How to solve (x+3)/(x-5) > 8

1. Factor the equation (x + 2)(x + 3) = 0. 2. Set each factor equal to zero: x + 2 = 0 and x + 3 = 0. 3. Solve for x: x = -2 and x = -3.

Solve (x+3)/(x-5) > 8 | Equatix

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Question

$$\frac{ x+3 }{ x-5 } > 8$$

Solve for x

$x\in \left(5,\frac{43}{7}\right)$

Solution Steps

Denominator $x-5$ cannot be zero since division by zero is not defined. There are two cases.

$$x-5>0$$ $$x-5<0$$

Consider the case when $x-5$ is positive. Move $-5$ to the right hand side.

$$x>5$$

The initial inequality does not change the direction when multiplied by $x-5$ for $x-5>0$.

$$x+3>8\left(x-5\right)$$

Multiply out the right hand side.

$$x+3>8x-40$$

Move the terms containing $x$ to the left hand side and all other terms to the right hand side.

$$x-8x>-3-40$$

Combine like terms.

$$-7x>-43$$

Divide both sides by $-7$. Since $-7$ is negative, the inequality direction is changed.

$$x<\frac{43}{7}$$

Consider condition $x>5$ specified above.

$$x\in \left(5,\frac{43}{7}\right)$$

Now consider the case when $x-5$ is negative. Move $-5$ to the right hand side.

$$x<5$$

The initial inequality changes the direction when multiplied by $x-5$ for $x-5<0$.

$$x+3<8\left(x-5\right)$$

Multiply out the right hand side.

$$x+3<8x-40$$

Move the terms containing $x$ to the left hand side and all other terms to the right hand side.

$$x-8x<-3-40$$

Combine like terms.

$$-7x<-43$$