$2x-\dfrac{x+1}{2} + \dfrac{1}{3}(x+3)= \dfrac{7}{3}$
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)
$2x-\dfrac{x+1}{2} + \dfrac{1}{3}(x+3)= \dfrac{7}{3}$
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)
$\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 $11x=5$).
$\eqalign{12x -3(x+1)&=12x +(-1)\cdot3 (x+1) \cr &=12x +(-3)(x+1)\cr &=12x+(-3)x+(-3)\cdot1\cr&=12x-3x-3.}$
Of course, once you're accustomed to it, you just "distribute the negative sign".
$\begin{align*} && 2x-\frac{x+1}{2}+\frac{x+3}{3} &= \frac{7}{3} & \cdot 6 \\ &\Leftrightarrow& 12x - 3x - 3 + 2x +6 &= 14 & \text{rearrange} \\ &\Leftrightarrow& 11x&=11 \end{align*}$
Multiply by $6$ to clear fractions:
$12x - 3(x +1) +2(x +3) = 14$
Eventually you'll get
$11x = 11$