This document discusses solving rational equations by using the cross product property. It provides three examples of solving rational equations:
1) Solving a simple rational equation with a single rational expression on each side. The solutions are -6 and 2.
2) Solving a rational equation with rational expressions on both sides. The solution appears to be 2, but it is an extraneous solution.
3) Solving a rational equation by factoring the denominators to find the least common denominator (LCD).
This is a power point that I use to explain to students how to solve an equation where the variables are on both sides. Multiple steps using inverse operations are used to move the terms on either side of the "river."
Este archivo contiene problemas resueltos del curso de estadística y probabilidades, cada problema esta resuelto paso a paso para su mejor comprensión del lector
This is a power point that I use to explain to students how to solve an equation where the variables are on both sides. Multiple steps using inverse operations are used to move the terms on either side of the "river."
Este archivo contiene problemas resueltos del curso de estadística y probabilidades, cada problema esta resuelto paso a paso para su mejor comprensión del lector
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2. Vocabulary
Rational Equation - an equation that contains one or
more rational expressions.
When both sides of the equation are single rational
expressions, you can solve the equation by using the
cross products property.
Extraneous Solutions – A solution of a transformed
equation that is not a solution of the original equation.
3. Example 1
Use the cross products property
6
x
Solve
= . Check your solution.
2
x +4
6
x
=
2
x +4
Write original equation.
12 = x2 + 4x
Cross products property
0 = x2 + 4x – 12
Subtract 12 from each side.
0 = (x + 6) (x – 2)
Factor polynomial.
x + 6 = 0 or x – 2 = 0
x = – 6 or
x = 2
Zero-product property
Solve for x.
4. Example 1
ANSWER
CHECK
Use the cross products property
The solutions are – 6 and 2.
If x = – 6:
If x = 2:
6
? –6
=
–6 + 4
2
6 ? 2
=
2 +4 2
–3 = –3
1= 1
5. Example 2
Solve
Multiply by the LCD
x
x – 2
2
1
+
=
5
x – 2
. Check your solution.
Multiply each side of the equation by the LCD to
transform the rational equation into a polynomial equation.
x
x – 2
x
x – 2
• 5 (x – 2 ) +
1
5
2
1
+
=
5
• 5 (x – 2 ) =
Write
original
equation.
x – 2
2
x – 2
• 5 (x – 2 )
Multiply
by LCD,
5 ( x – 2).
6. Example 2
Multiply by the LCD
x • 5 (x – 2 )
x – 2
5 (x – 2 )
2 • 5 (x – 2 )
=
+
x – 2
5
Multiply, then divide out
common factors.
5x + x – 2 =
10
Simplify.
6x – 2 =
10
Combine like terms.
2
Solve for x.
x =
The solution appears to be 2, but the expressions
and
2
x – 2
are undefined when x
=
2.
So, 2 is an
extraneous solution.
ANSWER
The rational equation has no solution.
x
x – 2
7. Example 3
Solve
3
x – 7
Factor to find the LCD
+ 1 =
8
x2 – 9x + 14
. Check your solution.
Write each denominator in factored form. The LCD is
( x – 2 ) ( x – 7 ).
3
8
x – 7
3
x – 7
+ 1 =
(x – 2 ) (x – 7 )
• (x – 2 ) (x – 7 ) + 1 • (x – 2 ) (x – 7 )
8
=
(x – 2 ) (x – 7 )
• (x – 2 ) (x – 7 )
