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Solve for b (93)b=9(3b)

$(9 \cdot 3) \cdot b = 9 \cdot (3 \cdot b)$

Step 1

Multiply $9$ by $3$ .

$27 b = 9 \cdot 3 \cdot b$

Step 2

Multiply $9$ by $3$ .

$27 b = 27 b$

Step 3

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

Subtract $27 b$ from both sides of the equation.

$27 b - 27 b = 0$

Subtract $27 b$ from $27 b$ .

$0 = 0$

$0 = 0$

Step 4

Since $0 = 0$ , the equation will always be true for any value of $b$ .

All real numbers

Step 5

The result can be shown in multiple forms.

All real numbers

Interval Notation:

$(- \infty, \infty)$

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