# Literal fraction

• Apr 24th 2009, 05:12 AM
strunz
Literal fraction
[Math]\frac{10}{r-3}+\frac{4}{3-r}=6[/tex]

[MATh]\frac{10(3-r)}{(r-3)(3-r)}+\frac{4(r-3)}{(r-3)(3-r)}=\frac{6(r-3)(3-r)}{(r-3)(3-r)}[/tex]

Is it correct to this point? The latest fraction is 0. Then 30 -10r +4r -12 = 0, r= 3
My book says r=4 so at some point I must have made a mistake.
• Apr 24th 2009, 06:02 AM
stapel
Quote:

Originally Posted by strunz
[Math]\frac{10}{r-3}+\frac{4}{3-r}=6[/tex]

Note that 3 - r = -r + 3 = -(r - 3). So the common denominator is just r - 3:

. . . . .$\displaystyle \frac{10}{r\, -\, 3}\, -\, \frac{4}{r\, -\, 3}\, =\, 6$

A useful method for solving rational equations is to multiply through by the common denominator. In this case, you would end up with:

. . . . .$\displaystyle 10\, -\, 4\, =\, 6(r\, -\, 3)$

. . . . .$\displaystyle 1\, =\, r\, -\, 3$

Solve the linear equation. Remember to check the value of "r" in the original equation, to make sure there are no "division by zero" problems. (Wink)
• Apr 24th 2009, 06:03 AM
red_dog
$\displaystyle \frac{10}{r-3}+\frac{4}{3-r}=\frac{10}{r-3}-\frac{4}{r-3}=\frac{6}{r-3}$

The equation becomes $\displaystyle \frac{6}{r-3}=6\Rightarrow r-3=1\Rightarrow r=4$