Download 2-5B Types of Solutions

Teacher’s please delete personal pictures in slides
6, 7, and 8.
How do you know when to give a decimal answer?
The instructions will tell you what decimal position you
will need to round.
Otherwise, if dividing does NOT produce a positive or
negative whole number (known as an integer), then
leave the answer fractional, but in lowest terms.

36
 6
6

36
 5.142857143...
7
2-5 Solving Equations with the
Variable on Each Side
2-A10 Pages 101–103 #31–34,37–42,49 and
PRACTICE Wkbk. Page 14 #14–24.
(Complete PRACTICE Workbook problems on binder paper.)
Algebra 1
Glencoe McGraw-Hill
Linda Stamper
STEPS FOR SOLVING LINEAR EQUATIONS
1. Simplify each side by distributing and/or combining
like terms.
2. Collect variable terms on the side where the
coefficient is greater.
3. Isolate the variable using inverse operations.
4. Check your solution in the original equation.
Solve.
Write problem.
Change subtraction
to addition.
Distribute and
combine like terms.
Collect variable
terms on one side.
Isolate the variable using
inverse operations.
Copy in your
spiral
notebook!
1
21x  18  23 - 2x - 5
3
1
21x  18  23 +––2x +–– 5
3
7x  6  23   2x  10
7x  6  2x  33
 2x
 2x
9x  6  33
6 6
9x  27
9
9
x 3
Please take a highlighter or red pencil and draw a
line through the equal signs of the problem you just
copied. Do you have a straight line?
Example 1
1
42r  8  49r  70
7
8r  32  7r  10
 7r
 7r
r  32  10
 32  32
r  42
Solve.
Example 2
2
 24d  27   8 d  4   6
3
2
 24d   27  8  d   4   6
3
 16d  18  8d  32   6
 16d  18  8d  26
 16d
 16d



18  24d  26
 26
 26
 8  24d
24 24
8

d
24
1
 d
3

Linear equations (equations that have a variable to the
first power) can have one solution, no solution, or it is an
identity and all real numbers are solutions.
The equation 2x = 2x is called an identity.
One side is identical to the other side.
Solve the equation, if possible. Determine if the equation
has no solution, all real numbers are solutions, or one
solution (if the equation has one solution, give the
solution).
x  12  40
 12  12
x  28
3x  12  3x
 3x
 3x
12  0
no solution
3x  2  3x  6
3x  6  3x  6
 3x
 3x
66
all real numbers
True statement
means
all equation
real numbers are solutions.
This
type of
is called an Identity!
False statementThe
means
no solution.
variable
can
represent all real
numbers.
Solve the equation, if possible. Determine if the equation
has no solution, all real numbers are solutions, or one
solution (if the equation has one solution, give the
solution).
Example 3
5x  3  3x  15  2x
5x  15  5x  15
 5x
 5x
15  15
Example 4
5x  3  3x  10  2x
5x  15  5x  10
 5x
 5x
15  10
all real numbers
no solution
Example 5
5x  3  3x  5  x
5x  15  4x  5
 4x
 4x
x  15  5
 15  15
x  10
Solve the equation, if possible. Determine if the equation
has no solution, all real numbers are solutions, or one
solution (if the equation has one solution, give the
solution).
Example 6
1
62m  7   18  12m Example 7
3
12m  42  6  4m 85c  2  1032  4c
 4m
 4m 40c  16  320  40c
Example 8
8m  42  6
 40c
 40c
1
 42  42


20y  400
4
y

20

16  320
5
8m  48
no solution
4 y  80  4 y  80
8 8
 4y
 4y
m6
80  80
all real numbers
Example 9 Find the value of x so that the figures have the
same perimeter.
5x + 1
x+4
2x
2x + 5
x+9
P = a+ b + c
P = 2l + 2w
P  x  4   2x  5  5x  1
P  2x  9  22x 
2x  9  22x   x  4   2x  5  5x  1
2x  18  4x  x  4  2x  5  5x  1
6xformula
 18  8xfor
 10each geometric figure.
Write the perimeter
 6x
 6x
Substitute.
18
 2x  10 then . . .
If perimeter is equal to
perimeter
 10
 10
8  2x
4x
2-A10 Pages 101–103 #31–34,37–42,49 and
PRACTICE Wkbk. Page 14 #14–24.
(Complete PRACTICE Workbook problems on binder paper.)