Solve for k (2k-3)/-5>7

$\frac{2 k - 3}{- 5} > 7$

Step 1

Multiply both sides by $- 5$ .

$\frac{2 k - 3}{- 5} \cdot - 5 < 7 \cdot - 5$

Step 2

Simplify.

Simplify the left side.

Simplify $\frac{2 k - 3}{- 5} \cdot - 5$ .

Cancel the common factor of $5$ .

Factor $5$ out of $- 5$ .

$\frac{2 k - 3}{5 (- 1)} \cdot - 5 < 7 \cdot - 5$

Factor $5$ out of $- 5$ .

$\frac{2 k - 3}{5 \cdot - 1} \cdot (5 \cdot - 1) < 7 \cdot - 5$

Cancel the common factor.

$\frac{2 k - 3}{5 \cdot - 1} \cdot (5 \cdot - 1) < 7 \cdot - 5$

Rewrite the expression.

$\frac{2 k - 3}{- 1} \cdot - 1 < 7 \cdot - 5$

Combine $\frac{2 k - 3}{- 1}$ and $- 1$ .

$\frac{(2 k - 3) \cdot - 1}{- 1} < 7 \cdot - 5$

Simplify the expression.

Move the negative one from the denominator of $\frac{(2 k - 3) \cdot - 1}{- 1}$ .

$- 1 \cdot ((2 k - 3) \cdot - 1) < 7 \cdot - 5$

Rewrite $- 1 \cdot ((2 k - 3) \cdot - 1)$ as $- ((2 k - 3) \cdot - 1)$ .

$- ((2 k - 3) \cdot - 1) < 7 \cdot - 5$

Apply the distributive property.

$- (2 k \cdot - 1 - 3 \cdot - 1) < 7 \cdot - 5$

Multiply.

Multiply $- 1$ by $2$ .

$- (- 2 k - 3 \cdot - 1) < 7 \cdot - 5$

Multiply $- 3$ by $- 1$ .

$- (- 2 k + 3) < 7 \cdot - 5$

Apply the distributive property.

$- (- 2 k) - 1 \cdot 3 < 7 \cdot - 5$

Multiply.

Multiply $- 2$ by $- 1$ .

$2 k - 1 \cdot 3 < 7 \cdot - 5$

Multiply $- 1$ by $3$ .

$2 k - 3 < 7 \cdot - 5$

Simplify the right side.

Multiply $7$ by $- 5$ .

$2 k - 3 < - 35$

Step 3

Solve for $k$ .

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

Add $3$ to both sides of the inequality.

$2 k < - 35 + 3$

Add $- 35$ and $3$ .

$2 k < - 32$

Divide each term in $2 k < - 32$ by $2$ and simplify.

Divide each term in $2 k < - 32$ by $2$ .

$\frac{2 k}{2} < \frac{- 32}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 k}{2} < \frac{- 32}{2}$

Divide $k$ by $1$ .

$k < \frac{- 32}{2}$

Simplify the right side.

Divide $- 32$ by $2$ .

$k < - 16$

Step 4

The result can be shown in multiple forms.

Inequality Form:

$k < - 16$

Interval Notation:

$(- \infty, - 16)$

Solve for k (2k-3)/-5>7

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