Linear Algebra Examples

$- 7 y^{2} + z y - x = 0$

Step 1

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 2

Substitute the values $a = - 7$ , $b = z$ , and $c = - x$ into the quadratic formula and solve for $y$ .

$\frac{- z \pm \sqrt{z^{2} - 4 \cdot (- 7 \cdot (- x))}}{2 \cdot - 7}$

Step 3

Simplify.

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Step 3.1

Multiply $- 4 \cdot - 7 \cdot - 1$ .

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Step 3.1.1

Multiply $- 4$ by $- 7$ .

$y = \frac{- z \pm \sqrt{z^{2} + 28 \cdot (- 1 x)}}{2 \cdot - 7}$

Step 3.1.2

Multiply $28$ by $- 1$ .

$y = \frac{- z \pm \sqrt{z^{2} - 28 x}}{2 \cdot - 7}$

Step 3.2

Multiply $2$ by $- 7$ .

$y = \frac{- z \pm \sqrt{z^{2} - 28 x}}{- 14}$

Step 3.3

Simplify $\frac{- z \pm \sqrt{z^{2} - 28 x}}{- 14}$ .

$y = \frac{z \pm \sqrt{z^{2} - 28 x}}{14}$

Step 4

Simplify the expression to solve for the $+$ portion of the $\pm$ .

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Step 4.1

Multiply $- 4 \cdot - 7 \cdot - 1$ .

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Step 4.1.1

Multiply $- 4$ by $- 7$ .

$y = \frac{- z \pm \sqrt{z^{2} + 28 \cdot (- 1 x)}}{2 \cdot - 7}$

Step 4.1.2

Multiply $28$ by $- 1$ .

$y = \frac{- z \pm \sqrt{z^{2} - 28 x}}{2 \cdot - 7}$

Step 4.2

Multiply $2$ by $- 7$ .

$y = \frac{- z \pm \sqrt{z^{2} - 28 x}}{- 14}$

Step 4.3

Simplify $\frac{- z \pm \sqrt{z^{2} - 28 x}}{- 14}$ .

$y = \frac{z \pm \sqrt{z^{2} - 28 x}}{14}$

Step 4.4

Change the $\pm$ to $+$ .

$y = \frac{z + \sqrt{z^{2} - 28 x}}{14}$

Step 5

Simplify the expression to solve for the $-$ portion of the $\pm$ .

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Step 5.1

Multiply $- 4 \cdot - 7 \cdot - 1$ .

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Step 5.1.1

Multiply $- 4$ by $- 7$ .

$y = \frac{- z \pm \sqrt{z^{2} + 28 \cdot (- 1 x)}}{2 \cdot - 7}$

Step 5.1.2

Multiply $28$ by $- 1$ .

$y = \frac{- z \pm \sqrt{z^{2} - 28 x}}{2 \cdot - 7}$

Step 5.2

Multiply $2$ by $- 7$ .

$y = \frac{- z \pm \sqrt{z^{2} - 28 x}}{- 14}$

Step 5.3

Simplify $\frac{- z \pm \sqrt{z^{2} - 28 x}}{- 14}$ .

$y = \frac{z \pm \sqrt{z^{2} - 28 x}}{14}$

Step 5.4

Change the $\pm$ to $-$ .

$y = \frac{z - \sqrt{z^{2} - 28 x}}{14}$

Step 6

The final answer is the combination of both solutions.

$y = \frac{z + \sqrt{z^{2} - 28 x}}{14}$

$y = \frac{z - \sqrt{z^{2} - 28 x}}{14}$

Step 7

Set the radicand in $\sqrt{z^{2} - 28 x}$ greater than or equal to $0$ to find where the expression is defined.

$z^{2} - 28 x \geq 0$

Step 8

Solve for $z$ .

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Step 8.1

Add $28 x$ to both sides of the inequality.

$z^{2} \geq 28 x$

Step 8.2

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

$\sqrt{z^{2}} \geq \sqrt{28 x}$

Step 8.3

Simplify the equation.

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Step 8.3.1

Simplify the left side.

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Step 8.3.1.1

Pull terms out from under the radical.

$| z | \geq \sqrt{28 x}$

Step 8.3.2

Simplify the right side.

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Step 8.3.2.1

Simplify $\sqrt{28 x}$ .

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Step 8.3.2.1.1

Rewrite $28 x$ as $2^{2} \cdot (7 x)$ .

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Step 8.3.2.1.1.1

Factor $4$ out of $28$ .

$| z | \geq \sqrt{4 (7) x}$

Step 8.3.2.1.1.2

Rewrite $4$ as $2^{2}$ .

$| z | \geq \sqrt{2^{2} \cdot 7 x}$

Step 8.3.2.1.1.3

Add parentheses.

$| z | \geq \sqrt{2^{2} \cdot (7 x)}$

Step 8.3.2.1.2

Pull terms out from under the radical.

$| z | \geq | 2 | \sqrt{7 x}$

Step 8.3.2.1.3

The absolute value is the distance between a number and zero. The distance between $0$ and $2$ is $2$ .

$| z | \geq 2 \sqrt{7 x}$

Step 8.4

Write $| z | \geq 2 \sqrt{7 x}$ as a piecewise.

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Step 8.4.1

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

$z \geq 0$

Step 8.4.2

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

$z \geq 2 \sqrt{7 x}$

Step 8.4.3

Find the domain of $z \geq 2 \sqrt{7 x}$ and find the intersection with $z \geq 0$ .

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Step 8.4.3.1

Find the domain of $z \geq 2 \sqrt{7 x}$ .

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Step 8.4.3.1.1

Set the radicand in $\sqrt{7 x}$ greater than or equal to $0$ to find where the expression is defined.

$7 x \geq 0$

Step 8.4.3.1.2

Divide each term in $7 x \geq 0$ by $7$ and simplify.

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Step 8.4.3.1.2.1

Divide each term in $7 x \geq 0$ by $7$ .

$\frac{7 x}{7} \geq \frac{0}{7}$

Step 8.4.3.1.2.2

Simplify the left side.

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Step 8.4.3.1.2.2.1

Cancel the common factor of $7$ .

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Step 8.4.3.1.2.2.1.1

Cancel the common factor.

$\frac{7 x}{7} \geq \frac{0}{7}$

Step 8.4.3.1.2.2.1.2

Divide $x$ by $1$ .

$x \geq \frac{0}{7}$

Step 8.4.3.1.2.3

Simplify the right side.

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Step 8.4.3.1.2.3.1

Divide $0$ by $7$ .

$x \geq 0$

Step 8.4.3.1.3

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

$[0, \infty)$

Step 8.4.3.2

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

$z \geq 0$

Step 8.4.4

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

$z < 0$

Step 8.4.5

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

$- z \geq 2 \sqrt{7 x}$

Step 8.4.6

Find the domain of $- z \geq 2 \sqrt{7 x}$ and find the intersection with $z < 0$ .

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Step 8.4.6.1

Find the domain of $- z \geq 2 \sqrt{7 x}$ .

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Step 8.4.6.1.1

Set the radicand in $\sqrt{7 x}$ greater than or equal to $0$ to find where the expression is defined.

$7 x \geq 0$

Step 8.4.6.1.2

Divide each term in $7 x \geq 0$ by $7$ and simplify.

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Step 8.4.6.1.2.1

Divide each term in $7 x \geq 0$ by $7$ .

$\frac{7 x}{7} \geq \frac{0}{7}$

Step 8.4.6.1.2.2

Simplify the left side.

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Step 8.4.6.1.2.2.1

Cancel the common factor of $7$ .

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Step 8.4.6.1.2.2.1.1

Cancel the common factor.

$\frac{7 x}{7} \geq \frac{0}{7}$

Step 8.4.6.1.2.2.1.2

Divide $x$ by $1$ .

$x \geq \frac{0}{7}$

Step 8.4.6.1.2.3

Simplify the right side.

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Step 8.4.6.1.2.3.1

Divide $0$ by $7$ .

$x \geq 0$

Step 8.4.6.1.3

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

$[0, \infty)$

Step 8.4.6.2

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

$No solution$

Step 8.4.7

Write as a piecewise.

${\begin{cases} z \geq 2 \sqrt{7 x} & z \geq 0 \end{cases}$

Step 8.5

Find the intersection of $z \geq 2 \sqrt{7 x}$ and $z \geq 0$ .

$z \geq 2 \sqrt{7 x}$ and $z \geq 0$

Step 8.6

Find the union of the solutions.

$z \geq N o (Max i m u m)$

Step 9

The domain is all real numbers.

Interval Notation:

$(- \infty, \infty)$

Set-Builder Notation:

${z | z \in ℝ}$

Step 10