduction to Problem Solving
Example
The measure of the second angle of a triangle is twice
the measure of the smallest angle. The measure of the
third angle of the triangle is three times the measure of
the smallest angle. Find the measures of the angles.
Let x = degree measure of smallest angle
2x = degree measure of second angle
3x = degree measure of third angle
xº
2xº
3xº
2.2 - An Introduction to Problem Solving
The sum of the measures of the angles of a triangle
equals 180.
x + 2x + 3x
=
180
6x = 180
6 x 180

6
6
x = 30º
2x = 60º
3x = 90º
2.2 - An Introduction to Problem Solving
Example
The sum of three consecutive even integers is 252. Find
the integers.
x = the first even integer
x + 2 = next even integer
x + 4 = next even integer
x + x + 2 + x + 4 = 252
3x + 6 = 252
3x = 246
3x 246
The integers are:
x  82

82, 84 and 86.
3
3
2.3 – Formulas and Problem Solving
A formula:
An equation that states a known relationship among
multiple quantities (has more than one variable in it).
A = lw
I = PRT
d = rt
V = lwh
(Area of a rectangle = length · width)
(Simple Interest = Principal · Rate · Time)
(distance = rate · time)
(Volume of a rectangular solid = length·width·height)
2.3 – Formulas and Problem Solving
Solving Equations for a Specified Variable
Step 1: Multiply on both sides to clear the equation of fractions if
they appear.
Step 2: Use the distributive property to remove parentheses if
they appear.
Step 3: Simplify each side of the equation by combining like
terms.
Step 4: Get all terms containing the specified variable on one side
and all other terms on the other side by using the
addition property of equality.
Step 5: Get the specified variable alone by using the
multiplication property of equality.
2.3 – Formulas and Problem Solving
Example
Solve: T = mnr
T
mnr

mr mr
T
n
mr
for n.
2.3 – Formulas and Problem Solving
Example
Solve: 3y – 2x = 7
for y.
3y  2x  2x  7  2x
3y  7  2x
3y 7  2x

3
3
7  2x
7 2
y
or y   x
3
3 3
2.3 – Formulas and Problem Solving
Example
Solve: for A = P + PRT
A  P  P  P  PRT
A  P  PRT
A  P PRT

PR
PR
A P
T
PR
for T.
2.3 – Formulas and Problem Solving
Example
Solve: A  P  PRT
for P.
Factor out P on the right side.
A  P(1  RT )
Divide both sides by 1 + RT.
A
P(1  RT )

1  RT
1  RT
Simplify.
A
P
1  RT
2.3 – Formulas and Problem Solving
Example
A flower bed is in the shape of a triangle with one side
twice the length of the shortest side, and the third side is
30 feet more than the length of the shortest side. Find
the dimensions if the perimeter is 102 feet.
x = the length of the shortest side
2x = the length of the second side
x + 30 = the length of the third side
x
2x
x + 30
2.3 – Formulas and Problem Solving
x
2x
Perimeter is 102 feet
x + 30
P=a+b+c
102 = x + 2x + x + 30
102 = 4x + 30
102 – 30 = 4x + 30 – 30
72 = 4x
72 4 x

4
4
x = 18 feet
2(18) = 36 feet
18 + 30 = 48 feet