Linear Algebra Examples

$\frac{9 w^{2} - 49}{3 w^{2} - 1 w - 14} \cdot \frac{7 w^{2} + 6 w - 16}{49 w^{2} - 64}$

Step 1

Set the denominator in $\frac{9 w^{2} - 49}{3 w^{2} - 1 w - 14}$ equal to $0$ to find where the expression is undefined.

$3 w^{2} - 1 w - 14 = 0$

Step 2

Solve for $w$ .

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Step 2.1

Rewrite $- 1 w$ as $- w$ .

$3 w^{2} - w - 14 = 0$

Step 2.2

Factor by grouping.

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Step 2.2.1

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 3 \cdot - 14 = - 42$ and whose sum is $b = - 1$ .

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Step 2.2.1.1

Factor $- 1$ out of $- w$ .

$3 w^{2} - (w) - 14 = 0$

Step 2.2.1.2

Rewrite $- 1$ as $6$ plus $- 7$

$3 w^{2} + (6 - 7) w - 14 = 0$

Step 2.2.1.3

Apply the distributive property.

$3 w^{2} + 6 w - 7 w - 14 = 0$

Step 2.2.2

Factor out the greatest common factor from each group.

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Step 2.2.2.1

Group the first two terms and the last two terms.

$(3 w^{2} + 6 w) - 7 w - 14 = 0$

Step 2.2.2.2

Factor out the greatest common factor (GCF) from each group.

$3 w (w + 2) - 7 (w + 2) = 0$

Step 2.2.3

Factor the polynomial by factoring out the greatest common factor, $w + 2$ .

$(w + 2) (3 w - 7) = 0$

Step 2.3

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$w + 2 = 0$

$3 w - 7 = 0$

Step 2.4

Set $w + 2$ equal to $0$ and solve for $w$ .

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Step 2.4.1

Set $w + 2$ equal to $0$ .

$w + 2 = 0$

Step 2.4.2

Subtract $2$ from both sides of the equation.

$w = - 2$

Step 2.5

Set $3 w - 7$ equal to $0$ and solve for $w$ .

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Step 2.5.1

Set $3 w - 7$ equal to $0$ .

$3 w - 7 = 0$

Step 2.5.2

Solve $3 w - 7 = 0$ for $w$ .

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Step 2.5.2.1

Add $7$ to both sides of the equation.

$3 w = 7$

Step 2.5.2.2

Divide each term in $3 w = 7$ by $3$ and simplify.

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Step 2.5.2.2.1

Divide each term in $3 w = 7$ by $3$ .

$\frac{3 w}{3} = \frac{7}{3}$

Step 2.5.2.2.2

Simplify the left side.

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Step 2.5.2.2.2.1

Cancel the common factor of $3$ .

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Step 2.5.2.2.2.1.1

Cancel the common factor.

$\frac{3 w}{3} = \frac{7}{3}$

Step 2.5.2.2.2.1.2

Divide $w$ by $1$ .

$w = \frac{7}{3}$

Step 2.6

The final solution is all the values that make $(w + 2) (3 w - 7) = 0$ true.

$w = - 2, \frac{7}{3}$

Step 3

Set the denominator in $\frac{7 w^{2} + 6 w - 16}{49 w^{2} - 64}$ equal to $0$ to find where the expression is undefined.

$49 w^{2} - 64 = 0$

Step 4

Solve for $w$ .

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Step 4.1

Add $64$ to both sides of the equation.

$49 w^{2} = 64$

Step 4.2

Divide each term in $49 w^{2} = 64$ by $49$ and simplify.

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Step 4.2.1

Divide each term in $49 w^{2} = 64$ by $49$ .

$\frac{49 w^{2}}{49} = \frac{64}{49}$

Step 4.2.2

Simplify the left side.

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Step 4.2.2.1

Cancel the common factor of $49$ .

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Step 4.2.2.1.1

Cancel the common factor.

$\frac{49 w^{2}}{49} = \frac{64}{49}$

Step 4.2.2.1.2

Divide $w^{2}$ by $1$ .

$w^{2} = \frac{64}{49}$

Step 4.3

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

$w = \pm \sqrt{\frac{64}{49}}$

Step 4.4

Simplify $\pm \sqrt{\frac{64}{49}}$ .

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Step 4.4.1

Rewrite $\sqrt{\frac{64}{49}}$ as $\frac{\sqrt{64}}{\sqrt{49}}$ .

$w = \pm \frac{\sqrt{64}}{\sqrt{49}}$

Step 4.4.2

Simplify the numerator.

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Step 4.4.2.1

Rewrite $64$ as $8^{2}$ .

$w = \pm \frac{\sqrt{8^{2}}}{\sqrt{49}}$

Step 4.4.2.2

Pull terms out from under the radical, assuming positive real numbers.

$w = \pm \frac{8}{\sqrt{49}}$

Step 4.4.3

Simplify the denominator.

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Step 4.4.3.1

Rewrite $49$ as $7^{2}$ .

$w = \pm \frac{8}{\sqrt{7^{2}}}$

Step 4.4.3.2

Pull terms out from under the radical, assuming positive real numbers.

$w = \pm \frac{8}{7}$

Step 4.5

The complete solution is the result of both the positive and negative portions of the solution.

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Step 4.5.1

First, use the positive value of the $\pm$ to find the first solution.

$w = \frac{8}{7}$

Step 4.5.2

Next, use the negative value of the $\pm$ to find the second solution.

$w = - \frac{8}{7}$

Step 4.5.3

The complete solution is the result of both the positive and negative portions of the solution.

$w = \frac{8}{7}, - \frac{8}{7}$

Step 5

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

Interval Notation:

$(- \infty, - 2) \cup (- 2, - \frac{8}{7}) \cup (- \frac{8}{7}, \frac{8}{7}) \cup (\frac{8}{7}, \frac{7}{3}) \cup (\frac{7}{3}, \infty)$

Set-Builder Notation:

${w | w \neq \frac{8}{7}, - \frac{8}{7}, - 2, \frac{7}{3}}$

Step 6