# Math Help - Inequalities

1. ## Inequalities

Solve the inequality. State the solution set using interval notation and graph.

x-3/x+4>7

x^2-5x-7<0

2. Originally Posted by pashah
Solve the inequality. State the solution set using interval notation and graph.
x-3/x+4>7
x^2-5x-7<0
Hello, pashah,

I assume that your first problem reads: $\frac{x-3}{x+4}>7$

Multiply by denominator. The next steps depends on the sign of the denominator!
$x-3>7x+28 \ \wedge\ x>-4$
$-31>6x \ \wedge\ x>-4$
$-\frac{31}{6}>x \ \wedge\ x>-4$. that means $x\notin \mathbb{R}$
or
$x-3<7x+28 \ \wedge\ x<-4$
$-31<6x \ \wedge\ x<-4$
$-\frac{31}{6}. So the solution is:
$-\frac{31}{6}

2nd problem:
Solve this equation: $x^2-5x-7=0$. You'll get:
$x=\frac{5}{2}-\frac{\sqrt{53}}{2}\ \vee\ x=\frac{5}{2}+\frac{\sqrt{53}}{2}$
Because this equation belongs to a quadratic funtion which open upward, the part of the function between the zeros is below the x-axis, that means it is smaller than zero. Therefore your solution is:
$\frac{5}{2}-\frac{\sqrt{53}}{2}

Greetings

EB

3. Originally Posted by pashah
Solve the inequality. State the ... graph....
Hello, pashah,

I've attached the graph of the 1rst problem.

Greetings

EB

4. Originally Posted by pashah
Solve the inequality. State the ...graph.
x^2-5x-7<0
Hello, pashah,

I've attached the graph of your 2nd problem.

Greetings

EB