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$$2x-\dfrac{x+1}{2} + \dfrac{1}{3}(x+3)= \dfrac{7}{3}$$

When I solve this I always end up with 11x = 5, which is wrong, no matter which way I solve it. Does anyone know how to solve it? Steps? (Because I know the answer should be x=1)

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uhm -- is this $(1/3) (x+3)$ or $\frac{1}{3(x+3)}$ – user20266 Dec 12 '11 at 18:35
Added a picture – Friend of Kim Dec 12 '11 at 18:38
I removed the picture after clearing up the equation. – The Chaz 2.0 Dec 12 '11 at 18:45
You do not show your work, so I am guessing. But when we bring the left side to the common denominator $6$, we get $\frac{12x-3(x+1)+2(x+3)}{6}$. The top simplifies to $11x+3$. You probably simplified it wrongly to $11x+9$. Minus signs are evil. – André Nicolas Dec 12 '11 at 18:46
I know, they are! I didn't realize that it was one term, so it got -3 instead of +3. Thanks everyone! – Friend of Kim Dec 12 '11 at 18:47
up vote 3 down vote accepted

$$\eqalign{&2x -{x+1\over 2}+{x+3\over 3 }={7\over 3}\cr &\iff12x \color{red}{- 3}(x+1) +{2 (x+3)}={14}\cr &\iff12x-3x\color{red}{-3}+2x+6 ={14}\cr &\iff 11x =11\cr &\iff x=1 } $$

You most likely forgot to "distribute the negative" (since you said you obtained $11x=5$).

To see what's going on there: we are using the rule that subtraction of a quantity is the same adding $(-1)$ times the quantity.

$$\eqalign{12x -3(x+1)&=12x +(-1)\cdot3 (x+1) \cr &=12x +(-3)(x+1)\cr &=12x+(-3)x+(-3)\cdot1\cr&=12x-3x-3.}$$

Of course, once you're accustomed to it, you just "distribute the negative sign".

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$$\begin{align*} && 2x-\frac{x+1}{2}+\frac{x+3}{3} &= \frac{7}{3} & \cdot 6 \\ &\Leftrightarrow& 12x - 3x - 3 + 2x +6 &= 14 & \text{rearrange} \\ &\Leftrightarrow& 11x&=11 \end{align*}$$

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Thanks, but 12x-3x, then you wrote -3, why isn't it +3? it says x+1 and not -1. Hopefully this is where I left of. Edit: Is x+1 one term? So it's like -(x+1) = -x-1 to remove the brackets? – Friend of Kim Dec 12 '11 at 18:43

Multiply by $6$ to clear fractions:

$12x - 3(x +1) +2(x +3) = 14$

Eventually you'll get

$11x = 11$

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