Gateway Quiz Reminders:
• The next Gateway will be given in class next week
(check your course calendar.)
• Before then, each student who did not score 8/8 on
the first Gateway must go over that quiz one-onone with a teacher or TA and get a signature from
them on the bottom of the second page of the quiz.
This can be done either in the open lab or in any class
session after lecture.
• You must show your classroom teacher the signed
quiz before you will be allowed to take the next
version.
NOW PLEASE
CLOSE
YOUR LAPTOPS
and prepare to
take notes.
Section 2.2
An 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.
Equivalent equations are equations with the
same solutions.
A linear equation in one variable
can be written in the form ax + b = c, a  0
Examples:
3x + 6 = -10
½x – 2 = ¼
4x = 0
7y + 10 = 0
Note: “Linear” means that the variable is not raised to a power.
So x3 – 2x + 1 = 0 is NOT a linear equation.
A solution to an equation is a number that can
be substituted in place of the variable that makes the
equation a true statement.
Examples:
Is x = 5 a solution of the equation 2x – 10 = 0?
Check: Sub 5 in for x:
2·5 – 10 =
10 – 10 = 0 
So x = 5 IS a solution of 2x – 10 = 0.
Is y = -3 a solution of the equation 3y + 10 = -2?
Check: Sub -3 in for y:
3·(-3) + 10 =
-9 + 10 = 1, which is not equal to -2
So x = -3 IS NOT a solution of 3y + 10 = -2.
Equivalent equations are equations with
the same solutions.
Example:
x – 1 = 5 and 3x + 1 = 19 are
equivalent equations
because plugging in x = 6
makes both of them true.
In this section we will study two properties that can be applied to
linear equations in order to find a solution (a number that makes the
statement true when it is plugged in for the variable). These
properties allow us to make complicated linear equations into
simpler equivalent linear equations.
NOTE: Our final goal in solving linear equations is to obtain an
equivalent equation of the very simple form “ x = a number”.
Addition Property of Equality:
If a = b then a + c = b + c and a – c = b – c .
Examples: If x = y, then x + 3 = y + 3
If x = y, then x - 7= y - 7
Multiplication Property of Equality:
If a = b then ac = bc and a/c = b/c.
Examples: If x = y, then 5x= 5y
If x = y, then x/6= y/6
Solving linear equations in one variable
1)
2)
3)
4)
Simplify each side of equation by distributing
where necessary and then combining like
terms.
Get all variable terms on one side and number
terms on the other side of equation (addition
property of equality).
Get variable alone with no coefficient (i.e. an
understood coefficient of 1) in front of it
(multiplication property of equality).
ALWAYS check solution by substituting into
original problem.
Examples of Solving Equations
3a + 2a = 7 + 4a
5a = 7 + 4a
(Simplify the left side)
5a + -4a = 7 + 4a + -4a
a=7
(Add –4a to each side)
(Simplify both sides)
Now CHECK your answer in 3a + 2a = 7 + 4a:
Sub 7 in for a in the original equation:
First side: 3·7 + 2·7 = 21 + 14 = 35
Second side: 7 + 4·7 = 7 + 28 = 35
35 = 35 
Problem from today’s homework:
Don’t forget:
Always check your answer by plugging it back
in for the variable in the original equation given
in the problem to see if both sides of the equation
come out to the same number.
Example of Solving Equation
15 – (6 – 7k) = 2 + 6k
15 – 6 + 7k = 2 + 6k (Use distributive property)
9 + 7k = 2 + 6k (Simplify the left side)
9 + 7k + -6k = 2 + 6k + -6k (Add –6k to each side)
9 + k = 2 (Simplify both sides)
9 + -9 + k = 2 + -9
k = -7
(Add –9 to both sides)
(Simplify both sides)
Remember to check your answer by plugging -7 in for k in the equation.
Problem from today’s homework:
Another example of solving an equation using the
multiplication property:
3
x6
8
83
8
  x  6 
38
3
x  16
(multiply both sides by fraction)
(simplify both sides)
Plan for solving more complicated linear equations:
Steps for solving linear equations in one variable:
1)
2)
3)
4)
5)
6)
7)
8)
Multiply to clear fractions (if necessary).
Use the distributive property as needed.
Simplify each side of the equation by combining like
terms.
Get all the variable (letter) terms on one side and the
constants (pure number terms) on the other side of
equation by using the addition property of equality.
Simplify each side again by combining like terms.
Get the variable completely by itself (get rid of the
coefficient (the number in front of the letter) by using
the multiplication property of equality.
Simplify the answer if needed.
Check solution by substituting into original problem.
Using these steps to solve a linear equation:
1. Multiply to clear fractions (if
Example: Solve the equation
2(x – 5) + 7x = 5 + 3(4 -x)
Step 1: There are no fractions to clear.
Step 2: Use the distributive property:
2∙x + 2∙(-5) + 7x = 5 + 3∙4 +3∙(-x)
2x – 10 + 7x = 5 + 12 – 3x
Step 3: Simplify both sides:
2x – 10 + 7x = 5 + 12 – 3x
9x – 10 = 17 – 3x
necessary).
2. Use the distributive property as
needed.
3. Simplify each side of the
equation by combining like terms.
4. Get all the variable (letter)
terms on one side and the
constants (pure number terms) on
the other side of equation by
using the addition property of
equality.
5. Simplify each side again by
combining like terms.
6. Get the variable completely by
itself (get rid of the coefficient
(the number in front of the letter)
by using the multiplication
property of equality.
7. Simplify the answer if needed.
8. Check solution by substituting
into original problem.
Example (continued): Solve 2(x – 5) + 7x = 5 + 3(4 -x)
Step 3 left us with this equation:
9x – 10 = 17 – 3x
Step 4: Get all variable terms on one
side and all pure numbers on other.
9x – 10 = 17 – 3x
add 3x to both sides
add 10 to both sides
+3x + 10 +10 + 3x
12x =
27
Step 5: Simplify each side again.
Step 6: Divide both sides by 12 to get x
27
by itself:
12x = 27
or x 
12
12
12
1. Multiply to clear fractions (if
necessary).
2. Use the distributive property as
needed.
3. Simplify each side of the
equation by combining like terms.
4. Get all the variable (letter)
terms on one side and the
constants (pure number terms) on
the other side of equation by
using the addition property of
equality.
5. Simplify each side again by
combining like terms.
6. Get the variable completely by
itself (get rid of the coefficient
(the number in front of the letter)
by using the multiplication
property of equality.
7. Simplify the answer if needed.
8. Check solution by substituting
into original problem.
Example (continued): Solve 2(x – 5) + 7x = 5 + 3(4 -x)
27
Step 6 left us with x 
12
Step 7: Simplify the answer:
27 3  9 3  3  3


12 2  6 2  2  3
333 33 9


2 23 2 2 4
ALWAYS DO THIS FINAL STEP:
Check the solution by substituting the
number back in for x in original equation.
1. Multiply to clear fractions (if
necessary).
2. Use the distributive property as
needed.
3. Simplify each side of the
equation by combining like terms.
4. Get all the variable (letter)
terms on one side and the
constants (pure number terms) on
the other side of equation by
using the addition property of
equality.
5. Simplify each side again by
combining like terms.
6. Get the variable completely by
itself (get rid of the coefficient
(the number in front of the letter)
by using the multiplication
property of equality.
7. Simplify the answer if needed.
8. Check solution by substituting
into original problem.
Example (continued): Solve 2(x – 5) + 7x = 5 + 3(4 -x)
ALWAYS DO THIS FINAL STEP:
Check the solution (9/4) by substituting the
number back in for x in original equation.
2(x – 5) + 7x = 5 + 3(4 - x)
41
Left side: 2(9/4 – 5) + 7∙9/4 =
4
Right side: 5 + 3(4 – 9/4) = 41
4
Use the order of operations and rules
for working with fractions to
calculate out both sides.
Since both sides of the equation come
out to the same number, we can be
certain that 9/4 is the correct solution
1. Multiply to clear fractions (if
necessary).
2. Use the distributive property as
needed.
3. Simplify each side of the
equation by combining like terms.
4. Get all the variable (letter)
terms on one side and the
constants (pure number terms) on
the other side of equation by
using the addition property of
equality.
5. Simplify each side again by
combining like terms.
6. Get the variable completely by
itself (get rid of the coefficient
(the number in front of the letter)
by using the multiplication
property of equality.
7. Simplify the answer if needed.
8. Check solution by substituting
into original problem.
Reminder:
This homework assignment
on section 2.2 is due
at the start of next class period.
Remember to go over your
Gateway Quiz with a teacher or
TA if you haven’t done so yet.
You may now
OPEN
your LAPTOPS
and log in to CourseCompass.