Which value of $x$ satisfies the equation $\frac{4}{3}\left(x-\frac{1}{4}\right)=-\frac{37}{3}$ ? $\newline$ $9$ $\newline$ $-8$ $\newline$ $8$ $\newline$ $-9$

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Q. Which value of

x

satisfies the equation

\frac{4}{3}\left(x-\frac{1}{4}\right)=-\frac{37}{3}

\newline

9

\newline

-8

\newline

8

\newline

-9

Isolate variable x: First, we need to isolate the variable $x$ by getting rid of the fraction on the left side of the equation. To do this, we can multiply both sides of the equation by the reciprocal of the fraction's coefficient, which is $\frac{3}{4}$ . So we multiply both sides by $\frac{3}{4}$ to get: $\frac{3}{4} \times \frac{4}{3}(x - \frac{1}{4}) = \frac{3}{4} \times -\frac{37}{3}$
Simplify left side: Now we simplify both sides of the equation. On the left side, the $(\frac{3}{4})$ and $(\frac{4}{3})$ will cancel each other out, leaving us with just $x - \frac{1}{4}$ . On the right side, we multiply the numerators and the denominators separately: $\frac{3 \times -37}{4 \times 3}$
Simplify right side: Simplifying the right side further, we get: $\newline$ $-\frac{111}{12}$ $\newline$ Now the equation looks like this: $\newline$ $x - \frac{1}{4} = -\frac{111}{12}$
Get x by itself: Next, we need to get x by itself. To do this, we add $\frac{1}{4}$ to both sides of the equation to cancel out the $-\frac{1}{4}$ on the left side: $\newline$ $x - \frac{1}{4} + \frac{1}{4} = -\frac{111}{12} + \frac{1}{4}$
Add fractions on right: We need to add the fractions on the right side. To do this, we need a common denominator. The least common multiple of $12$ and $4$ is $12$ , so we convert $\frac{1}{4}$ to $\frac{3}{12}$ : $\newline$ $x = -\frac{111}{12} + \frac{3}{12}$
Add numerators: Now we add the numerators on the right side: $\newline$ $x = (-111 + 3) / 12$ $\newline$ $x = -108 / 12$
Simplify fraction: Finally, we simplify the fraction on the right side by dividing $-108$ by $12$ : $x = \frac{-108}{12}$ $x = -9$