Hi, I am having trouble understanding how to answer questions like this:

$\displaystyle

-2=

\frac {1}{5}y-1

$

or

$\displaystyle

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

$

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- Oct 28th 2009, 01:25 AMhenryWSolving Rational Expressions
Hi, I am having trouble understanding how to answer questions like this:

$\displaystyle

-2=

\frac {1}{5}y-1

$

or

$\displaystyle

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

$ - Oct 28th 2009, 02:26 AMDebsta
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 - Oct 28th 2009, 02:30 AMrowe
I assume you want to solve for y?

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

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

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

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

$\displaystyle -5= y$

For the second,

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

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

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

Take the reciprocal of both sides:

$\displaystyle 2y = -20$

$\displaystyle y = -10$

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

- Oct 28th 2009, 07:32 AMHallsofIvy
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. - Oct 28th 2009, 07:36 AMaidan
Debsta's advice is good.

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.

.