Precalculus Examples

$y < x^{2}$ , $6 x - 5 y < 30$

Step 1

Simplify the first inequality.

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Step 1.1

Rewrite so $x$ is on the left side of the inequality.

$x^{2} > y$ and $6 x - 5 y < 30$

Step 1.2

Take the specified root of both sides of the inequality to eliminate the exponent on the left side.

$\sqrt{x^{2}} > \sqrt{y}$ and $6 x - 5 y < 30$

Step 1.3

Simplify the left side.

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Step 1.3.1

Pull terms out from under the radical.

$| x | > \sqrt{y}$ and $6 x - 5 y < 30$

Step 1.4

Write $| x | > \sqrt{y}$ as a piecewise.

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Step 1.4.1

To find the interval for the first piece, find where the inside of the absolute value is non-negative.

$x \geq 0$

Step 1.4.2

In the piece where $x$ is non-negative, remove the absolute value.

$x > \sqrt{y}$

Step 1.4.3

Find the domain of $x > \sqrt{y}$ and find the intersection with $x \geq 0$ .

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Step 1.4.3.1

Find the domain of $x > \sqrt{y}$ .

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Step 1.4.3.1.1

Set the radicand in $\sqrt{y}$ greater than or equal to $0$ to find where the expression is defined.

$y \geq 0$

Step 1.4.3.1.2

The domain is all values of $x$ that make the expression defined.

$[0, \infty)$

Step 1.4.3.2

Find the intersection of $x \geq 0$ and $[0, \infty)$ .

$x \geq 0$

Step 1.4.4

To find the interval for the second piece, find where the inside of the absolute value is negative.

$x < 0$

Step 1.4.5

In the piece where $x$ is negative, remove the absolute value and multiply by $- 1$ .

$- x > \sqrt{y}$

Step 1.4.6

Find the domain of $- x > \sqrt{y}$ and find the intersection with $x < 0$ .

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Step 1.4.6.1

Find the domain of $- x > \sqrt{y}$ .

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Step 1.4.6.1.1

Set the radicand in $\sqrt{y}$ greater than or equal to $0$ to find where the expression is defined.

$y \geq 0$

Step 1.4.6.1.2

The domain is all values of $x$ that make the expression defined.

$[0, \infty)$

Step 1.4.6.2

Find the intersection of $x < 0$ and $[0, \infty)$ .

$No solution$

Step 1.4.7

Write as a piecewise.

${\begin{cases} x > \sqrt{y} & x \geq 0 \end{cases}$ and $6 x - 5 y < 30$

Step 1.5

Solve $x > \sqrt{y}$ when $x \geq 0$ .

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Step 1.5.1

Solve $x > \sqrt{y}$ for $y$ .

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Step 1.5.1.1

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

$\sqrt{y} < x$ and $6 x - 5 y < 30$

Step 1.5.1.2

To remove the radical on the left side of the inequality, square both sides of the inequality.

${\sqrt{y}}^{2} < x^{2}$ and $6 x - 5 y < 30$

Step 1.5.1.3

Simplify each side of the inequality.

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Step 1.5.1.3.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{y}$ as $y^{\frac{1}{2}}$ .

${(y^{\frac{1}{2}})}^{2} < x^{2}$ and $6 x - 5 y < 30$

Step 1.5.1.3.2

Simplify the left side.

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Step 1.5.1.3.2.1

Simplify ${(y^{\frac{1}{2}})}^{2}$ .

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Step 1.5.1.3.2.1.1

Multiply the exponents in ${(y^{\frac{1}{2}})}^{2}$ .

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Step 1.5.1.3.2.1.1.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$y^{\frac{1}{2} \cdot 2} < x^{2}$ and $6 x - 5 y < 30$

Step 1.5.1.3.2.1.1.2

Cancel the common factor of $2$ .

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Step 1.5.1.3.2.1.1.2.1

Cancel the common factor.

$y^{\frac{1}{2} \cdot 2} < x^{2}$ and $6 x - 5 y < 30$

Step 1.5.1.3.2.1.1.2.2

Rewrite the expression.

$y < x^{2}$ and $6 x - 5 y < 30$

Step 1.5.1.3.2.1.2

Simplify.

$y < x^{2}$ and $6 x - 5 y < 30$

Step 1.5.2

Find the intersection of $y < x^{2}$ and $x \geq 0$ .

$0 \leq x < x^{2}$ and $6 x - 5 y < 30$

Step 1.6

Find the union of the solutions.

$0 \leq x < x^{2}$ and $6 x - 5 y < 30$

Step 2

Simplify the second inequality.

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Step 2.1

Add $5 y$ to both sides of the inequality.

$0 \leq x < x^{2}$ and $6 x < 30 + 5 y$

Step 2.2

Divide each term in $6 x < 30 + 5 y$ by $6$ and simplify.

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Step 2.2.1

Divide each term in $6 x < 30 + 5 y$ by $6$ .

$0 \leq x < x^{2}$ and $\frac{6 x}{6} < \frac{30}{6} + \frac{5 y}{6}$

Step 2.2.2

Simplify the left side.

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Step 2.2.2.1

Cancel the common factor of $6$ .

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Step 2.2.2.1.1

Cancel the common factor.

$0 \leq x < x^{2}$ and $\frac{6 x}{6} < \frac{30}{6} + \frac{5 y}{6}$

Step 2.2.2.1.2

Divide $x$ by $1$ .

$0 \leq x < x^{2}$ and $x < \frac{30}{6} + \frac{5 y}{6}$

Step 2.2.3

Simplify the right side.

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Step 2.2.3.1

Divide $30$ by $6$ .

$0 \leq x < x^{2}$ and $x < 5 + \frac{5 y}{6}$

Step 3