Question

$\frac{2}{3} (4x - 1) - (2x - \frac{1 + x}{3} )= \frac{1}{3} x + \frac{4}{3}$
Please solve this equation.
urgent!!​

Answer 1

$\sf{\frac{2}{3} (4x - 1) - (2x - \frac{1 + x}{3} )= \frac{1}{3} x + \frac{4}{3}}$

$\frac{2}{3}(4x-1)-2x+\frac{1+x}{3}=\frac{1}{3}x+\frac{4}{3}$

$\frac{2(4x-1)}{3}-2x+\frac{1+x}{3}=\frac{1}{3}x+\frac{4}{3}$

$\frac{2(4x-1)}{3}-2x+\frac{1+x}{3}=\frac{x}{3}+\frac{4}{3}$

2(4x-1)-6x+1+x=x+4

2(4x−1)−5x+1=x+4

8x−2−5x+1=x+4

3x−1=x+4

3x=x+4+1

3x=x+5

3x−x=5

2x=5

$x=\frac{5}{2}$

Answer 2

Step-by-step explanation:

$\mathtt{\frac{2}{3} (4x - 1) - (2x - \frac{1 + x}{3} )= \frac{1}{3} x + \frac{4}{3}}$

$\mathtt{ \frac{2(4x - 1)}{3} - \frac{6x - 1 +x}{3} = \frac{x}{3} + \frac{4}{3} }$

$\mathtt{\frac{8x - 2}{3} - \frac{5x - 1}{3} = \frac{x + 4}{3}}$

$\mathtt{8x - 2 - 5x - 1 = x + 4}$

$\mathtt{3x - 3 = x + 4}$

$\mathtt{2x = 7}$

$\boxed {\mathtt \red{x = \frac{7}{2} }}$

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