Solve for b 5(b-7)>b

$5 (b - 7) > b$

Step 1

Simplify $5 (b - 7)$ .

Rewrite.

$0 + 0 + 5 (b - 7) > b$

Simplify by adding zeros.

$5 (b - 7) > b$

Apply the distributive property.

$5 b + 5 \cdot - 7 > b$

Multiply $5$ by $- 7$ .

$5 b - 35 > b$

Step 2

Move all terms containing $b$ to the left side of the inequality.

Subtract $b$ from both sides of the inequality.

$5 b - 35 - b > 0$

Subtract $b$ from $5 b$ .

$4 b - 35 > 0$

Step 3

Add $35$ to both sides of the inequality.

$4 b > 35$

Step 4

Divide each term in $4 b > 35$ by $4$ and simplify.

Divide each term in $4 b > 35$ by $4$ .

$\frac{4 b}{4} > \frac{35}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 b}{4} > \frac{35}{4}$

Divide $b$ by $1$ .

$b > \frac{35}{4}$

Step 5

The result can be shown in multiple forms.

Inequality Form:

$b > \frac{35}{4}$

Interval Notation:

$(\frac{35}{4}, \infty)$