# Inequalities

• Aug 2nd 2006, 09:03 PM
pashah
Inequalities
Solve the inequality. State the solution set using interval notation and graph.

x-3/x+4>7

x^2-5x-7<0
• Aug 3rd 2006, 12:44 AM
earboth
Quote:

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
• Aug 3rd 2006, 11:36 AM
earboth
Quote:

Originally Posted by pashah
Solve the inequality. State the ... graph....

Hello, pashah,

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

Greetings

EB
• Aug 3rd 2006, 11:59 AM
earboth
Quote:

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