Basic Math Examples

- 2 (j - 3) = - 2 j + 6

$- 2 (j - 3) = - 2 j + 6$

Step 1

Simplify

- 2 (j - 3)

$- 2 (j - 3)$ .

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Step 1.1

Rewrite.

0 + 0 - 2 (j - 3) = - 2 j + 6

$0 + 0 - 2 (j - 3) = - 2 j + 6$

Step 1.2

Simplify by adding zeros.

- 2 (j - 3) = - 2 j + 6

$- 2 (j - 3) = - 2 j + 6$

Step 1.3

Apply the distributive property.

- 2 j - 2 \cdot - 3 = - 2 j + 6

$- 2 j - 2 \cdot - 3 = - 2 j + 6$

Step 1.4

Multiply

- 2

$- 2$ by

- 3

$- 3$ .

- 2 j + 6 = - 2 j + 6

$- 2 j + 6 = - 2 j + 6$

- 2 j + 6 = - 2 j + 6

$- 2 j + 6 = - 2 j + 6$

Step 2

Move all terms containing

j

$j$ to the left side of the equation.

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Step 2.1

Add

2 j

$2 j$ to both sides of the equation.

- 2 j + 6 + 2 j = 6

$- 2 j + 6 + 2 j = 6$

Step 2.2

Combine the opposite terms in

- 2 j + 6 + 2 j

$- 2 j + 6 + 2 j$ .

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Step 2.2.1

Add

- 2 j

$- 2 j$ and

2 j

$2 j$ .

0 + 6 = 6

$0 + 6 = 6$

Step 2.2.2

Add

0

$0$ and

6

$6$ .

6 = 6

$6 = 6$

6 = 6

$6 = 6$

6 = 6

$6 = 6$

Step 3

Since

6 = 6

$6 = 6$ , the equation will always be true for any value of

j

$j$ .

All real numbers

Step 4

The result can be shown in multiple forms.

All real numbers

Interval Notation:

(- \infty, \infty)

$(- \infty, \infty)$

⎡ ⎢ ⎣ \begin{matrix} x^{2} & \frac{1}{2} \sqrt{π} & \int x d x \end{matrix} ⎤ ⎥ ⎦

$[\begin{array}{c} x^{2} & \frac{1}{2} \\ \sqrt{π} & \int x d x \end{array}]$