Download Algebra I: Section 3

CC MATH I STANDARDS UNIT 3
4.6 Solving Equations with VARIABLE ON BOTH SIDES: PART 1
WARM UP: Solve for the given variable.
1)
4)
7  3z  13
5
 8  3x
2
y
7
3
2)
4 
5)
7 x  2  4 x  10
3) 2  5 x  17
6)
4(5  2 x )  16
INTRODUCTION: GUESS VALUES and CHECK by Substitution
Example: 5x = x – 12
GUESS:
CHECK:
#1: x = 4
5(4) = 20 and 4 – 12 = -8
NO; 20 and -8 are 28 apart
#2: x = 6
5(6) = 30 and 6 – 12 = -6
NO; 30 and 06 are 36 apart
#3: x = 2
5(2) = 10 and 2 – 12 = -10
NO; 10 and -10 are 20 apart
#4: x = 0
5(0) = 0 and 0 – 12 = -12 NO
NO; 0 and -12 are 12 apart
#5: x = -2
5(-2) = -10 and -2 – 12 = -14 NO NO; -10 and -14 are 4 apart
#6: x = -3
5(-3) = -15 and -3 – 13 = -15
YES; x = -3 is a solution
1) 2a + 7 = 3a + 4
2) 4 + 6y – 6 = 6y – 2
3) 2z + 2 = -6 + 2z
TYPES OF ANSWERS IN VARIABLE ON BOTH SIDES PROBLEMS
 Variables on both sides can create 3 different type of solution scenarios
Type #1:
ONE SOLUTION
EXACTLY ONE number
can be used as an answer
Example:
2a + 7 = 3a + 4
Solution:
a = 3 is only answer
Type #2: ALL NUMBERS
(Identity)
ANY NUMBER (pos, neg,
fraction, whole, etc) can be
used as an answer
Example:
4 + 6y – 6 = 6y – 2
Solution:
y = 0, -1, 3 ALL WORK
Type #3:
NO SOLUTION
NO POSSIBLE NUMBER
can be used as an answer
Example:
2z + 2 = -6 + 2z
Solution:
No value will ever work
What is the process for solving with VARIABLES on BOTH SIDES?
You don’t want guess and check if the answer is NO SOLUTION, DECIMAL, or
FRACTION answer. The following steps will lead you to the correct solution.
Step #1: If needed, SIMPLIFY each side of equation separately
 Distribute Property AND Combine any like terms
Step #2: MOVE by adding or subtracting VARIABLE LIKE TERMS to one
side of the equation
Step #3: SOLVE the remaining equation (SADMEP)
How do we identify when solving what type of equations have
ONE Solution, NO Solution, or All NUMBERS?
 Always check which type of solution your equation satisfies.
Example #1:
Example #2:
Example #3:
2x  1  4x  3
7 x  21  5 x  12 x  20 6 x  9 x  2  6  15 x  8
BEGINNING EXAMPLES of MOVING VARIABLE TERMS:
1) a + 9 = 4a
2) -5b = 7b + 36
3) 9x = -12 + 9x
4) 4x – 15 = 7x
5) 7r + 56 = 15r
2
1
n

n2
7)
3
3
8) 5 
3
3
n n
8
8
6) 40 + 10y = - 10y
9) 
3
1
y   y6
4
4
INTERMEDIATE EXAMPLES of MOVING VARIABLE TERMS:
It doesn’t matter where you move the variable if you watch the positive and negative signs
GENERAL RULE OF THUMB: Move the SMALLER Variable to the BIGGER VARIABLE
1) 10x + 3 = 8x + 9
2) 7 – 2z = 22 + 3z
3) 3y - 13 = 3y + 16
4) 7r - 6 = - 3r + 44
5) 5x – 9 = -9 + 5x
6) 13 – 3y = 6y - 59
7) – 5y + 15 = -4y – 12
8) -2z + 12 = - 12 + 2z
9) - 6x + 8 = - 5 - 6x
ADVANCED EXAMPLES of MOVING VARIABLE TERMS:
SIMPLIFY EACH SIDE BEFORE MOVING TERMS
Use a separate piece of paper as needed.
1) 7p – 2 + 3p = 8p – 1
2) 7r + 10 = -10 + 8r – 22
3) 2m + 5 = 5m – 35 – 3m
4) 3r + 3 – 5 = 9r – 2 – 6r
5) -3s + 8 + 8s = 7s – 2
6) 4q + 6 = 5q – 42 + 7q
7) 13t – 9t + 80 = 4t + 100 – 20
8) 40c + 8 – 13 = 5 + 40c
9) 13x – 21x + 17 = -3x – 5 – 8
10) 12 – 7x – 2x = 8x + 12 – 17x
11) 3 + 3t – 7 = 3t + 4
12) 8 – 12z = 15 – 3z + 20
13) 15m – 7m + 11 = 12 – 5m + 2 + 7 m
14) 8 – 3n – 4n + 16 = 5n – 6 + 2n – 12
15) 8p + 9 = 3 + 8p + 2
16) 7r – 4r + 12 = -6r + 12 + 9r
CC MATH I STANDARDS UNIT 3
4.6 Solving Equations with VARIABLE ON BOTH SIDES: PART 2
1)
3 z  8  9z
WARM UP: Solve for the given variable.
3) 9a  2  6  12a  3a
2) 8  2 x  5 x  13
4)  6b  7  7  10b  4b
5)
27  5 x  11x  3
6)
2(5  3 x )  14
REVIEW: SIMPLIFYING
#1) 7(3x + 6) – 5x
#2) 12z – 7 + 9z – 8
#3) 6y – 4(5 – 7y)
#4) – 5r + 9 – 3r – 7r
#5) 5 – 2 (8 + 11n)
#6) 3(x + 7) + 4(2x – 5)
VARIABLE ON BOTH SIDES: SOLUTION TYPES
What are the 3 types of solutions you can have in an equation?
1) ___________________
2) ___________________
3) ___________________
PRACTICE VARIABLE ON BOTH SIDES: Combine Like Terms First
1) 10p – 7p + 5 = 5p + 21
2) 46 + 6x = 2 – 3x – 2x
3) -9s + 3s + 8 = 12 + 5s – 11s
4) 12x – 7x + 9 = 3x + 6
5) 17 + 6r – 5 = 13r + 12 – 7r
6) 11m + 12 + 5m = 47 + 7m – 8
5
3
6

n

2

n7
7)
4
4
2
1
18

7

n

n5
8)
3
2
PRACTICE VARIABLE ON BOTH SIDES: DISTRIBUTIVE PROPERTY
1) 2(5x + 3) = 7x – 8
2) 7(m – 2) = 4(3m + 4)
3) 4r – 17 = 2(r – 8)
4) -2(b – 7) = 14 – 2b
5) 3x + 12 = 6(3 – x)
7) 3(2y + 4) = 4(y + 5)
8) 4(2p + 9) = 8(5+ p)
6) -2(7 – 3s) = 4s+ 2
PRACTICE VARIABLE ON BOTH SIDES: SIMPLIFYING EXPRESSIONS
1) 3(2x – 7) = 5x – x – 11
2) -2x + 15 + 13x = 10 – 5(7 – x)
3) 6x + 7(x + 2) = 23 + 13x –9
4) 13 – 9(2x + 3) = -4(5x + 4) + 2x
5) 2(9 + 4x) + 6(2 – x) = 7x
6) 6(2x + 5) – 2(7x + 8) = 4(x + 8)