+0  
 
0
158
2
avatar

https://latex.artofproblemsolving.com/9/8/8/98894e96aa6b82ba4786614fbac149dcb1e489ff.png

Guest Dec 29, 2017

Best Answer 

 #2
avatar+7155 
+1

 

First let's factor all the numerators and denominators.

 

\(\frac{r^2-5r+4}{r^2-8r+7}\,=\,\frac{r^2-2r-15}{r^2-r-20} \\~\\ \frac{(r-4)(r-1)}{(r-7)(r-1)}\,=\,\frac{(r-5)(r+3)}{(r-5)(r+4)} \)

 

Now we can reduce both fractions.

 

\(\frac{(r-4)}{(r-7)}\,=\,\frac{(r+3)}{(r+4)}\)                 and      r ≠ 1 ,  r ≠ 5

 

Multiply both sides of the equation by  (r - 7) ,  and multiply both sides by  (r + 4) .

 

(r - 4)(r + 4)   =   (r + 3)(r - 7)         and      r ≠ 7 ,  r ≠ -4

 

r2 - 16   =   r2 - 4r - 21

                                         Subtract  r2  from both sides.

-16  =  -4r - 21

                                         Add  21  to both sides.

5   =   -4r

                                         Divide both sides by  -4 .

\(-\frac54\)   =   r                         This is not a restrited value, so this is the solution.  smiley

hectictar  Dec 30, 2017
 #1
avatar+139 
+1

cross multiply

azsun  Dec 29, 2017
edited by Guest  Dec 29, 2017
 #2
avatar+7155 
+1
Best Answer

 

First let's factor all the numerators and denominators.

 

\(\frac{r^2-5r+4}{r^2-8r+7}\,=\,\frac{r^2-2r-15}{r^2-r-20} \\~\\ \frac{(r-4)(r-1)}{(r-7)(r-1)}\,=\,\frac{(r-5)(r+3)}{(r-5)(r+4)} \)

 

Now we can reduce both fractions.

 

\(\frac{(r-4)}{(r-7)}\,=\,\frac{(r+3)}{(r+4)}\)                 and      r ≠ 1 ,  r ≠ 5

 

Multiply both sides of the equation by  (r - 7) ,  and multiply both sides by  (r + 4) .

 

(r - 4)(r + 4)   =   (r + 3)(r - 7)         and      r ≠ 7 ,  r ≠ -4

 

r2 - 16   =   r2 - 4r - 21

                                         Subtract  r2  from both sides.

-16  =  -4r - 21

                                         Add  21  to both sides.

5   =   -4r

                                         Divide both sides by  -4 .

\(-\frac54\)   =   r                         This is not a restrited value, so this is the solution.  smiley

hectictar  Dec 30, 2017

11 Online Users

avatar
avatar
avatar
avatar

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.