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

Dec 29, 2017

#2
+8394
+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

5   =   -4r

Divide both sides by  -4 .

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

Dec 30, 2017

#1
+191
+1

cross multiply

Dec 29, 2017
edited by Guest  Dec 29, 2017
#2
+8394
+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

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