Transcript both sides - StCeciliaHonorsMath

Solve Multistep
Equations
Honors Math – Grade 8
Get Ready for the Lesson
In 2003, about 46.9 million U.S. households had
dial-up Internet service and about 26 million had
broadband service. During the next five years, it
was projected that the number of dial-up users
would decrease an average of 3 million per year
and the number of broadband users would
increase an average of 8 million per year. The
following expressions represent the number of
dial-up and broadband users x years after 2003.
Dial-up Users: 46.9 – 3x
Broadband: 26 + 8x
To find the time at which Dial-up users and
broadband users are the same number, set the
equations equal to each other.
46.9  3x  26  8x
To solve an equation the variable on
both sides, use the Addition &
Subtraction Properties of Equality to
write an equation with all the variables
on one side and the constants on the
other.
Solve each equation.
First, move one of the
variables.
Since 10p is greater than
8p, leave 10p on the left
and move 8p.
Subtract 8p from both
sides of the equation.
Add 2 to both sides of the
equation.
Divide both sides of the
equation by 2.
Answers may be left in
fractional form or written as
a decimal.
Don’t forget to check your
solution!
 2  10 p  8 p  1
 8 p  8 p
 2  2 p  1
2
 2
2p 1
2
2
1
p
2
Solve each equation.
First, move one of the
variables.
Since 7w is greater than 3w,
leave 7w on the right and
move 39.
Subtract 3p from both
sides of the equation.
Divide both sides of the
equation by 2.
Answers may be left in
fractional form or written as
a decimal.
Don’t forget to check your
solution!
3w  2  7w
 3w  3w
2  4w
4 4
1
w
2
Solve each equation.
To eliminate the
denominators, think what is
the LCD of ½ and ¼ ?
The LCD is 4.
Multiply both sides of the
equation by 4.
Distribute the 4 on each
side of the equation.
Since 2s is greater than 1s,
move 1s to the other side of
the equation.
Subtract s from both sides
of the equation.
Subtract 4 from both sides
of the equation.
Don’t forget to check your
solution!

s  
1


4
4  1 
s

6


2  
4

2s  4  s  24
s
 s
s  4  24
 4  4
s  28
Solve each equation.
If an equation contains
grouping symbols, first use
the Distributive Property to
remove the grouping
symbols.
4  2r  ? 4  8  ?
1
1
 49r  ?  70  ?
7
7
Since 8r is greater than 7r,
move 7r to the other side of
the equation.
Subtract 7r from both sides
of the equation.
Add 32 to both sides of the
equation.
1
4(2r  8)  (49r  70)
7
8r  32  7r 10
 7r
 7r
r  32  10
 32  32
r  42
Solve each equation.
If an equation contains
grouping symbols, first use
the Distributive Property to
remove the grouping
symbols.
7  n  ? 7 1  ?
 23  ?  2 n  ?
Rewrite plus a negative!
Since 7n is greater than 2n,
move 2n to the other side of
the equation.
Add 2n to both sides of the
equation.
Add 7 to both sides of the
equation.
7(n  1)  2(3  n)
7 n  7  6  (2n)
7n  7  6  2n
 2n

 2n
9n  7  6
 7  7
9n  1
9 9
1
n
9
Solve each equation.
If an equation contains
grouping symbols, first use
the Distributive Property to
remove the grouping
symbols.
3  6  ? 3  2s  ?
Since 8s is greater than 6s,
move 6s to the other side of
the equation.
Add 6s to both sides of the
equation.
Add 10 to both sides of the
equation.
8s  10  3(6  2 s )
8s 10  18  6s
 6s

14s 10  18
 10  10
14s  28
14 14
s2
 6s
Solve each equation.
Before collecting unknown
terms, you may first have to
group like terms.
Since 7v is greater than 3v,
leave 7v on the left and
move 3v.
Add 3v to both sides of the
equation.
Subtract 9 from both sides
of the equation.
Divide both sides of the
equation by 10.
Answers may be left in
fractional form or written as
a decimal. REDUCE!
Don’t forget to check your
solution!
6v  9  v  4  3v
7v  9  4  3v
 3v  3v
10v  9  4
 9  9
10v  5
1
10 10 v  
2
Solve each equation.
If an equation contains
grouping symbols, first use
the Distributive Property to
remove the grouping
symbols.
5 m  ?
5 7  ?
2m  5  5(m  7)  3m
2m  5  5m  35  3m
2m  5  2m  35
 2m
 2m
Group like terms.
Subtract 2m from both
sides of the equation.

5  35
Since 5 = -35 is a false statement, the
original equation is a contradiction.
The solution set is the null set.
Solve each equation.
If an equation contains
grouping symbols, first use
the Distributive Property to
remove the grouping
symbols.
3 r  ?
3 1  ?
Group like terms.
3(r  1)  5  3r  2
3r  3  5  3r  2
3r  2  3r  2
Since the expressions on each side of the equation are the same,
this equation illustrates the Reflexive Property of Equality. This
equation is called an identity. It is true for all values of r. The
solution set is all real numbers!