Transcript 127 LCD

Week 9, Day Two
HW # 32- Videos for section 3-4 AND p. 132 # 1-11 all
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keyword: MT8CA 3-4
Warm up
a) 5𝑛 + 3𝑛 − 𝑛 + 5 = 26
b) −81 = 7𝑘 + 19 + 3𝑘
c) 37 = 15𝑎 − 5𝑎 − 3
d) Lydia rode 243 miles in a three-day bike trip. On the first
day, Lydia rode 67 miles. On the second day, she rode 92
miles. How many miles per hour did she average on the
third day if she rode for 7 hours?
http://www.mvschools.org/Page/2248
Warm Up Response
a) 5𝑛 + 3𝑛 − 𝑛 + 5 = 26 n=3
b) −81 = 7𝑘 + 19 + 3𝑘 k= -10
c) 37 = 15𝑎 − 5𝑎 − 3 a=4
d) Lydia rode 243 miles in a three-day bike trip. On the
first day, Lydia rode 67 miles. On the second day, she
rode 92 miles. How many miles per hour did she
average on the third day if she rode for 7 hours?
12 miles/hour
Homework Videos
Homework Check
p. 126 # 12-23 all AND
12) n=3
22) b=4
13) k=-10 23) x=-12
14)c=2
15)w=3
16)a=4
17)y=5
18)p=22
19) h=6
20)g=3
21) m=2
p. 127 # 41-48
41) C
42) 92
43) n=-3
44) y=-2
45) x=121
46) 15
47) 6t+3k-15
48) 5a-b+4
p. 126 # 12-23 all
20.)
AND
22.)
p. 127 # 41-48
• Solving Equations with Variables on Both Sides
(3-4) {AltOpener 3-4 Solv Eq w Variab on Both
Sides.pdf}
• Simplifying Expressions #3 Worksheet
Slides for extra practice at home
• Note, some of the formatting may be off and
the ppt moves from Mac to PC.
Additional Example 1A: Solving Equations with Variables on Both Sides
Solve.
4x + 6 = x
4x + 6 = x
– 4x
To collect the variable terms on one side, subtract 4x
from both sides.
– 4x
6 = –3x
6
–3x
=
–3
–3
–2 = x
Since x is multiplied by -3, divide both sides by –3.
Helpful Hint
You can always check your solution by substituting the value back into the
original equation.
Additional Example 1B: Solving Equations with Variables on Both Sides
Solve.
9b – 6 = 5b + 18
9b – 6 = 5b + 18
– 5b
– 5b
To collect the variable terms on one side, subtract 5b from
both sides.
4b – 6 = 18
+6 +6
Since 6 is subtracted from 4b, add 6 to both
sides.
4b = 24
4b
24
=
4
4
b=6
Since b is multiplied by 4, divide both sides by 4.
Additional Example 1C: Solving Equations with Variables on Both Sides
Solve.
9w + 3 = 9w + 7
9w + 3 = 9w + 7
– 9w
– 9w
3≠
To collect the variable terms on one side, subtract
9w from both sides.
7
There is no solution. There is no number that can be substituted for the variable w to make
the equation true.
Helpful Hint
if the variables in an equation are eliminated and the resulting statement is
false, the equation has no solution.
Check It Out! Example 1A
Solve.
5x + 8 = x
5x + 8 = x
– 5x
– 5x
To collect the variable terms on one side, subtract 5x
from both sides.
8 = –4x
8
–4x
=
–4
–4
–2 = x
Since x is multiplied by –4, divide both sides by –4.
Check It Out! Example 1B
Solve.
3b – 2 = 2b + 12
3b – 2 = 2b + 12
– 2b
– 2b
b–2=
+2
b
=
To collect the variable terms on one side,
subtract 2b from both sides.
12
+2
14
Since 2 is subtracted from b, add 2 to both
sides.
Check It Out! Example 1C
Solve.
3w + 1 = 3w + 8
3w + 1 = 3w + 8
– 3w
– 3w
1≠
To collect the variable terms on one side, subtract
3w from both sides.
8
No solution. There is no number that can be substituted for the variable w to
make the equation true.
To solve more complicated equations, you may need to first simplify by
combining like terms or clearing fractions. Then add or subtract to collect
variable terms on one side of the equation. Finally, use properties of equality
to isolate the variable.
Additional Example 2A: Solving Multi-Step Equations with Variables
on Both Sides
Solve.
10z – 15 – 4z = 8 – 2z – 15
10z – 15 – 4z = 8 – 2z – 15
6z – 15 = –2z – 7
+ 2z
Combine like terms.
+ 2z
8z – 15 =
+ 15
Add 2z to both sides.
–7
+15
Add 15 to both sides.
8z = 8
8z
8
8
=
z=1
8
Divide both sides by 8.
Additional Example 2B: Solving Multi-Step Equations with Variables
on Both Sides
y
5
+
y
5
y
5
3y
3
–5 = y –
4
+
3y
3
–5 = y –
4
3y
–5
= 20
(
) ( )
() ( ) ( )
20
20
+
y
5+ 20
7
10
7
10
7
10
3
y–
4
Multiply by the LCD, 20.
7
10
( )
3
3y
– 20 5 = 20(y) – 20 4
4y + 12y – 15 = 20y – 14
16y – 15 = 20y – 14
Combine like terms.
Additional Example 2B Continued
16y – 15 = 20y – 14
– 16y
– 16y
Subtract 16y from both sides.
–15 = 4y – 14
+ 14
+ 14
Add 14 to both sides.
–1 = 4y
–1
=
4
–1
=y
4
4y
4
Divide both sides by 4.
Check It Out! Example 2A
Solve.
12z – 12 – 4z = 6 – 2z + 32
12z – 12 – 4z = 6 – 2z + 32
8z – 12 = –2z + 38
+ 2z
Combine like terms.
+ 2z
10z – 12 =
+ 12
Add 2z to both sides.
38
+12
Add 12 to both sides.
10z = 50
10z
10
50
=
z=5
10
Divide both sides by 10.
Check It Out! Example 2B
y
4
y
4
(
24
y
4
+
5y
3
+6 = y –
4
6
8
+
5y
3
+6 = y –
4
6
8
5y
+6
= 24
) ( )
() ( ) ( )
24
+
y
4+ 24
6
8
3
y–
4
Multiply by the LCD, 24.
6
8
( )
5y
+ 24 6 = 24(y) – 24
3
4
6y + 20y + 18 = 24y – 18
26y + 18 = 24y – 18
Combine like terms.
Check It Out! Example 2B Continued
26y + 18 = 24y – 18
– 24y
– 24y
2y + 18 =
Subtract 24y from both sides.
– 18
– 18
– 18
Subtract 18 from both sides.
2y = –36
2y
=
2
–36
2
y = –18
Divide both sides by 2.
Additional Example 3: Business Application
Daisy’s Flowers sells a rose bouquet for $39.95 plus $2.95 for every rose. A
competing florist sells a similar bouquet for $26.00 plus $4.50 for every rose. Find
the number of roses that would make both florists' bouquets cost the same price.
What is the price?
Write an equation for each service. Let c represent the total cost and r represent
the number of roses.
total cost is flat fee plus cost for each rose
Daisy’s: c
= 39.95 + 2.95
r
Other:
= 26.00 +
r
c
4.50
Additional Example 3 Continued
Now write an equation showing that the costs are equal.
39.95 + 2.95r = 26.00 + 4.50r
– 2.95r
39.95
– 26.00
13.95
– 2.95r
Subtract 2.95r from both
sides.
= 26.00 + 1.55r
– 26.00
=
1.55r
13.95
1.55
=
1.55r 1.55
Subtract 26.00 from both
sides.
Divide both sides by 1.55.
9=r
The two bouquets from either florist would cost the same when
purchasing 9 roses.
Additional Example 3 Continued
To find the cost, substitute 9 for r into either equation.
Daisy’s:
Other florist:
c = 39.95 + 2.95r
c = 26.00 + 4.50r
c = 39.95 + 2.95(9)
c = 26.00 + 4.50(9)
c = 39.95 + 26.55
c = 26.00 + 40.50
c = 66.5
c = 66.5
The cost for a bouquet with 9 roses at either florist is $66.50.
Check It Out! Example 3
Marla’s Gift Baskets sells a muffin basket for $22.00 plus $2.25 for every balloon. A
competing service sells a similar muffin basket for $16.00 plus $3.00 for every balloon.
Find the number of balloons that would make both baskets cost the same price.
Write an equation for each service. Let c represent the total cost and b represent the
number of balloons.
total cost is flat fee plus cost for each balloon
Marla’s: c
Other:
c
= 22.00
= 16.00
+ 2.25
+ 3.00
b
b
Check It Out! Example 3 Continued
Now write an equation showing that the costs are equal.
22.00 + 2.25b = 16.00 + 3.00b
– 2.25b
22.00
– 16.00
6.00
– 2.25b
Subtract 2.25b from both
sides.
= 16.00 + 0.75b
– 16.00
=
6.00
0.75
=
0.75b
0.75b 0.75
Subtract 16.00 from both
sides.
Divide both sides by 0.75.
8=b
The two services would cost the same when purchasing a muffin
basket with 8 balloons.