Class notes
Lecture notes
- Course
- 201-015-RE
- Institution
- Dawson College ( )
Lecture notes of 3 pages for the course 201-015-RE at (Sec v math)
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Equations Containing fractions-
Such
equations involve ✗ terms -
in fractions /denominators ,
so
in order to solve
for × we need to free it from the denominator
"
clearing by
"
by denominators ,
i.e
, multiply the common
denominator on both sides to cancel all denominators .
-
AB : The
original equation has restrictions
can't which
we ÷
by 0, so
any
✗ value would cause
-
division
by 0 must be
rejected as a
potential solution to the
need to check over solutions in the
equation .
Therefore ,
we either
original equation or note the excluded values at the
beginning .
Eg Solve the equation
:
Clear fractions
A. Xtl - ✗ = 1 :
multiply by
✗ Xts ✗1-5 common denominator ✗ 1×1-5)
✗ 1×+5) ✗+1 _
✗ = ✗ 1×+5 ) 1-
✗ ✗+5 its
✗lxt 5) 1×+1) -
✗ tx☒x = ✗ 1×1/-5)
* xts its
1×+571×+1) -
xʰ=✗
✗21-6×1-5 -
x2 -
✗ = 0
5×+5=0
SX = -5 ✗ =
-
I
B I -
12 = 3
✗2-4 Xt2
Factor denominator : I -
12 = 3
1×-2/1×+21 ✗-2
Multiply both sides by
common den tx- 2) 1×+21 : I 1×-211×+21 -
121×1-211×+121=31×-211*21
1×1-211×421 ¥2 only
✗2- 4 12=3×-6 >
✗ -5=0--311--5 solution
(11-5/1×+2)=0 {
-
2-
✗ 3×1-16+6=0
Rejected)
(cause of's by 0
✗ 1-2--0=31,1*2
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