# Solving Rational Expressions

• October 28th 2009, 01:25 AM
henryW
Solving Rational Expressions
Hi, I am having trouble understanding how to answer questions like this:

$
-2=
\frac {1}{5}y-1
$

or

$
\frac {-5}{2y}= \frac {1}{4}
$
• October 28th 2009, 02:26 AM
Debsta
1. Get rid of the -1 by adding 1 on bothe sides, the get rid of the x 1/5 by x5 on both sides
2. If a/b = c/d then ad=bc (called cross multiplying) then you have a simpler equation to solve
• October 28th 2009, 02:30 AM
rowe
Quote:

Originally Posted by henryW
Hi, I am having trouble understanding how to answer questions like this:

$
-2=
\frac {1}{5}y-1
$

or

$
\frac {-5}{2y}= \frac {1}{4}
$

I assume you want to solve for y?

$-2=\frac {1}{5}y-1$

$-1=\frac {1}{5}y$

$5\cdot(-1)=5\cdot\frac {1}{5}y$

$-5=\frac {5 \cdot 1}{5}y$

$-5= y$

For the second,

$\frac {-5}{2y}= \frac {1}{4}$

$\frac{1}{-5}\cdot\frac {-5}{2y}= \frac {1}{4}\cdot\frac{1}{-5}$

$\frac{1}{2y}= \frac {1}{-20}$

Take the reciprocal of both sides:

$2y = -20$

$y = -10$

That's one way of doing it, you might not be comfortable with using reciprocals, though.

• October 28th 2009, 07:32 AM
HallsofIvy
In general, the simplest way to handle equations with "rational fractions" is to multiply both sides by the least common denominator of all the fractions, thus elminating the fractions.

For example, to solve 3- x/5= 1/3, multiply both sides of the equation by 3(5)= 15 to get 3(15)- (x/5)(15)= (1/3)(15) or 45- 3x= 5. Then -3x= 5- 45= -40 and x= 40/3.

Warning, if your fractions have the "unknown", x, in a denominator, you must check any solutions in the original equation to see if any make a denominator 0. If so, that solution must be thrown out.
• October 28th 2009, 07:36 AM
aidan
You want to simplify the equations.
The critical part is that whatever you do to the left hand side you must also do the equivalent to the right hand side.

.