How do you solve $11 c - 2 (2 c + 1) = 3 (c - 6)$ $11c - 2( 2c + 1) = 3( c - 6)$ ?

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How do you solve $11 c - 2 (2 c + 1) = 3 (c - 6)$ $11c - 2( 2c + 1) = 3( c - 6)$ ?

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Answer 1 · 2018-03-13T03:26:42

$c = \frac{8}{9}$ $c = 8/9$

Explanation:

Expand brackets:
$11 c - 4 c - 2 = 3 c - 18$ $11c - 4c - 2 = 3c - 18$
Move variable to one side, numbers to the other:
$11 c - 4 c - 3 c = 2 - 18$ $11c - 4c - 3c = 2 - 18$
Solve for c:
$18 c = 16$ $18c = 16$
$c = \frac{16}{18}$ $c = 16/18$
$c = \frac{8}{9}$ $c = 8/9$

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How do you solve $11 c - 2 (2 c + 1) = 3 (c - 6)$ $11c - 2( 2c + 1) = 3( c - 6)$ ?

1 Answer

Explanation:

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How do you solve 11c - 2( 2c + 1) = 3( c - 6)11c−2(2c+1)=3(c−6)11c - 2( 2c + 1) = 3( c - 6)?

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Explanation:

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How do you solve $11 c - 2 (2 c + 1) = 3 (c - 6)$ $11c - 2( 2c + 1) = 3( c - 6)$ ?