Algebra Demystified 2nd Ed

(Marvins-Underground-K-12) #1
Chapter 7 linear equaTionS 211

Anytime we multiply (or divide) both sides of the equation by an expression
with a variable in it, we must check our solution(s) in the original equation.
When we cross-multiply, we are implicitly multiplying both sides of the equa-
tions by the denominators of each fraction, so we must check our solution in
this case as well. The reason that we must check our solution is that a solution
to the converted equation might cause a zero to be in a denominator of the
original equation. Such solutions are called extraneous solutions. Let us see what
happens in the next example.


EXAMPLES

1
2

3
2

6
22

22
xx

x
x

xx

=
+


−+

−−
()()

TheLCDis()().


()()^1 ()()
()()

xx
x

xx
x

x
x

−+

=− +
+


−+






22
2

223
2

6
22 x 

Multiply each side by the LCD.

xxx
x

xx x
xx

+= −+
+

−− +
−+

222 3
2

226
22

()(()( )
()()

)

Distribute the LCD.
xxx
xxx
xx
xx
x

+= −−
+= −−
+=−−
++
+=−

23 26
23 66
236
33
426

()

222
48
2


=−
=−

x
x

But x = −2 leads to a zero in a denominator of the original equation, so
x = −2 is not a solution to the original equation. The original equation, then,
has no solution.

EXAMPLES EXAMPLES
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