How do you solve $\frac{r - 8}{- 2} = \frac{11 - 4 r}{11}$ $\frac { r - 8} { - 2} = \frac { 11- 4r } { 11}$ ?

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Answer 1 · 2017-08-13T16:32:59

r = 22

First, you multiply both sides by 22 to get rid of the denominator.

$\frac{r - 8}{- 2} = \frac{11 - 4 r}{11} \to$ $(r-8)/-2=(11-4r)/11 ->$

$22 \cdot \frac{r - 8}{- 2} = 22 \cdot \frac{11 - 4 r}{11} \to$ $22*(r-8)/-2=22*(11-4r)/11 ->$

$- 11 (r - 8) = 2 (11 - 4 r)$ $-11(r-8)=2(11-4r)$

Next, you want to expand both sides of the equation.

$- 11 (r - 8) = 2 (11 - 4 r) \to$ $-11(r-8)=2(11-4r) ->$

$- 11 r + 88 = 22 - 8 r$ $-11r+88=22-8r$

Then you want to put like terms together.

$- 11 r + 88 = 22 - 8 r \to$ $-11r+88=22-8r ->$

$- 11 r + 88 - 22 + 11 r = 22 - 8 r - 22 + 11 r \to$ $-11r+88-22+11r=22-8r-22+11r ->$

$66 = 3 r$ $66=3r$

Finally, solve for r.

$66 = 3 r \to$ $66=3r ->$

$\frac{66}{3} = (3 \frac{r}{3}) \to$ $66/3=(3r/3) ->$

$22 = r$ $22=r$

If you have time, you can double check to make sure your answer makes sense by pugging it the value for r.

$\frac{22 - 8}{- 2} = \frac{11 - 4 (22)}{11} \to$ $(22-8)/-2=(11-4(22))/11 ->$

$\frac{14}{- 2} = \frac{11 - 88}{11} \to$ $(14)/-2=(11-88)/11 ->$

$\frac{14}{- 2} = \frac{- 77}{11} \to$ $(14)/-2=(-77)/11 ->$

$- 7 = - 7$ $-7=-7$ Which is a true statement, so $r = 22$ $r=22$ is correct.

How do you solve r−8−2=11−4r11\frac { r - 8} { - 2} = \frac { 11- 4r } { 11}?