Download SOLVING EQUATIONS W/ FRACTIONS USING LCM!

SOLVING EQUATIONS
W/ FRACTIONS USING
LCM!
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…
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Ex 1: Solve the equation
•  Solve
3
7
a −1 = a + 9
11
11
"3
% "7
%
11$ a −1' = $ a + 9 '11
# 11
& # 11
&
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
! 2x $ ! 4 $
3# & = # & 3
" 3 % " 3%
2x = 4
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!
x=2
Note: We could also have solved this problem by multiplying both sides by
3/2, the reciprocal of 2/3!
Pt. 1 Practice…
•  1.
•  3.
3
7
x +1 =
4
4
2.
1
3
x + 3 = x −1
7
7
5
15
x=−
8
8
4.
9
1
2x + = 3+ x
4
4
1) x=1; 2) x=14 3) x= -3 4) x= 3/7
Ex 2: Different Denominators
•  Solve
2 1 1
− b+ =
3 9 18
" 2 1% " 1 %
18 $ − b + ' = $ '18
# 3 9 & # 18 &
−12b + 2 =1
−12b = −1
−12b −1
=
−12 −12
b=
1
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!
Ex 2B: Different Denominators
•  Solve
−3 2
= x
4 5
" 3% "2 %
20 $ − ' = $ x ' 20
# 4& #5 &
−15 = 8x
−15 8
= x
8 8
−15
=x
8
4 and 5 have a LCM of 20, so
multiply each side by 20!
Ex 2C: Different Denominators
•  Solve
x +8 4+ x
=
3
4
! x +8$ ! 4 + x $
12 #
&=#
&12
" 3 % " 4 %
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!
Pt. 2 Practice
•  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
x − 8 15
=
12
3
6.
x − 7 2x + 3
=
4
2
•  5.
1) 71/4
2) 10/3
3) -3/4
4) 10/17 5) 68
6) -13/3
Ex 3: 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
Distributing the 2/3 would give
us more fractions…
We can undo the fraction by
multiplying by 3, then only have
to distribute the 2!
Ex 3: OR……
•  Solve
2
( x + 4) =16
3
! 3$2
! 3$
# & ( x + 4) =16 # &
"2%3
"2%
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!
Pt. 3 Practice…
1
( x + 4) =10
4
2)
3) 1 10x −10 = 5x −17
(
)
4)
1)
5
1) 36,
2) 7,
3) 5,
4) 0
3 1
= (10 − x )
2 2
2
1
(10x + 6) = (15x +12)
3
3
Ex 4: Variables in Denominators
•  Solve
2 −8
=
3 x
! 2 $ ! −8 $
3x # & = # & 3x
" 3% " x %
2x = −24
2x −24
=
2
2
x = −12
The Denominators 3 and x can
be eliminated by multiplying both
sides each one!
Ex 4B: Variables in Denominators?
•  Solve
x, 3x and 9 have a LCM of 9x...
2 4 2
− =
Multiply by 9x! Clear those demoninators!
x 3x 9
"2 4 % "2%
9x $ − ' = $ ' 9x
# x 3x & # 9 &
18 −12 = 2x
6 = 2x
3= x
Ex 4C: Variables in Denominators
•  Solve
9
−63
=
x −1 28
" 9 % " −63 %
( x −1) (28)$# '& = $# '& ( x −1) (28)
x −1
28
(28) ( 9) = (−63) ( x −1)
252 = −63x + 63
189 = −63x
−3 = x
The Denominators 28 and x – 1
can be eliminated by multiplying
both sides each one!
Pt. 4 Practice…
1)
4 −12
=
5
x
2)
2
−26
=
x − 5 65
3)
3 12
− =
4 2x
4)
3
−18
=
x + 4 −5x − 30
5)
5
17
+6 =
x
x
6)
x + 2 x −1
+
=5
3
6
1) -15
2) 0
3) -8
4) 6
5) 2
6) 9
Boom!
•  Solve
2 − x x + 3 −1
−
=
x
3x
3