Solve for p 9(4-(3-p))+3p=2(6p+5)-1

$9 (4 - (3 - p)) + 3 p = 2 (6 p + 5) - 1$

Simplify $9 (4 - (3 - p)) + 3 p$ .

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Simplify each term.

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Simplify each term.

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Apply the distributive property.

$9 (4 - 1 \cdot 3 - - p) + 3 p = 2 (6 p + 5) - 1$

Multiply $- 1$ by $3$ .

$9 (4 - 3 - - p) + 3 p = 2 (6 p + 5) - 1$

Multiply $- - p$ .

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Multiply $- 1$ by $- 1$ .

$9 (4 - 3 + 1 p) + 3 p = 2 (6 p + 5) - 1$

Multiply $p$ by $1$ .

$9 (4 - 3 + p) + 3 p = 2 (6 p + 5) - 1$

Subtract $3$ from $4$ .

$9 (1 + p) + 3 p = 2 (6 p + 5) - 1$

Apply the distributive property.

$9 \cdot 1 + 9 p + 3 p = 2 (6 p + 5) - 1$

Multiply $9$ by $1$ .

$9 + 9 p + 3 p = 2 (6 p + 5) - 1$

Add $9 p$ and $3 p$ .

$9 + 12 p = 2 (6 p + 5) - 1$

Simplify $2 (6 p + 5) - 1$ .

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Simplify each term.

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Apply the distributive property.

$9 + 12 p = 2 (6 p) + 2 \cdot 5 - 1$

Multiply $6$ by $2$ .

$9 + 12 p = 12 p + 2 \cdot 5 - 1$

Multiply $2$ by $5$ .

$9 + 12 p = 12 p + 10 - 1$

Subtract $1$ from $10$ .

$9 + 12 p = 12 p + 9$

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

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Subtract $12 p$ from both sides of the equation.

$9 + 12 p - 12 p = 9$

Combine the opposite terms in $9 + 12 p - 12 p$ .

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Subtract $12 p$ from $12 p$ .

$9 + 0 = 9$

Add $9$ and $0$ .

$9 = 9$

Since $9 = 9$ , the equation will always be true for any value of $p$ .

All real numbers

The result can be shown in multiple forms.

All real numbers

Interval Notation:

$(- \infty, \infty)$

Solve for p 9(4-(3-p))+3p=2(6p+5)-1

Solving MATH problems