Solve the Inequality for x (x^2+2x-3)/(3x^2-7x-6)>0

$\frac{x^{2} + 2 x - 3}{3 x^{2} - 7 x - 6} > 0$

Step 1

Find all the values where the expression switches from negative to positive by setting each factor equal to $0$ and solving.

$x^{2} + 2 x - 3 = 0$

$3 x^{2} - 7 x - 6 = 0$

Step 2

Factor $x^{2} + 2 x - 3$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 3$ and whose sum is $2$ .

$- 1, 3$

Write the factored form using these integers.

$(x - 1) (x + 3) = 0$

Step 3

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

$x - 1 = 0$

$x + 3 = 0$

Step 4

Set $x - 1$ equal to $0$ and solve for $x$ .

Set $x - 1$ equal to $0$ .

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

Step 5

Set $x + 3$ equal to $0$ and solve for $x$ .

Set $x + 3$ equal to $0$ .

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

Step 6

The final solution is all the values that make $(x - 1) (x + 3) = 0$ true.

$x = 1, - 3$

Step 7

Factor by grouping.

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 $a \cdot c = 3 \cdot - 6 = - 18$ and whose sum is $b = - 7$ .

Factor $- 7$ out of $- 7 x$ .

$3 x^{2} - 7 x - 6 = 0$

Rewrite $- 7$ as $2$ plus $- 9$

$3 x^{2} + (2 - 9) x - 6 = 0$

Apply the distributive property.

$3 x^{2} + 2 x - 9 x - 6 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(3 x^{2} + 2 x) - 9 x - 6 = 0$

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

$x (3 x + 2) - 3 (3 x + 2) = 0$

Factor the polynomial by factoring out the greatest common factor, $3 x + 2$ .

$(3 x + 2) (x - 3) = 0$

Step 8

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

$3 x + 2 = 0$

$x - 3 = 0$

Step 9

Set $3 x + 2$ equal to $0$ and solve for $x$ .

Set $3 x + 2$ equal to $0$ .

$3 x + 2 = 0$

Solve $3 x + 2 = 0$ for $x$ .

Subtract $2$ from both sides of the equation.

$3 x = - 2$

Divide each term in $3 x = - 2$ by $3$ and simplify.

Divide each term in $3 x = - 2$ by $3$ .

$\frac{3 x}{3} = \frac{- 2}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} = \frac{- 2}{3}$

Divide $x$ by $1$ .

$x = \frac{- 2}{3}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{2}{3}$

Step 10

Set $x - 3$ equal to $0$ and solve for $x$ .

Set $x - 3$ equal to $0$ .

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

Step 11

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

$x = - \frac{2}{3}, 3$

Step 12

Solve for each factor to find the values where the absolute value expression goes from negative to positive.

$x = 1, - 3$

$x = - \frac{2}{3}, 3$

Step 13

Consolidate the solutions.

$x = 1, - 3, - \frac{2}{3}, 3$

Step 14

Find the domain of $\frac{x^{2} + 2 x - 3}{3 x^{2} - 7 x - 6}$ .

Set the denominator in $\frac{x^{2} + 2 x - 3}{3 x^{2} - 7 x - 6}$ equal to $0$ to find where the expression is undefined.

$3 x^{2} - 7 x - 6 = 0$

Solve for $x$ .

Factor by grouping.

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 $a \cdot c = 3 \cdot - 6 = - 18$ and whose sum is $b = - 7$ .

Factor $- 7$ out of $- 7 x$ .

$3 x^{2} - 7 x - 6 = 0$

Rewrite $- 7$ as $2$ plus $- 9$

$3 x^{2} + (2 - 9) x - 6 = 0$

Apply the distributive property.

$3 x^{2} + 2 x - 9 x - 6 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(3 x^{2} + 2 x) - 9 x - 6 = 0$

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

$x (3 x + 2) - 3 (3 x + 2) = 0$

Factor the polynomial by factoring out the greatest common factor, $3 x + 2$ .

$(3 x + 2) (x - 3) = 0$

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

$3 x + 2 = 0$

$x - 3 = 0$

Set $3 x + 2$ equal to $0$ and solve for $x$ .

Set $3 x + 2$ equal to $0$ .

$3 x + 2 = 0$

Solve $3 x + 2 = 0$ for $x$ .

Subtract $2$ from both sides of the equation.

$3 x = - 2$

Divide each term in $3 x = - 2$ by $3$ and simplify.

Divide each term in $3 x = - 2$ by $3$ .

$\frac{3 x}{3} = \frac{- 2}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} = \frac{- 2}{3}$

Divide $x$ by $1$ .

$x = \frac{- 2}{3}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{2}{3}$

Set $x - 3$ equal to $0$ and solve for $x$ .

Set $x - 3$ equal to $0$ .

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

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

$x = - \frac{2}{3}, 3$

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

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

Step 15

Use each root to create test intervals.

$x < - 3$

$- 3 < x < - \frac{2}{3}$

$- \frac{2}{3} < x < 1$

$1 < x < 3$

$x > 3$

Step 16

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

Test a value on the interval $x < - 3$ to see if it makes the inequality true.

Choose a value on the interval $x < - 3$ and see if this value makes the original inequality true.

$x = - 6$

Replace $x$ with $- 6$ in the original inequality.

$\frac{{(- 6)}^{2} + 2 (- 6) - 3}{3 {(- 6)}^{2} - 7 \cdot - 6 - 6} > 0$

The left side $0.1458\overline{3}$ is greater than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $- 3 < x < - \frac{2}{3}$ to see if it makes the inequality true.

Choose a value on the interval $- 3 < x < - \frac{2}{3}$ and see if this value makes the original inequality true.

$x = - 2$

Replace $x$ with $- 2$ in the original inequality.

$\frac{{(- 2)}^{2} + 2 (- 2) - 3}{3 {(- 2)}^{2} - 7 \cdot - 2 - 6} > 0$

The left side $- 0.15$ is not greater than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $- \frac{2}{3} < x < 1$ to see if it makes the inequality true.

Choose a value on the interval $- \frac{2}{3} < x < 1$ and see if this value makes the original inequality true.

$x = 0$

Replace $x$ with $0$ in the original inequality.

$\frac{{(0)}^{2} + 2 (0) - 3}{3 {(0)}^{2} - 7 \cdot 0 - 6} > 0$

The left side $0.5$ is greater than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $1 < x < 3$ to see if it makes the inequality true.

Choose a value on the interval $1 < x < 3$ and see if this value makes the original inequality true.

$x = 2$

Replace $x$ with $2$ in the original inequality.

$\frac{{(2)}^{2} + 2 (2) - 3}{3 {(2)}^{2} - 7 \cdot 2 - 6} > 0$

The left side $- 0.625$ is not greater than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $x > 3$ to see if it makes the inequality true.

Choose a value on the interval $x > 3$ and see if this value makes the original inequality true.

$x = 6$

Replace $x$ with $6$ in the original inequality.

$\frac{{(6)}^{2} + 2 (6) - 3}{3 {(6)}^{2} - 7 \cdot 6 - 6} > 0$

The left side $0.75$ is greater than the right side $0$ , which means that the given statement is always true.

True

Compare the intervals to determine which ones satisfy the original inequality.

$x < - 3$ True

$- 3 < x < - \frac{2}{3}$ False

$- \frac{2}{3} < x < 1$ True

$1 < x < 3$ False

$x > 3$ True

$x < - 3$ True

$- 3 < x < - \frac{2}{3}$ False

$- \frac{2}{3} < x < 1$ True

$1 < x < 3$ False

$x > 3$ True

Step 17

The solution consists of all of the true intervals.

$x < - 3$ or $- \frac{2}{3} < x < 1$ or $x > 3$

Step 18

The result can be shown in multiple forms.

Inequality Form:

$x < - 3 or - \frac{2}{3} < x < 1 or x > 3$

Interval Notation:

$(- \infty, - 3) \cup (- \frac{2}{3}, 1) \cup (3, \infty)$

Step 19

Solve the Inequality for x (x^2+2x-3)/(3x^2-7x-6)>0

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