Lecture notes

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