How do you solve $\frac{2 y}{2 y + 6} + \frac{9 y - 12}{3 y + 9} = \frac{9 y + 11}{y + 3}$ $(2y) / (2y+6) + (9y-12) / (3y+9) = (9y+11) / (y+3)$ ?

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How do you solve $\frac{2 y}{2 y + 6} + \frac{9 y - 12}{3 y + 9} = \frac{9 y + 11}{y + 3}$ $(2y) / (2y+6) + (9y-12) / (3y+9) = (9y+11) / (y+3)$ ?

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Answer 1 · 2015-10-19T11:00:51

Some of these fractions can be simplified.

Explanation:

$\frac{2 y}{2 y + 6} : \frac{2}{2} = \frac{y}{y + 3}$ $(2y)/(2y+6):2/2=y/(y+3)$

$\frac{9 y - 12}{3 y + 9} : \frac{3}{3} = \frac{3 y - 4}{y + 3}$ $(9y-12)/(3y+9):3/3=(3y-4)/(y+3)$

So now we have:
$\frac{y}{y + 3} + \frac{3 y - 4}{y + 3} = \frac{9 y + 11}{y + 3}$ $y/(y+3)+(3y-4)/(y+3)=(9y+11)/(y+3)$

Since the numerators are all equal we may leave them out, on the condition $y \neq - 3$ $y!=-3$ for this would make the numerators $= 0$ $=0$ and the whole equation would be senseless.

$y + 3 y - 4 = 9 y + 11 \to$ $y+3y-4=9y+11->$

All the $y$ $y$ 's to one side, all the numbers to the other:

$- 5 y = 15 \to y = - 3$ $-5y=15->y=-3$

And this is exactly the solution that was forbidden.

Answer : no sensible solution.

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How do you solve $\frac{2 y}{2 y + 6} + \frac{9 y - 12}{3 y + 9} = \frac{9 y + 11}{y + 3}$ $(2y) / (2y+6) + (9y-12) / (3y+9) = (9y+11) / (y+3)$ ?

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Explanation:

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How do you solve 2y2y+6+9y−123y+9=9y+11y+3(2y) / (2y+6) + (9y-12) / (3y+9) = (9y+11) / (y+3)?

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How do you solve $\frac{2 y}{2 y + 6} + \frac{9 y - 12}{3 y + 9} = \frac{9 y + 11}{y + 3}$ $(2y) / (2y+6) + (9y-12) / (3y+9) = (9y+11) / (y+3)$ ?