Download 3.1 Solving Equations Using Addition and Subtraction

2-4 Solving Multi-Step Equations
and Consecutive Integer Problems
Algebra 1
Glencoe McGraw-Hill
Linda Stamper
Solving Multi-Step Equations
The prefix “multi” means “more than one”. A multi-step
equation is solved by transforming the equation more than
one time. Always remember the basic rule:
Whatever you do to one side of the equal sign, you
must also do to the other side of the equal sign.
When solving multi-step equations, undo the addition or
subtraction first before undoing multiplication or division.
Solve.
Write the problem.
Undo subtraction.
Undo fraction using
the reciprocal.
Remember: In
algebra work
downward. Line up
the equal signs.
Skip one line after
the answer.
1
x  5  10
2
5 5
1
 2  1 x  15 2 
 
1
 1 2
 
x  30
Solve.
x
 25  5
Write the problem.
5
 25  25
Undo addition.
1 x
5  5  205 
Undo division.
x  100
The 5 must be in
the numerator
position or midway.
It should not be
written in the
denominator
position!
Solve.
Write the problem.
Change subtraction to
addition and distribute.
Combine like terms.
Undo double sign.
Undo subtraction.
Undo multiplication.
8x - 2x  7   16
8x +–– 2x  7   16
8x   2x   14  16
6x   14  16
6x  14  16
 14  14
1
6x  30
6
6
x5
Example 1
2y  5  1
5 5
2y   4
2
2
y  2
Are your
equal signs in
a line?
Solve.
Example 2
Example 3
2x  9x  17   4
2x - 5x - 9  27
 7 x  17   4
– – 5x +–– 9  27
2x +
 17  17

2
x

5x  45  27
 7x  21
Combinelike
7 terms.
7
Distribute
3x  45  27
x 3
 45  45
 3x  18
3 3
x6
Example 4
4
x  2  8
5
1
2
4
5
  x  2  8 5 
 4 5
4
 
x  2  10
2 2
x  12
Multiply both sides
by the reciprocal.
Solve.
Example 5
Example 6
3
x  15
12  x  2
 6
10
9
1
4
1
10
3


10
  12    x  2 9  x  15
 6 9 
3
3
10


 
9
40  x  2
x  15  54
2
2
 15  15
38  x
x  39
The study of numbers and the relationship
between them is called number theory.
Consecutive integers are integers in counting order, such
as 7, 8, and 9.
Beginning with an odd integer and counting by two will
result in consecutive odd integers, such as -3, -1, 1, 3, 5
Beginning with an even integer and counting by two will
result in consecutive even integers, such as -4, -2, 0, 2, 4
Two consecutive integers have a sum of 77. Find the integers.
Assign Labels.
Verbal Model.
Algebraic Model.
Solve.
Sentence.
Let f = first integer
Let f + 1 = second integer
first integer + second integer = total
f + ( f + 1) = 77
f  f  1  77
2f  1  77
1 1
2f  76
2
2
f  38
The integers are 38 and 39.
Check

38 + 39 = 77
Two consecutive odd integers have a sum of 92. Find the integers.
Assign Labels.
Verbal Model.
Algebraic Model.
Solve.
Sentence.
Let f = first integer
Let f + 2 = second integer
first integer + second integer = total
f + ( f + 2) = 92
f  f  2  92
2f  2  92
2 2
2f  90
2
2
f  45
The integers are 45 and 47.
Check

45 + 47 = 92
Example 7 Find three consecutive integers whose sum is 195.
Example 8 Find two consecutive odd integers whose sum is –96.
Example 9 Find three consecutive even integers whose sum is -42.
Assign Labels.
Verbal Model.
Algebraic Model.
Solve.
Sentence.
Example 7 Find three consecutive integers whose sum is 195.
Assign Labels.
Verbal Model.
Algebraic Model.
Solve.
Sentence.
Let f = first integer
Let ff + 1 = second integer
Let ff + 2 = third integer
first # + second # + third # = total
f + ( f +1) + ( f + 2) = 195
f  f  1  f  2  195
3f  3  195
3 3
3f  192
3
3
f  64
64   1
65
The integers are 64, 65 and 66.
Check

64 + 65 + 66 = 195
64   2
66
Example 8 Find two consecutive odd integers whose sum is –96.
Assign Labels.
Verbal Model.
Algebraic Model.
Solve.
Sentence.
Let f = first integer
Let f 
+ 2 = second integer
first integer + second integer = total
f + ( f +2) = -96
f  f  2  96
2f  2  96
2
2
2f  98
2
2
f  49
 49   2
The integers are -49 and -47.
Check

-49 + -47 = -96
 47
Example 9 Find three consecutive even integers whose sum is -42.
Assign Labels.
Let f = first number
Let f + 2 = second number
Let f + 4 = third number
first # + second # + third # = total
f + ( f +2) + ( f + 4) = -42
Algebraic Model.
f  f  2  f  4  42
Solve.
3f  6  42
6
6
f2
3f  48
 16  2
3
3
f  16
 14
Verbal Model.
Sentence.
The integers are -16, -14 and -12.
Check

-16 + (-14) + (-12) = -42
f4
 16  4
 12
2-A6 Pages 95-97 #11-26,32-38,47.