solve the equation for 2q+3/3 + 3q-4/6 = q-2/2

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- Dec 16th 2009, 04:54 PM #1

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- Dec 16th 2009, 05:01 PM #2

- Dec 16th 2009, 05:05 PM #3

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$\displaystyle \frac{2q+3}{3} + \frac{3q-4}{6} = \frac{q-2}{2}$

Multiply both sides by 6 to get rid of the fractions.

$\displaystyle 2(2q + 3) + 3q - 4 = 3(q - 2)$

Distribute 2 into (2q + 3) and 3 into (q - 2).

$\displaystyle 4q + 6 + 3q - 4 = 3q - 6$

Combine like terms.

$\displaystyle 7q + 2 = 3q - 6$

Subtract 3q from both sides and subtract 2 from both sides to isolate q.

$\displaystyle 4q = -8$

Divide by 4 on both sides.

$\displaystyle q = -2$

- Dec 16th 2009, 05:06 PM #4

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