Download 6-3B Solving Multi-Step Inequalities

6-3B Solving Multi-Step
Inequalities
Algebra 1
Glencoe McGraw-Hill
Linda Stamper
Solving Multi–Step Inequalities
A multi-step inequality is solved by transforming the
inequality more than one time. Undo addition or
subtraction before undoing multiplication or division or
you may make the problem more complicated to solve.
Always remember the basic rule when isolating the
variable:
Whatever you do to one
side of the inequality
sign, you must also do to
the other side of the
inequality sign.
To solve inequalities that have variables on both sides of
the inequality symbol, first collect the variable terms on
one side.
Solve  6x  22  3x  31.
Write the problem.
Collect variable terms
on one side.
Undo addition and
simplify.
Undo multiplication
and simplify.
 6x  22  3x  31
 6x
 6x
22  3x  31
 31
 31
 9  3x
3 3
3  x
x  3
To solve inequalities that have variables on both sides of
the inequality symbol, first collect the variable terms on
one side.
Solve  6x  22  3x  31.
Write the problem.
Collect variable terms
on one side.
Undo addition and
simplify.
Undo multiplication
and simplify.
 6x  22  3x  31
 3x
 3x
 3x  22  31
 22  22
 3x /
 9
 3 < 3
x  3
When solving inequalities that contain grouping symbols,
first use the distributive property.
Solve 6x  3  3x  3x  2.
Write problem.
Distribute and
combine like terms.
Collect variable
terms on one side.
Isolate the variable using
inverse operations.
6x  3  3x  3x  2
6x  18  3x  3x  6
9x  18  3x  6
 3x
 3x
6x  18   6
 18  18
6x  24
6
6
x  4
Example 1
64  12w  6w
 12w  12w
64  18w
18 18
32
w
9
32
w
9
Solve.
Example 2
 6x  19  2x  55
 2x
 2x
 4x  19  55
 19  19
 4x 
/ 36
 4 > 4
x  9
Leave the Use good
fractionform – place
improper. the
It negative
must be inout front!
lowest
terms.
Example 3
5x  16  2x  9x  15
7 x  16  9x  15
 7x
 7x
 16  2x  15
 15
 15
 1  2x
2 2
1
 x
2
1
x
2
If solving an inequality results in a statement that is
always true, the solution is the set of all real numbers.
If solving an inequality results in a statement that is
never true, the solution is an empty set, written as .
Solve 8x  2  3x  4  5x  7   8.
Write problem.
8x  2  3x  4   5x  7   8
Distribute and
8x  16  3x  12  5x  35  8
combine like terms.
5x  28  5x  27
Collect variable
 5x
 5x
terms on one side.
28  27
Result is a false statement.

Copy in your
spiral
notebook!
Solve 3x  4  x  12  2x  12.
Write problem.
3x  4   x  12  2x  12
Distribute and
3x  12  x  12  2x  24
combine like terms.
2x  24  2x  24
Collect variable
 2x
 2x
terms on one side.
24  24
all real numbers
Result is a true statement.
Copy in your
spiral
notebook!
Solve 3x  4  x  12  2x  12.
Write problem.
3x  4   x  12  2x  12
Distribute and
3x  12  x  12  2x  24
combine like terms.
2x  24  2x  24
Collect variable
Subtract
24 from
 24
 24
termsside.
on one side.
2x  2x
each
xx
Result is a true statement.
all real numbers
Be sure to simplify the inequality to determine whether
it is a true statement. If the inequality is true, the
solution is the set of all real numbers. If the inequality
is false, the solution is the empty set, written as .
Solve.
1)  32  3b  2
 6  9b  2
6
6
 9b 
/ 8
9 > 9
8
b 
9
2)  m  81   9m  7
m
m
 81   8m  7
 7
7
 88 
/  8m
8 < 8
11  m
m  11
Solve.
3) 8x  15  4x  5x  6
12x  15  5x  6
 5x
 5x
7x  15  6
 15  15
7x  21
7
7
x  3
4) 7x  2  3x  2
7 x  2  3x  6
 3x
 3x
10x  2  6
2 2
10x  8
10 10
4
x
5
Solve.
5)  7x  4  11x  8x  22x  1
 7x  28  11x  8x  4x  2
4x  28  4x  2
 4x
 4x
 28  2

Solve.
6) 4 y  3  2y  3y  2  y  12
4 y  3  2y  3y  6  y  12
4 y  3  4 y  18
 4y
 4y
 3  18
all real numbers
Write an inequality and then solve.
7) One half of a number increased by one is greater than
or equal to six less than one fourth of the number.
1
1
n 1  n 6
2
4
2
1
n 1  n 6
4
4
1
1
 n
 n
4
4
1
n  1  6
4
1 1
 4 1
4
n


7
 
 
 1 4
1
n  28
Write an inequality and then solve.
8) Sixteen more than three fourths of a number is
more than one eighth of the number less than 2.
3
1
n  16  2  n
4
8
3
6
1
n  16  2  n
4
8
8
1
1
 n
 n
8
8
7
n  16  2
8
 16  16
-2
 8 7 n  14 8 
 
 
7
 8
7
n  16
6-A4 Pages 311–312 #26–33, 39–42, 44–47.