1. Algebra 1 09/21/16
EQ: How do I solve equations with variables on both sides?
HW: Due Friday pg. 95 # 1-33 all
Bring textbooks tomorrow
Quiz on Friday 2.1 – chapter 1
Warm up: see board

2. Check It Out! Example 1a
Solve the equation. Check your answer.
4b + 2 = 3b
4b + 2 = 3b
To collect the variable terms on one side, subtract 3b from both sides.
–3b –3b
b + 2 = 0
Since 2 is added to b, subtract 2 from both sides.
– 2 – 2
b = –2

3. Additional Example 1: Solving Equations with Variables on Both Sides
Solve 7n – 2 = 5n + 6.
7n – 2 = 5n + 6
To collect the variable terms on one side, subtract 5n from both sides.
–5n –5n
2n – 2 =
Since 2 is subtracted from 2n, add 2 to both sides.
2n =
Since n is multiplied by 2, divide both sides by 2 to undo the multiplication.
n = 4

4. Another Way: It does not matter
Solve 7n – 2 = 5n + 6.
7n – 2 = 5n + 6
Since 2 is subtracted from 2n, add 2 to both sides.
7n = 5n
To collect the variable terms on one side, subtract 5n from both sides.
-5n -5n
2n =
Since n is multiplied by 2, divide both sides by 2 to undo the multiplication.
n = 4

5. To solve more complicated equations, you may need to first simplify by using the Distributive Property or combining like terms.

6. Additional Example 2: Simplifying Each Side Before Solving Equations
Solve the equation.
4 – 6a + 4a = –1 – 5(7 – 2a)
4 – 6a + 4a = –1 –5(7 – 2a)
Distribute –5 to the expression in parentheses.
4 – 6a + 4a = –1 –5(7) –5(–2a)
4 – 6a + 4a = –1 – a
Combine like terms.
4 – 2a = – a
Since –36 is added to 10a, add 36 to both sides.
40 – 2a = a

7. Additional Example 2 Continued
Solve 4 – 6a + 4a = –1 – 5(7 – 2a).
40 – 2a = 10a
To collect the variable terms on one side, add 2a to both sides.
+ 2a +2a
= 12a
Since a is multiplied by 12, divide both sides by 12.

8. Check It Out! Example 2a
Solve the equation. Check your answer.
Distribute to the expression in parentheses. 1 2
To collect the variable terms on one side, subtract b from both sides. 1 2
3 = b – 1
Since 1 is subtracted from b, add 1 to both sides.
4 = b

9. Check It Out! Example 2b
Solve the equation. Check your answer.
3x + 15 – 9 = 2(x + 2)
3x + 15 – 9 = 2(x + 2)
Distribute 2 to the expression in parentheses.
3x + 15 – 9 = 2(x) + 2(2)
3x + 15 – 9 = 2x + 4
Combine like terms.
3x + 6 = 2x + 4
To collect the variable terms on one side, subtract 2x from both sides.
–2x –2x
x + 6 =
Since 6 is added to x, subtract 6 from both sides.
– – 6
x = –2

10. An identity is an equation that is always true, no matter what value is substituted for the variable. The solution set of an identity is all real numbers. Some equations are always false. Their solution sets are empty. In other words, their solution sets contain no elements.

11. Additional Example 3A: Infinitely Many Solutions or No Solutions
Solve the equation.
10 – 5x + 1 = 7x + 11 – 12x
Identify like terms.
10 – 5x + 1 = 7x + 11 – 12x
Combine like terms on the left and the right.
11 – 5x = 11 – 5x
All values of x will make the equation true. In other words, all real numbers are solutions.

12. Additional Example 3B: Infinitely Many Solutions or No Solutions
Solve the equation.
12x – 3 + x = 5x – 4 + 8x
12x – 3 + x = 5x – 4 + 8x
Identify like terms.
Combine like terms on the left and the right.
13x – 3 = 13x – 4
–13x –13x
Subtract 13x from both sides. 
–3 = –4
False statement; the solution set is .
The equation 12x – 3 + x = 5x – 4 + 8x is always false. There is no value of x that will make the equation true. There are no solutions.

13. The empty set can be written as  or {}.
Writing Math

14.  Check It Out! Example 3a Solve the equation. 4y + 7 – y = 10 + 3y
Identify like terms.
3y + 7 = 3y + 10
–3y –3y
Subtract 3y from both sides. 
False statement; the solution set is .
7 = 10
The equation 4y + 7 – y = y is always false. There is no value of y that will make the equation true. There are no solutions.

15. Check It Out! Example 3b
Solve the equation.
2c c = –14 + 3c + 21
2c c = –14 + 3c + 21
Identify like terms.
Combine like terms on the left and the right.
3c + 7 = 3c + 7
The statement 3c + 7 = 3c + 7 is true for all values of c. The equation 2c c = –14 + 3c + 21 is an identity. All values of c will make the equation true. In other words, all real numbers are solutions.