Solve the Inequality for u 9x-71>3(3x-7u)

$9 x - 71 > 3 (3 x - 7 u)$

Step 1

Rewrite so $u$ is on the left side of the inequality.

$3 (3 x - 7 u) < 9 x - 71$

Step 2

Divide each term in $3 (3 x - 7 u) < 9 x - 71$ by $3$ and simplify.

Divide each term in $3 (3 x - 7 u) < 9 x - 71$ by $3$ .

$\frac{3 (3 x - 7 u)}{3} < \frac{9 x}{3} + \frac{- 71}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 (3 x - 7 u)}{3} < \frac{9 x}{3} + \frac{- 71}{3}$

Divide $3 x - 7 u$ by $1$ .

$3 x - 7 u < \frac{9 x}{3} + \frac{- 71}{3}$

Simplify the right side.

Simplify each term.

Cancel the common factor of $9$ and $3$ .

Factor $3$ out of $9 x$ .

$3 x - 7 u < \frac{3 (3 x)}{3} + \frac{- 71}{3}$

Cancel the common factors.

Factor $3$ out of $3$ .

$3 x - 7 u < \frac{3 (3 x)}{3 (1)} + \frac{- 71}{3}$

Cancel the common factor.

$3 x - 7 u < \frac{3 (3 x)}{3 \cdot 1} + \frac{- 71}{3}$

Rewrite the expression.

$3 x - 7 u < \frac{3 x}{1} + \frac{- 71}{3}$

Divide $3 x$ by $1$ .

$3 x - 7 u < 3 x + \frac{- 71}{3}$

Move the negative in front of the fraction.

$3 x - 7 u < 3 x - \frac{71}{3}$

Step 3

Move all terms not containing $u$ to the right side of the inequality.

Subtract $3 x$ from both sides of the inequality.

$- 7 u < 3 x - \frac{71}{3} - 3 x$

Combine the opposite terms in $3 x - \frac{71}{3} - 3 x$ .

Subtract $3 x$ from $3 x$ .

$- 7 u < 0 - \frac{71}{3}$

Subtract $\frac{71}{3}$ from $0$ .

$- 7 u < - \frac{71}{3}$

Step 4

Divide each term in $- 7 u < - \frac{71}{3}$ by $- 7$ and simplify.

Divide each term in $- 7 u < - \frac{71}{3}$ by $- 7$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- 7 u}{- 7} > \frac{- \frac{71}{3}}{- 7}$

Simplify the left side.

Cancel the common factor of $- 7$ .

Cancel the common factor.

$\frac{- 7 u}{- 7} > \frac{- \frac{71}{3}}{- 7}$

Divide $u$ by $1$ .

$u > \frac{- \frac{71}{3}}{- 7}$

Simplify the right side.

Multiply the numerator by the reciprocal of the denominator.

$u > - \frac{71}{3} \cdot \frac{1}{- 7}$

Move the negative in front of the fraction.

$u > - \frac{71}{3} (- \frac{1}{7})$

Multiply $- \frac{71}{3} (- \frac{1}{7})$ .

Multiply $- 1$ by $- 1$ .

$u > 1 (\frac{71}{3}) \frac{1}{7}$

Multiply $\frac{71}{3}$ by $1$ .

$u > \frac{71}{3} \cdot \frac{1}{7}$

Multiply $\frac{71}{3}$ by $\frac{1}{7}$ .

$u > \frac{71}{3 \cdot 7}$

Multiply $3$ by $7$ .

$u > \frac{71}{21}$

Step 5

The result can be shown in multiple forms.

Inequality Form:

$u > \frac{71}{21}$

Interval Notation:

$(\frac{71}{21}, \infty)$

Solve the Inequality for u 9x-71>3(3x-7u)

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