How do you isolate $r$ $r$ in $4 (x - 3) - r + 2 x = 5 (3 x - 7) - 9 x$ $4(x-3)-r+2x=5(3x-7)-9x$ ?

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Answer 1 · 2016-08-21T14:32:20

$r = 23$ $r=23$

$4 (x - 3) - r + 2 x = 5 (3 x - 7) - 9 x$ $4(x-3)-r+2x=5(3x-7)-9x$

or

$4 x - 12 - r + 2 x = 15 x - 35 - 9 x$ $4x-12-r+2x=15x-35-9x$

or

$6 x - 12 = r + 6 x - 35$ $6x-12=r+6x-35$

or

$r - 35 = - 12$ $r-35=-12$

or

$r = 35 - 12$ $r=35-12$

or

$r = 23$ $r=23$

Answer 2 · 2016-08-21T14:44:07

$r = 23$ $r=23$

$4 (x - 3) - r + 2 x = 5 (3 x - 7) - 9 x$ $4(x-3)-r+2x=5(3x-7)-9x$

$\Leftrightarrow 4 x - 12 - r + 2 x = 15 x - 35 - 9 x$ $hArr4x-12-r+2x=15x-35-9x$

$\Leftrightarrow 6 x - 12 - r = 6 x - 35$ $hArr6x-12-r=6x-35$

or $cancel(6x)-12-r=cancel(6x)-35$

or $-12-r=-35$ or

$35=12+r$ or

$r=35-12=23$

How do you isolate rrr in 4(x-3)-r+2x=5(3x-7)-9x4(x−3)−r+2x=5(3x−7)−9x4(x-3)-r+2x=5(3x-7)-9x?