Basic Math Examples

$3 (7 c + 2) = 5 (4 c + 1) + 17$

Step 1

Simplify $3 (7 c + 2)$ .

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Step 1.1

Rewrite.

$0 + 0 + 3 (7 c + 2) = 5 (4 c + 1) + 17$

Step 1.2

Simplify by adding zeros.

$3 (7 c + 2) = 5 (4 c + 1) + 17$

Step 1.3

Apply the distributive property.

$3 (7 c) + 3 \cdot 2 = 5 (4 c + 1) + 17$

Step 1.4

Multiply.

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Step 1.4.1

Multiply $7$ by $3$ .

$21 c + 3 \cdot 2 = 5 (4 c + 1) + 17$

Step 1.4.2

Multiply $3$ by $2$ .

$21 c + 6 = 5 (4 c + 1) + 17$

Step 2

Simplify $5 (4 c + 1) + 17$ .

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Step 2.1

Simplify each term.

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Step 2.1.1

Apply the distributive property.

$21 c + 6 = 5 (4 c) + 5 \cdot 1 + 17$

Step 2.1.2

Multiply $4$ by $5$ .

$21 c + 6 = 20 c + 5 \cdot 1 + 17$

Step 2.1.3

Multiply $5$ by $1$ .

$21 c + 6 = 20 c + 5 + 17$

Step 2.2

Add $5$ and $17$ .

$21 c + 6 = 20 c + 22$

Step 3

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

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Step 3.1

Subtract $20 c$ from both sides of the equation.

$21 c + 6 - 20 c = 22$

Step 3.2

Subtract $20 c$ from $21 c$ .

$c + 6 = 22$

Step 4

Move all terms not containing $c$ to the right side of the equation.

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Step 4.1

Subtract $6$ from both sides of the equation.

$c = 22 - 6$

Step 4.2

Subtract $6$ from $22$ .

$c = 16$