1. Skill Check
Lesson Presentation
Lesson Quiz

2. Skill Check
Simplify.
1. 5x x
7x + 6
2. 1 – 3y – 5 + 4y
y – 4
3. –2(k + 9)
–2k – 18
4. –3(17 – 2y)
–51 + 6y

3. Spanish Club The Spanish club is arranging a trip to a Mexican restaurant in a nearby city. Those who go must share the $60 cost of using a school bus for the trip. The restaurant’s buffet costs $5 per person. How many students must sign up for this trip in order to limit the cost to $10 per student? You will see how to use an equation to answer this question.

4. Every equation in this lesson has variables on both sides of the equation. You can solve such an equation by getting the variable terms on one side of the equation and the constant terms on the other side.

5. EXAMPLE 1
Solving an Equation with the Variable on Both Sides
Solve 7n – 5 = 10n + 13.
SOLUTION
7n – 5 = 10n + 13
Write original equation.
7n – = 10n + 13
– 7n – 7n
Subtract 7n from each side.
–5 = 3n + 13
Simplify.
– = 3n + 13
– – 13
Subtract 13 from each side.
–18 = 3n
Simplify.
–18 = 3n =
Divide each side by 3.
–6 = n
Simplify.
ANSWER
The solution is –6.

6. EXAMPLE 2
Writing and Solving an Equation
In order for the cost per student to be $10, how many students must go on the Spanish club trip to the Mexican restaurant if the school bus fee is $60 and buffet is $5 per person?
SOLUTION
Let s represent the number of students. Write a verbal model.
Cost per student
Number of students
Cost of buffet
Cost of school bus • + =
10s = 5s + 60
Substitute.
10s = 5s
– 5s –5s
Subtract 5s from each side.
5s = 60
Simplify.

7. 2 5s = 60 = 5 s = 12 Writing and Solving an Equation
EXAMPLE 2
Writing and Solving an Equation
In order for the cost per student to be $10, how many students must go on the Spanish club trip to the Mexican restaurant if the school bus fee is $60 and buffet is $5 per person?
SOLUTION
Let s represent the number of students. Write a verbal model.
Cost per student
Number of students
Cost of buffet
Cost of school bus • + =
5s = 60
Divide each side by 5. 5 =
s = 12
Simplify.
ANSWER
The club needs 12 students to go on the trip.

8. Number of Solutions When you solve an equation, you may find that it has no solution or that every number is a solution.

9. EXAMPLE 3
An Equation with No Solution
Solve 5(2x + 1) = 10x.
SOLUTION
5(2x + 1) = 10x
Write original equation.
10x + 5 = 10x
Distributive property
Notice that 10x + 5 = 10x is not true because the number 10x cannot be equal to 5 more than itself. The equation has no solution. As a check, you can continue solving the equation.
10x = 10x
– 10x – 10x
Subtract 10x from each side.
5 = 0
Simplify.
The statement 5 = 0 is not true, so the equation has no solution.

10. EXAMPLE 4
Solving an Equation with All Numbers as Solutions
Solve 6x + 2 = 2(3x + 1).
6x + 2 = 2(3x + 1)
Write original equation.
6x + 2 = 6x + 2
Distributive property.
Notice that for all values of x, the statement 6x + 2 = 6x + 2 is true. The equation has every number as a solution.

11. EXAMPLE 5
Solving an Equation to Find a Perimeter
Geometry Find the perimeter of the square. 2x
x + 4 1
A square has four sides of equal length. Write an equation and solve for x.
2x = x + 4
Write equation.
2x = x + 4
– x – x
Subtract x from each side.
x = 4
Simplify. 2
Find the length of one side by substituting 4 for x in either expression.
2x = 2( ) 4
= 8
Substitute 4 for x and multiply. 3
To find the perimeter, multiply the length of one side by 4.
4 • 8 = 32
ANSWER
The perimeter of the square is 32 units.

12. Lesson Quiz
Solve the equation.
1. 13x + 9 = 11x + 13 2
2. –3k – 25 = 5k – 1 –3
no solution
3. 6(1 + 5y) = 30y – 2
4. –14 + 7m = 7(m – 2)
all numbers
5. Challenge For what value(s) of a does the equation 5y + 10 = 5(2 + ay) have all numbers as a solution? 1