How do you divide $\frac{3 x^{2}}{6 x - 3}$ $(3x^2)/(6x - 3)$ ?

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Answer 1 · 2015-11-22T17:55:21

$\frac{x^{2}}{2 x - 1}$ $x^2/(2x-1)$

You can factor in the denominator.

$6$ $6$ and $3$ $3$ both share the common factor of $3$ $3$ , so we can extract $3$ $3$ from both $6$ $6$ and $3$ $3$ .

$3 (\frac{6 x}{3} - \frac{3}{3})$ $3((6x)/3-3/3)$ $\Rightarrow$ $rArr$ $3 (2 x - 1)$ $3(2x-1)$

Original equation: $\frac{3 x^{2}}{6 x - 3}$ $(3x^2)/(6x-3)$

New equation: $\frac{3 (x^{2})}{3 (2 x - 1)}$ $(3(x^2))/(3(2x-1))$

The two $3$ $3$ 's cancel out

$(cancel(3)(x^2))/(cancel(3)(2x-1))$

Final equation: $x^2/(2x-1)$

How do you divide (3x^2)/(6x - 3)3x26x−3(3x^2)/(6x - 3)?