Transcript Course 3

Solving Equations with
11-3 Variables on Both Sides
Warm Up
Solve.
1. 2x + 9x – 3x + 8 = 16
2. –4 = 6x + 22 – 4x
3. 2 + x = 5 1
7 7
7
4. 9x – 2x = 3 1
16 4
8
Course 3
Solving Equations with
11-3 Variables on Both Sides
Learn to solve equations with variables
on both sides of the equal sign.
Determine if the equation has a
solution and the type of solution.
Course 3
Solving Equations with
11-3 Variables on Both Sides
Some problems produce equations that have
variables on both sides of the equal sign.
Solving an equation with variables on both
sides is similar to solving an equation with a
variable on only one side. You can add or
subtract a term containing a variable on both
sides of an equation.
Course 3
Solving Equations with
11-3 Variables on Both Sides
Additional Example 1A: Solving Equations with
Variables on Both Sides
Solve.
4x + 6 = x
4x + 6 = x
– 4x
– 4x
6 = –3x
6 = –3x
–3
–3
–2 = x
Course 3
Subtract 4x from both sides.
Divide both sides by –3.
Solving Equations with
11-3 Variables on Both Sides
Helpful Hint
Check your solution by substituting the value
back into the original equation. For example,
4(-2) + 6 = -2 or -2 = -2.
Course 3
Solving Equations with
11-3 Variables on Both Sides
Additional Example 1B: Solving Equations with
Variables on Both Sides
Solve.
9b – 6 = 5b + 18
9b – 6 = 5b + 18
– 5b
– 5b
Subtract 5b from both sides.
4b – 6 = 18
+6 +6
4b = 24
4b = 24
4
4
b=6
Course 3
Add 6 to both sides.
Divide both sides by 4.
Solving Equations with
11-3 Variables on Both Sides
Additional Example 1C: Solving Equations with
Variables on Both Sides
Solve.
9w + 3 = 9w + 7
9w + 3 = 9w + 7
– 9w
– 9w
3≠
Subtract 9w from both sides.
7
No solution. There is no number that can be
substituted for the variable w to make the
equation true.
Course 3
Solving Equations with
11-3 Variables on Both Sides
Helpful Hint
if the variables in an equation are eliminated
and the resulting statement is false, the
equation has no solution.
Course 3
Solving Equations with
11-3 Variables on Both Sides
Check It Out: Example 1A
Solve.
5x + 8 = x
5x + 8 = x
– 5x
– 5x
8 = –4x
8 = –4x
–4
–4
–2 = x
Course 3
Subtract 5x from both sides.
Divide both sides by –4.
Solving Equations with
11-3 Variables on Both Sides
Check It Out: Example 1B
Solve.
3b – 2 = 2b + 12
3b – 2 = 2b + 12
– 2b
– 2b
Subtract 2b from both sides.
b–2=
+2
b
=
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12
+ 2 Add 2 to both sides.
14
Solving Equations with
11-3 Variables on Both Sides
Check It Out: Example 1C
Solve.
3w + 1 = 3w + 8
3w + 1 = 3w + 8
– 3w
– 3w
1≠
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.
Course 3
Solving Equations with
11-3 Variables on Both Sides
To solve multi-step equations with variables on
both sides, first combine like terms and clear
fractions. Then add or subtract variable terms
to both sides so that the variable occurs on
only one side of the equation. Then use
properties of equality to isolate the variable.
Course 3
Solving Equations with
11-3 Variables on Both Sides
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 Combine like terms.
+ 2z
+ 2z
Add 2z to both sides.
8z – 15
+ 15
8z
8z
8
z
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=
=8
= 8
8
=1
–7
+15 Add 15 to both sides.
Divide both sides by 8.
Solving Equations with
11-3 Variables on Both Sides
Additional Example 2B: Solving Multi-Step Equations
with Variables on Both Sides
y
3y 3
7
+
–
=y–
5
5
10
4
y
3y
7
+
– 3 =y–
5
5
10
4
y
3y
7
3
20 5 + 5 –
= 20 y – 10 Multiply by the LCD,
4
20.
7
y
3
3y
20 5 + 20 5
– 20 4 = 20(y) – 20 10
(
()
)
( )
(
)
()
( )
4y + 12y – 15 = 20y – 14
16y – 15 = 20y – 14
Course 3
Combine like terms.
Solving Equations with
11-3 Variables on Both Sides
Additional Example 2B Continued
16y – 15 = 20y – 14
– 16y
– 16y
–15 = 4y – 14
+ 14
–1 = 4y
–1 = 4y
4
4
–1= y
4
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+ 14
Subtract 16y from both
sides.
Add 14 to both sides.
Divide both sides by 4.
Solving Equations with
11-3 Variables on Both Sides
Check It Out: Example 2A
Solve.
12z – 12 – 4z = 6 – 2z + 32
12z – 12 – 4z = 6 – 2z + 32
8z – 12 = –2z + 38 Combine like terms.
+ 2z
+ 2z
Add 2z to both sides.
10z – 12 =
38
+ 12
+12 Add 12 to both sides.
10z = 50
10z = 50
Divide both sides by 10.
10
10
z=5
Course 3
Solving Equations with
11-3 Variables on Both Sides
Check It Out: Example 2B
y
5y 3
+
+
=y–
4
6
4
y
5y
+
+3 =y–
4
6
4
y
5y
+
+3
= 24
4
6
4
6
8
6
8
by the LCD,
(
) (
) Multiply
24.
6
y
3
5y
24(4 ) + 24( 6 )+ 24( 4)= 24(y) – 24( 8 )
24
6
y–
8
6y + 20y + 18 = 24y – 18
26y + 18 = 24y – 18
Course 3
Combine like terms.
Solving Equations with
11-3 Variables on Both Sides
Check It Out: Example 2B Continued
26y + 18 = 24y – 18
– 24y
– 24y
2y + 18 =
– 18
– 18
– 18
2y = –36
2y = –36
2
2
y = –18
Course 3
Subtract 24y from both
sides.
Subtract 18 from both
sides.
Divide both sides by 2.
Solving Equations with
11-3 Variables on Both Sides
Additional Example 3: Business Application
Daisy’s Flowers sell 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.
Course 3
Solving Equations with
11-3 Variables on Both Sides
Additional Example 3 Continued
39.95 + 2.95r = 26.00 + 4.50r
– 2.95r
39.95
– 2.95r
=
– 26.00
13.95
Subtract 2.95r from
both sides.
26.00 + 1.55r
Subtract 26.00 from
both sides.
– 26.00
=
Let r represent the
price of one rose.
1.55r
13.95
1.55r
Divide both sides by
=
1.55
1.55
1.55.
9=r
The two services would cost the same when
purchasing 9 roses.
Course 3
Solving Equations with
11-3 Variables on Both Sides
Check It Out: Example 3
Marla’s Gift Baskets sell 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 gift basket companies
muffin baskets cost the same price.
Course 3
Solving Equations with
11-3 Variables on Both Sides
Check It Out: Example 3 Continued
22.00 + 2.25b = 16.00 + 3.00b
– 2.25b
22.00
– 2.25b
Let b represent the
price of one balloon.
Subtract 2.25b from
both sides.
= 16.00 + 0.75b
– 16.00
– 16.00
6.00
=
6.00
.75
0.75b
Subtract 16.00 from
both sides.
0.75b
Divide both sides by
0.75
0.75.
8=b
The two services would cost the same when purchasing
8 balloons.
Course 3
=
Solving Equations with
11-3 Variables on Both Sides
Additional Example 4: Multi-Step Application
Jamie spends the same amount of
money each morning. On Sunday, he
bought a newspaper for $1.25 and also
bought two doughnuts. On Monday, he
bought a newspaper for fifty cents and
bought five doughnuts. On Tuesday, he
spent the same amount of money and
bought just doughnuts. How many
doughnuts did he buy on Tuesday?
Course 3
Solving Equations with
11-3 Variables on Both Sides
Additional Example 4 Continued
First solve for the price of one doughnut.
Let d represent the price
1.25 + 2d = 0.50 + 5d of one doughnut.
– 2d
– 2d Subtract 2d from both sides.
1.25
= 0.50 + 3d
Subtract 0.50 from both
– 0.50
– 0.50
sides.
0.75
=
3d
Course 3
0.75 = 3d
3
3
Divide both sides by 3.
0.25 = d
The price of one
doughnut is $0.25.
Solving Equations with
11-3 Variables on Both Sides
Additional Example 4 Continued
Now find the amount of money Jamie spends each
morning.
Choose one of the original
1.25 + 2d
expressions.
1.25 + 2(0.25) = 1.75 Jamie spends $1.75 each
morning.
Find the number of doughnuts Jamie buys on Tuesday.
Let n represent the
0.25n = 1.75
number of doughnuts.
0.25n 1.75
Divide both sides by 0.25.
0.25 = 0.25
n = 7; Jamie bought 7 doughnuts on Tuesday.
Course 3
Solving Equations with
11-3 Variables on Both Sides
Check It Out: Example 4
Helene walks the same distance every
day. On Tuesdays and Thursdays, she
walks 2 laps on the track, and then
walks 4 miles. On Mondays,
Wednesdays, and Fridays, she walks 4
laps on the track and then walks 2
miles. On Saturdays, she just walks
laps. How many laps does she walk on
Saturdays?
Course 3
Solving Equations with
11-3 Variables on Both Sides
Check It Out: Example 4 Continued
First solve for distance around the track.
Let x represent the distance
2x + 4 = 4x + 2
around the track.
– 2x
– 2x
Subtract 2x from both sides.
4 = 2x + 2
–2
–2
2 = 2x
2 = 2x
2
2
1=x
Course 3
Subtract 2 from both sides.
Divide both sides by 2.
The track is 1 mile
around.
Solving Equations with
11-3 Variables on Both Sides
Check It Out: Example 4 Continued
Now find the total distance Helene walks each day.
2x + 4
Choose one of the original
expressions.
2(1) + 4 = 6
Helene walks 6 miles each day.
Find the number of laps Helene walks on Saturdays.
1n = 6
n=6
Let n represent the
number of 1-mile laps.
Helene walks 6 laps on Saturdays.
Course 3
Solving Equations with
11-3 Variables
Insert Lesson
Title
Here
on Both
Sides
Try This!
Solve.
1. 4x + 16 = 2x
2. 8x – 3 = 15 + 5x
3. 2(3x + 11) = 6x + 4
1
1
4. 4 x = 2 x – 9
5. An apple has about 30 calories more than an
orange. Five oranges have about as many calories
as 3 apples. How many calories are in each?
Course 3
Homework:
Workbook pg. 85