Solve: \( \frac{x}{2}-\frac{1}{4}\left(x-\frac{1}{3}\right)=\frac{1}{6}(x+1)+\frac{1}{12} \)
Given:
\( \frac{x}{2}-\frac{1}{4}\left(x-\frac{1}{3}\right)=\frac{1}{6}(x+1)+\frac{1}{12} \)
To do:
We have to solve \( \frac{x}{2}-\frac{1}{4}\left(x-\frac{1}{3}\right)=\frac{1}{6}(x+1)+\frac{1}{12} \).
Solution:
$\frac{x}{2}-\frac{1}{4}\left(x-\frac{1}{3}\right)=\frac{1}{6}(x+1)+\frac{1}{12}$
$\Rightarrow \frac{x}{2} -\frac{1}{4}\left(\frac{3x-1}{3}\right) =\frac{x+1}{6} +\frac{1}{12}$
$\Rightarrow \frac{x}{2} -\frac{3x-1}{12} =\frac{2( x+1) +1}{12}$
$\Rightarrow \frac{6x-( 3x-1)}{12} =\frac{2x+2+1}{12}$
$\Rightarrow 6x-3x+1=2x+3$
$\Rightarrow 3x-2x=3-1$
$\Rightarrow x=2$
The value of $x$ is $2$.
Related Articles
- $\frac{x-1}{2}+\frac{2 x-1}{4}=\frac{x-1}{3}-\frac{2 x-1}{6}$.
- Solve \( \frac{2 x+1}{3 x-2}=1\frac{1}{4} \).
- Factorize:\( \left(x^{2}+\frac{1}{x^{2}}\right)-4\left(x+\frac{1}{x}\right)+6 \)
- Solve the following quadratic equation by factorization: $\frac{1}{(x-1)(x-2)}+\frac{1}{(x-2)(x-3)}+\frac{1}{(x-3)(x-4)}=\frac{1}{6}$
- Solve the following:\( \frac{3}{4}(7 x-1)-\left(2 x-\frac{1-x}{2}\right)=x+\frac{3}{2} \)
- Solve: $\frac{6 x+1}{3}-\frac{x-3}{6}=-1$.
- Solve for x: $\frac{1}{x+1} +\frac{3}{5x+1} =\frac{5}{x+4} ,\ x\neq 1,\ -\frac{1}{5} ,\ -4$
- Solve for $x$:$\frac{1}{x-3}-\frac{1}{x+5}=\frac{1}{6}, x≠3, -5$
- Simplify:\( \left(x^{4}+\frac{1}{x^{4}}\right)\left(x+\frac{1}{x}\right) \)
- Subtract the following:\( \left(4 x^{2}-\frac{1}{5} x+7\right)-\left(-2 x^{2}-\frac{1}{2} x+\frac{1}{3}\right) \)
- Solve the following$ 9 \frac{3}{4} \div\left[2 \frac{1}{6}+\left\{4 \frac{1}{3}-\left(1 \frac{1}{2}+1 \frac{3}{4}\right)\right\}\right]$
- If \( x+\frac{1}{x}=3 \), calculate \( x^{2}+\frac{1}{x^{2}}, x^{3}+\frac{1}{x^{3}} \) and \( x^{4}+\frac{1}{x^{4}} \).
- Solve for $x$:$\frac{x-1}{x-2}+\frac{x-3}{x-4}=3\frac{1}{3}, x≠2, 4$
- Find $x$, if(i) \( \left(\frac{1}{4}\right)^{-4} \times\left(\frac{1}{4}\right)^{-8}=\left(\frac{1}{4}\right)^{-4 x} \)(ii) \( \left(\frac{-1}{2}\right)^{-19}\div\left(\frac{-1}{2}\right)^{8}=\left(\frac{-1}{2}\right)^{-2 x+1} \)(iii) \( \left(\frac{3}{2}\right)^{-3} \times\left(\frac{3}{2}\right)^{5}=\left(\frac{3}{2}\right)^{2 x+1} \)(iv) \( \left(\frac{2}{5}\right)^{-3} \times\left(\frac{2}{5}\right)^{15}=\left(\frac{2}{5}\right)^{2+3 x} \)(v) \( \left(\frac{5}{4}\right)^{-x}\div\left(\frac{5}{4}\right)^{-4}=\left(\frac{5}{4}\right)^{5} \)(vi) \( \left(\frac{8}{3}\right)^{2 x+1} \times\left(\frac{8}{3}\right)^{5}=\left(\frac{8}{3}\right)^{x+2} \)
- Solve the following:$\frac{1}{x-2}+\frac{2}{x-1}=\frac{6}{x}$
Kickstart Your Career
Get certified by completing the course
Get Started