Question

\large -\frac{2}{3}x+3=\frac{2}{3}x+\frac{1}{3}

Answer 1

Answer:

$\boxed {x = 2}$

Step-by-step explanation:

Solve for the value of $x$:

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

-Take $\frac{2}{3}x$ and subtract it from $-\frac{2}{3}x$:

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

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

-Subtract both sides and convert $3$ to a fraction:

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

$-\frac{4}{3}x = \frac{1}{3} - \frac{9}{3}$

Since both $\frac{1}{3}$ and $\frac{9}{3}$ have the same denominator, then you would subtract the numerator:

$-\frac{4}{3}x = \frac{1 - 9}{3}$

$-\frac{4}{3}x = \frac{8}{3}$

-Multiply both sides by $-\frac{3}{4}$, which is the reciprocal of $-\frac{4}{3}$:

$x = \frac{8}{3} (-\frac{3}{4})$

$x = \frac{-8(-3)}{3 \times 4}$

$x = \frac{24}{12}$

-Divide $24$ by $12$:

$x = \frac{24}{12}$

$\boxed {x = 2}$

So, therefore, the value of $x$ is $2$.