Basic Math Examples

$5 (3 - 2 c) = - (10 c - 15)$

Step 1

Simplify $5 (3 - 2 c)$ .

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Step 1.1

Rewrite.

$0 + 0 + 5 (3 - 2 c) = - (10 c - 15)$

Step 1.2

Simplify by adding zeros.

$5 (3 - 2 c) = - (10 c - 15)$

Step 1.3

Apply the distributive property.

$5 \cdot 3 + 5 (- 2 c) = - (10 c - 15)$

Step 1.4

Multiply.

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Step 1.4.1

Multiply $5$ by $3$ .

$15 + 5 (- 2 c) = - (10 c - 15)$

Step 1.4.2

Multiply $- 2$ by $5$ .

$15 - 10 c = - (10 c - 15)$

Step 2

Simplify $- (10 c - 15)$ .

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Step 2.1

Apply the distributive property.

$15 - 10 c = - (10 c) - - 15$

Step 2.2

Multiply.

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Step 2.2.1

Multiply $10$ by $- 1$ .

$15 - 10 c = - 10 c - - 15$

Step 2.2.2

Multiply $- 1$ by $- 15$ .

$15 - 10 c = - 10 c + 15$

Step 3

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

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Step 3.1

Add $10 c$ to both sides of the equation.

$15 - 10 c + 10 c = 15$

Step 3.2

Combine the opposite terms in $15 - 10 c + 10 c$ .

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Step 3.2.1

Add $- 10 c$ and $10 c$ .

$15 + 0 = 15$

Step 3.2.2

Add $15$ and $0$ .

$15 = 15$

Step 4

Since $15 = 15$ , the equation will always be true for any value of $c$ .

All real numbers

Step 5

The result can be shown in multiple forms.

All real numbers

Interval Notation:

$(- \infty, \infty)$