Solve the inequality \( 6x + \frac{7}{3} > \frac{2}{7}x - 7 \).
Move all terms involving \( x \) to one side and constants to the other side.
Subtract \( \frac{2}{7}x \) from both sides:
\[
6x - \frac{2}{7}x + \frac{7}{3} > -7
\]
Combine like terms.
To combine \( 6x \) and \( -\frac{2}{7}x \), find a common denominator:
\[
6x = \frac{42}{7}x
\]
Thus:
\[
\frac{42}{7}x - \frac{2}{7}x = \frac{40}{7}x
\]
The inequality becomes:
\[
\frac{40}{7}x + \frac{7}{3} > -7
\]
Move the constant term to the other side.
Subtract \( \frac{7}{3} \) from both sides:
\[
\frac{40}{7}x > -7 - \frac{7}{3}
\]
Simplify the right-hand side.
Find a common denominator for \(-7\) and \(-\frac{7}{3}\):
\[
-7 = -\frac{21}{3}
\]
Thus:
\[
-7 - \frac{7}{3} = -\frac{21}{3} - \frac{7}{3} = -\frac{28}{3}
\]
The inequality becomes:
\[
\frac{40}{7}x > -\frac{28}{3}
\]
Solve for \( x \).
Multiply both sides by \( \frac{7}{40} \) (the reciprocal of \( \frac{40}{7} \)):
\[
x > -\frac{28}{3} \cdot \frac{7}{40}
\]
Simplify the multiplication:
\[
x > -\frac{196}{120}
\]
Reduce the fraction:
\[
x > -\frac{49}{30}
\]
The solution to the inequality is:
\[
\boxed{x > -\frac{49}{30}}
\]
The solution to the inequality is:
\[
\boxed{x > -\frac{49}{30}}
\]
Answered
