How do you solve $- 2 r + 2 (3 r - 2) = 17$ $-2r + 2( 3r - 2) = 17$ ?

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Answer 1 · 2016-10-31T09:34:55

Expand the equation before solving.

$= - 2 r + 2 (3 r - 2) = 17$ $= -2r + 2(3r-2) = 17$

$= - 6 r + 4 r + 6 r - 4 = 17$ $= -6r +4r +6r -4 =17$

$4 r = 21$ $4r=21$

$r = \frac{21}{4}$ $r=21/4$

Answer: $r = \frac{21}{4}$ $r=21/4$

How do you solve -2r + 2( 3r - 2) = 17−2r+2(3r−2)=17-2r + 2( 3r - 2) = 17?