Finite Math Examples

2 k + 1 > 5

$2 k + 1 > 5$

Step 1

Move all terms not containing

k

$k$ to the right side of the inequality.

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Step 1.1

Subtract

1

$1$ from both sides of the inequality.

2 k > 5 - 1

$2 k > 5 - 1$

Step 1.2

Subtract

1

$1$ from

5

$5$ .

2 k > 4

$2 k > 4$

2 k > 4

$2 k > 4$

Step 2

Divide each term in

2 k > 4

$2 k > 4$ by

2

$2$ and simplify.

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Step 2.1

Divide each term in

2 k > 4

$2 k > 4$ by

2

$2$ .

\frac{2 k}{2} > \frac{4}{2}

$\frac{2 k}{2} > \frac{4}{2}$

Step 2.2

Simplify the left side.

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Step 2.2.1

Cancel the common factor of

2

$2$ .

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Step 2.2.1.1

Cancel the common factor.

\frac{2 k}{2} > \frac{4}{2}

$\frac{2 k}{2} > \frac{4}{2}$

Step 2.2.1.2

Divide

k

$k$ by

1

$1$ .

k > \frac{4}{2}

$k > \frac{4}{2}$

k > \frac{4}{2}

$k > \frac{4}{2}$

k > \frac{4}{2}

$k > \frac{4}{2}$

Step 2.3

Simplify the right side.

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Step 2.3.1

Divide

4

$4$ by

2

$2$ .

k > 2

$k > 2$

k > 2

$k > 2$

k > 2

$k > 2$

Step 3

The result can be shown in multiple forms.

Inequality Form:

k > 2

$k > 2$

Interval Notation:

(2, \infty)

$(2, \infty)$

2 k + 1 > 5

$2 k + 1 > 5$

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

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