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Solving
Equations
withwith
Solving
Equations
10-3
10-3Variables on Both Sides
Variables on Both Sides
Warm Up
Lesson Presentation
Pre-Algebra
Pre-Algebra
Solving Equations with
10-3 Variables on Both Sides
Warm Up
Solve.
1. 2x + 9x – 3x + 8 = 16
x=1
2. –4 = 6x + 22 – 4x x = -13
3. 2 + x = 5 1 x = 34
7 7
7
4. 9x – 2x = 3 1
16 4
8
Pre-Algebra
x = 50
Solving Equations with
10-3 Variables on Both Sides
Learn to solve equations with variables on
both sides of the equal sign.
Pre-Algebra
Solving Equations with
10-3 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.
Pre-Algebra
Solving Equations with
10-3 Variables on Both Sides
Additional Example 1A: Solving Equations with
Variables on Both Sides
Solve.
A. 4x + 6 = x
4x + 6 = x
– 4x
– 4x
6 = –3x
6 = –3x
–3
–3
–2 = x
Pre-Algebra
Subtract 4x from both sides.
Divide both sides by –3.
Solving Equations with
10-3 Variables on Both Sides
Additional Example 1B: Solving Equations with
Variables on Both Sides
Solve.
B. 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
Pre-Algebra
Add 6 to both sides.
Divide both sides by 4.
Solving Equations with
10-3 Variables on Both Sides
Additional Example 1C: Solving Equations with
Variables on Both Sides
Solve.
C. 9w + 3 = 5w + 7 + 4w
9w + 3 = 5w + 7 + 4w
9w + 3 = 9w + 7
Combine like terms.
– 9w
– 9w
Subtract 9w from both sides.
3≠
7
No solution. There is no number that can be
substituted for the variable w to make the
equation true.
Pre-Algebra
Solving Equations with
10-3 Variables on Both Sides
Try This: Example 1A
Solve.
A. 5x + 8 = x
5x + 8 = x
– 5x
– 5x
8 = –4x
8 = –4x
–4
–4
–2 = x
Pre-Algebra
Subtract 4x from both sides.
Divide both sides by –4.
Solving Equations with
10-3 Variables on Both Sides
Try This: Example 1B
Solve.
B. 3b – 2 = 2b + 12
3b – 2 = 2b + 12
– 2b
– 2b
Subtract 2b from both sides.
b–2=
+2
b
=
Pre-Algebra
12
+ 2 Add 2 to both sides.
14
Solving Equations with
10-3 Variables on Both Sides
Try This: Example 1C
Solve.
C. 3w + 1 = 10w + 8 – 7w
3w + 1 = 10w + 8 – 7w
3w + 1 = 3w + 8
Combine like terms.
– 3w
– 3w
Subtract 3w from both sides.
1≠
8
No solution. There is no number that can be
substituted for the variable w to make the
equation true.
Pre-Algebra
Solving Equations with
10-3 Variables on Both Sides
To solve multistep 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.
Pre-Algebra
Solving Equations with
10-3 Variables on Both Sides
Additional Example 2A: Solving Multistep Equations
with Variables on Both Sides
Solve.
A. 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
Pre-Algebra
=
=8
= 8
8
=1
–7
+15 Add 15 to both sides.
Divide both sides by 8.
Solving Equations with
10-3 Variables on Both Sides
Additional Example 2B: Solving Multistep Equations
with Variables on Both Sides
B. y + 3y – 3 = y – 7
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
(
) (
)
7
y
3
3y
20(5 ) + 20( 5 ) – 20(4 )= 20(y) – 20( 10)
4y + 12y – 15 = 20y – 14
16y – 15 = 20y – 14
Pre-Algebra
Combine like terms.
Solving Equations with
10-3 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
Pre-Algebra
+ 14
Subtract 16y from both
sides.
Add 14 to both sides.
Divide both sides by 4.
Solving Equations with
10-3 Variables on Both Sides
Try This: Example 2A
Solve.
A. 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
Pre-Algebra
Solving Equations with
10-3 Variables on Both Sides
Try This: Example 2B
B. y + 5y + 3 = y –
4
6
4
y
5y
+
+3 =y–
4
6
4
y
5y
3
+
+
24 4
= 24
6
4
6
8
6
8
(
) (
) Multiply by the LCD.
6
y
3
5y
24(4 ) + 24( 6 )+ 24( 4)= 24(y) – 24( 8 )
6
y–
8
6y + 20y + 18 = 24y – 18
26y + 18 = 24y – 18
Pre-Algebra
Combine like terms.
Solving Equations with
10-3 Variables on Both Sides
Try This: Example 2B Continued
26y + 18 = 24y – 18
– 24y
– 24y
2y + 18 =
– 18
– 18
– 18
2y = –36
2y = –36
2
2
y = –18
Pre-Algebra
Subtract 24y from both
sides.
Subtract 18 from both
sides.
Divide both sides by 2.
Solving Equations with
10-3 Variables on Both Sides
Additional Example 3: Consumer 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?
Pre-Algebra
Solving Equations with
10-3 Variables on Both Sides
Additional Example 3 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
0.75 = 3d
3
3
Divide both sides by 3.
0.25 = d
The price of one
doughnut is $0.25.
Pre-Algebra
Solving Equations with
10-3 Variables on Both Sides
Additional Example 3 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.
Pre-Algebra
Solving Equations with
10-3 Variables on Both Sides
Try This: Example 3
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?
Pre-Algebra
Solving Equations with
10-3 Variables on Both Sides
Try This: Example 3 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
Pre-Algebra
Subtract 2 from both sides.
Divide both sides by 2.
The track is 1 mile
around.
Solving Equations with
10-3 Variables on Both Sides
Try This: Example 3 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.
Pre-Algebra
Solving Equations with
10-3 Variables on Both Sides
Lesson Quiz
Solve.
1. 4x + 16 = 2x x = –8
2. 8x – 3 = 15 + 5x x = 6
3. 2(3x + 11) = 6x + 4 no solution
1
1
x = 36
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?
An orange has 45 calories. An apple
has 75 calories.
Pre-Algebra