Transcript Course 3

Solving Equations with
11-3 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
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x = 50
Solving Equations with
11-3 Variables on Both Sides
Problem of the Day
An equilateral triangle and a regular
pentagon have the same perimeter.
Each side of the pentagon is 3 inches
shorter than each side of the triangle.
What is the perimeter of the triangle?
22.5 in.
Course 3
Solving Equations with
11-3 Variables on Both Sides
TB P. 593-597
Learn to solve equations with variables on
both sides of the equal sign.
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
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
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
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
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
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
Course 3
=
=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
y
3y 3
7
+
–
=y–
5
5
10
4
y
3y
7
+
– 3 =y–
5
5
10
4
y
3y
7
3
+
–
= 20 y – 10
5
5
4
by the LCD,
(
) (
) Multiply
20.
7
y
3
3y
20(5 ) + 20( 5 ) – 20(4 )= 20(y) – 20( 10)
20
4y + 12y – 15 = 20y – 14
16y – 15 = 20y – 14
Course 3
Combine like terms.
Solving Equations with
11-3 Variables on Both Sides
16y – 15 = 20y – 14
– 16y
– 16y
–15 = 4y – 14
+ 14
–1 = 4y
–1 = 4y
4
4
–1= y
4
Course 3
+ 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
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
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
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
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
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
Insert Lesson
Title
Here
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.
Course 3