1. Steps for Solving an Equations: Topic: Solving Complex Equations
Essential Question: How do we solve
Complex equations?
Standard: 8.EE.7
Question /
Key Words
Notes or Solutions to Problems
Simplify each side by distributing
and/or combining like terms
Steps for
Solving an
Equations:
2. Collect variables terms on the side
where the coefficient is greater.
Use inverse operations to isolate
the variable.
4. Check your solution in the original
equation.

2. Solve 3(2 – x) – x = -5(x+ 1) 6 – 3x – x = – 5x – 5 – 4x + 6 = – 5x
Question /
Key Words
Notes or Solutions to Problems
Solve
3(2 – x)
– x =
-5(x+ 1) 1)
3(2 – x) – x = -(5x + 1) 6
– 3x
– x =
– 5x
– 5
– 4x
+ 6 =
– 5x
– 5
+ 5x
+ 5x x
+ 6 =
– 5
– 6
– 6 x =
– 11

3. Solve 3(2 – 4x) + 3x = 9(x + 2) – 6x 6 – 12x +3x 9x + 18 – 6x = – 9x
Question /
Key Words
Notes or Solutions to Problems
Solve
3(2 – 4x)
+ 3x =
9(x + 2)
– 6x 2)
3(2 – 4x) + 3x = 9(x +2) –6x 6
– 12x
+3x 9x
+ 18
– 6x =
– 9x
+ 6 = 3x
+ 18
+ 9x
+9x 6 =
12x
+ 18
– 18
– 18 1
– 12 =
12x
– 1 = x

4. Solve 3) (15x + 20) = 6 – 2(x – 4) 3x + 4 = 6 – 2x + 8 3x + 4 = – 2x
Question /
Key Words
Notes or Solutions to Problems
Solve 1 5 3)
(15x + 20) = 6
– 2(x – 4) 1 5
(15x + 20) = 6 – 2(x – 4) 3x
+ 4 = 6
– 2x
+ 8 3x
+ 4 =
– 2x
+ 14
+ 2x
+ 2x 5x
+ 4 = 14
– 4
– 4 1 5x = 10 x = 2

5. 2(4x + 5) = 8x + 10 8x + 10 = 8x + 10 –8x –8x 10 = 10 Identity 6)
Question /
Key Words
Notes or Solutions to Problems
Solve the equation if
possible. Determine
whether it has one solution,
no solution, or is an identity.
2(4x + 5) = 8x
+ 10 8x
+ 10 = 8x
+ 10 6)
2(4x + 5) = 8x + 10
–8x
–8x 10 = 10
Answer: The equation 10 = 10 is always
true, so all values of x are solutions.
The original equation is an identity.
Identity

6. x – 1 = x + 7 –x –x -1 = 7 No solution. 7) x – 1 = x + 7 Question /
Key Words
Notes or Solutions to Problems
Solve the equation if
possible. Determine
whether it has one solution,
no solution, or is an identity. x
– 1 = x
+ 7 –x –x 7)
x – 1 = x + 7 -1 = 7
Answer: The equation -1 = 7 is never
true no matter what the value of x.
The original equation has no solution.
No solution.

7. Question /
Key Words
Notes or Solutions to Problems 4)
A rental car agency charges
$20 per day plus 11 cents
per mile to rent a certain
car. Another agency
charges $26 per day plus
5 cents per mile to rent
the same car. How many
miles per day will have to
be driven for the cost of a
car from the first agency
to equal the cost of a
car from the second
agency?
Cost of first rental Agency C =
0.11x
+ 20
Cost of second rental Agency C =
0.05x
+ 26
0.11x
+ 20 =
0.05x
+ 26
– 0.05x
– 0.05x
0.06x
+ 20 = 26
– 20
– 20
0.06x = 6

8. 4) 0.06x = 6 0.06 0.06 x = 100 miles Question / Key Words
Notes or Solutions to Problems 4)
A rental car agency charges
$20 per day plus 11 cents
per mile to rent a certain
Car. Another agency
Charges $26 per day plus
5 cents per mile to rent
the same car. How many
miles per day will have to
be driven for the cost of a
car from the first agency
to equal the cost of a
car from the second
agency? 1
0.06x = 6 x =
100
miles