Transcript 3.5 Solving Equations w/ Fractions

3.5 SOLVING
EQUATIONS W/
FRACTIONS
SWBAT:
Solve Equations with Fractions.
Fractions? Oh No!
• Equations involving fractions can be quite difficult to work
with.
• There are easy ways to change the way a problem looks
by using Multiplication and LCM’s!
What is an LCM again?
• It’s ok if you forgot…
CLICK HERE TO WATCH THIS VIDEO!!!
Ex 1: Solve the equation
• Solve
3
7
a -1 = a + 9
11
11
æ3
ö æ7
ö
11ç a -1÷ = ç a + 9 ÷11
è 11
ø è 11
ø
3a -11= 7a + 99
3a -11-3a = 7a + 99 -3a
-11= 4a + 99
-11- 99 = 4a + 99 - 99
-110 = 4a
-27.5 = a
Since both fractions have a
denominator of 11, multiply each
side by 11!
Distribute the 11 to both sides!
Now our equation is something
more familiar!
We can solve this!
Lets undo operations!
Ex 1B: Solve the equation
• Solve
2x 4
=
3 3
æ 2x ö æ 4 ö
3ç ÷ = ç ÷ 3
è 3 ø è3ø
2x = 4
x=2
Since both fractions have a
denominator of 3, multiply each
side by 3!
Now our equation is something
more familiar!
We can solve this!
Lets undo operations!
Note: We could also have solved this problem by multiplying both sides by
3/2, the reciprocal of 2/3!
You Try!
• 1.
• 3.
3
7
x +1 =
4
4
5
15
x=8
8
1) x=1; 2) x=5 3) x= -3 4) x= 3/7
2.
1
3
x + 3 = x -1
7
7
4.
9
1
2x + = 3+ x
4
4
Ex 2: Different Denominators
• Solve
2 1 1
- b+ =
3 9 18
æ 2 1ö æ 1 ö
18ç - b + ÷ = ç ÷18
è 3 9 ø è 18 ø
-12b + 2 =1
-12b + 2 - 2 =1- 2
-12b = -1
-12b -1
=
-12 -12
1
b=
12
3,9,and 18 have a LCM of 18, so
multiply each side by 18!
Distribute the 18 to both sides!
Now our equation is something
more familiar!
We can solve this!
Lets undo operations!
Ex 2B: Different Denominators
• Solve
-3 2
= x
4 5
æ 3ö æ2 ö
20 ç- ÷ = ç x ÷ 20
è 4ø è5 ø
-15 = 8x
-15 8
= x
8 8
-15
=x
8
4 and 5 have a LCM of 20, so
multiply each side by 20!
Distribute the 20 to both sides!
Now our equation is something
more familiar!
We can solve this!
Lets undo operations!
Ex 2C: Different Denominators
• Solve
x +8 4+ x
=
3
4
æ x +8ö æ 4 + x ö
12 ç
÷=ç
÷12
è 3 ø è 4 ø
4(x +8) = (4 + x)3
4x +32 =12 +3x
4x +32 -3x =12 +3x -3x
x +32 =12
x +32 -32 =12 -32
x = -20
3 and 4 have a LCM of 12, so
multiply each side by 12!
Use the 12 to undo the
denominators!
Distribute the leftover factors!
Now our equation is something
more familiar!
We can solve this!
You Try!
• 1.
1
7
x -8 =
2
8
2.
5 3
= x
8 16
• 3.
5 3
1
+ x=
8 4
16
4.
2
3
4
x + =1- x
5
7
7
• 5.
x - 8 15
=
12
3
6.
x - 7 2x + 3
=
4
2
1) 71/4
2) 10/3
3) -3/4
4) 10/17 5) 68
6) -13/3
Ex 2C: Different Denominators
• Solve
x +8 4+ x
=
3
4
æ x +8ö æ 4 + x ö
12 ç
÷=ç
÷12
è 3 ø è 4 ø
4(x +8) = (4 + x)3
4x +32 =12 +3x
4x +32 -3x =12 +3x -3x
x +32 =12
x +32 -32 =12 -32
x = -20
3 and 4 have a LCM of 12, so
multiply each side by 12!
Use the 12 to undo the
denominators!
Distribute the leftover factors!
Now our equation is something
more familiar!
We can solve this!
Ex 2C: Different Denominators
• Solve
x +8 4+ x
=
3
4
æ x +8ö æ 4 + x ö
12 ç
÷=ç
÷12
è 3 ø è 4 ø
4(x +8) = (4 + x)3
4x +32 =12 +3x
4x +32 -3x =12 +3x -3x
x +32 =12
x +32 -32 =12 -32
x = -20
3 and 4 have a LCM of 12, so
multiply each side by 12!
Use the 12 to undo the
denominators!
Distribute the leftover factors!
Now our equation is something
more familiar!
We can solve this!
Ex 3: Variables in Denominators
• Solve
2 -8
=
3 x
æ 2 ö æ -8 ö
3x ç ÷ = ç ÷ 3x
è 3ø è x ø
2x = -24
2x -24
=
2
2
x = -12
The Denominators 3 and x can
be eliminated by multiplying both
sides each one!
Ex 3B: Variables in Denominators
9
-63
=
• Solve
x -1 28
æ 9 ö æ -63 ö
( x -1) (28)çè ÷ø = çè ÷ø( x -1) (28)
x -1
28
(28) ( 9) = (-63) ( x -1)
252 = -63x + 63
252 - 63 = -63x + 63- 63
189 = -63x
189 -63x
=
-63 -63
-3 = x
The Denominators 28 and x – 1
can be eliminated by multiplying
both sides each one!
You try…
1)
4 -12
=
5
x
3 12
3) =
4 2x
1) -15, 2) 0
3) -8 4) 6
2)
4)
2
-26
=
x - 5 65
3
-18
=
x + 4 -5x - 30
Ex 4: Distribution of FRACTIONS??
• Solve
2
( x + 4) =16
3
2
(3) ( x + 4) =16(3)
3
2 ( x + 4) = 48
2x +8 = 48
2x +8-8 = 48-8
2x = 40
2x 40
=
2 2
x = 20
We can undo the fraction by
multiplying by 3, then only have
to distribute the 2!
Ex 4: OR……
• Solve
2
( x + 4) =16
3
æ3ö2
æ3ö
ç ÷ ( x + 4) =16 ç ÷
è2ø 3
è2ø
1( x + 4) = 24
x + 4 = 24
x = 20
We can also undo the fraction by
multiplying by its reciprocal!
When multiplying a fraction by
its Reciprocal, we always will get
a product of 1!
You try…
1
( x + 4) =10
4
1)
3) 1 10x -10 = 5x -17
(
)
5
1) 36,
3 1
= (10 - x )
2 2
2)
4) 2
3
2) 7,
3) 8,
4) 0
(10x + 6) =
1
(15x +12)
3