Algebra Examples

- 2 x - 3 \leq y

$- 2 x - 3 \leq y$

y - 1 < \frac{1}{2} x

$y - 1 < \frac{1}{2} x$

Step 1

Rewrite so

y

$y$ is on the left side of the inequality.

y \geq - 2 x - 3

$y \geq - 2 x - 3$

Step 2

Solve

y - 1 < \frac{1}{2} x

$y - 1 < \frac{1}{2} x$ for

y

$y$ .

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Step 2.1

Simplify

\frac{1}{2} x

$\frac{1}{2} x$ .

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Step 2.1.1

Rewrite.

y - 1 < 0 + 0 + \frac{1}{2} x

$y - 1 < 0 + 0 + \frac{1}{2} x$

Step 2.1.2

Simplify by adding zeros.

y - 1 < \frac{1}{2} x

$y - 1 < \frac{1}{2} x$

Step 2.1.3

Combine

\frac{1}{2}

$\frac{1}{2}$ and

x

$x$ .

y - 1 < \frac{x}{2}

$y - 1 < \frac{x}{2}$

y - 1 < \frac{x}{2}

$y - 1 < \frac{x}{2}$

Step 2.2

Add

1

$1$ to both sides of the inequality.

y < \frac{x}{2} + 1

$y < \frac{x}{2} + 1$

y < \frac{x}{2} + 1

$y < \frac{x}{2} + 1$

Step 3

Find the intersection of

y \geq - 2 x - 3

$y \geq - 2 x - 3$ and

y < \frac{x}{2} + 1

$y < \frac{x}{2} + 1$ .

- 2 x - 3 \leq y < \frac{x}{2} + 1

$- 2 x - 3 \leq y < \frac{x}{2} + 1$

Step 4

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

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