Precalculus: Mathematics for Calculus, 7th Edition

SOLUTION $x\lt\frac{-3}{2}$ INTERVAL $(\frac{-3}{2}, -1)$ GRAPH A line from $\frac{-3}{2}$ to -1. You have tp put an open circle in those two values and trace a line connecting them. In that way you are indicating that the values between those two solve the ineqaulity.
SOLVE 1 $\frac{x}{x+1}\gt3$ Cross multiply $x\gt3(x+1)$ $x\gt3x+3$ $x-3x\gt3$ $-2x\gt3$ You change signs because you are going to divide by -2 $x\lt\frac{-3}{2}$ So the solution is \frac{-3}{2} INTERVAL Now let’s analyze the inequality. It can’t be equal to 3 therefore we are going to use parenthesis. The inequality has a fraction therefore we have to analyze the denominator because it can’t equal zero otherwise it will be undefined. $$x+1=0$$ $$x=-1$$ So the solution can’t be negative one In order to know which one is the solution you have to think of the intervals that you have and test values that are between them $(\infty, \frac{-3}{2})$ = $y(-2)=\frac{-2}{-2+1}$ = $y(-2)=-2$ This interval gives values less than 3 so it is not the solution $(\frac{-3}{2}, -1)$ = $y(-1.2)=\frac{-1.2}{-1.2+1}$ = $y(-1.2)=6$ This interval gives us numbers greater than 3 therefore it is a solution. $(-1, \infty)$ = $y(3)=\frac{3}{3+1}$ = $y(3)=0.75$ This interval is not a solution The correct interval is $$(\frac{-3}{2},-1)$$ GRAPH You do a real number line and write the number from $\frac{-3}{2} to -1$. You have to use an open circle to indicate that you can’t use those values because they will give you 3 and that doesn’t make the inequality correct.