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Guest Dec 29, 2017

Best Answer 

 #2
avatar+5924 
+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
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2+0 Answers

 #1
avatar+64 
+1

cross multiply

azsun  Dec 29, 2017
edited by Guest  Dec 29, 2017
 #2
avatar+5924 
+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

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