College Algebra

Question

1 a) Solve: -5x - 4  > 4 (2x - 5)

 

b) Solve: x^2 < 9

 

c) Solve:     x -3      <  0

                 x + 5

Answer 1

I will help you with part b and c.

 

b)

 

Square-root both sides of the equation.

 

x = ±√(-9)

 

x = -3i          and        x = 3i

 

 

c)

 

(x - 3) / (x + 5) < 0

 

 

Notice that we have a rational expression that is negative.  If we find all the values of x in which the expression is undefined, we see that

 

x = -5    makes the expression undefined.  This is also the vertical asymptote when you graph it.  So we pick test points to evaluate the rational expression.  We want the result to be negative.

 

Also, the x-intercept is x=3

 

 

For x=-10:

 

(-10 - 3) / (-10 + 5) = -13 / -5 = 13 / 5

 

 

For x=-7:

 

(-7 - 3) / (-7 + 5) = -10 / -2 = 5

 

---------------------------------------------------------------------vertical asymptote

 

For x=-4

 

(-4 - 3) / (-4 + 5) = -7 / 1 = -7

 

 

For x=-3:

 

(-3 - 3) / (-3 + 5) = -6 / 2 = -3

 

 

For x=4:

 

(4 - 3) / (4 + 5) = 1 / 9

 

 

As you can see, the results are only negative between in the interval of (-5, 3).

 

Answer 2

a). -5x-4 > 4(2x-5)

 

     -5x-4 > 8x-20

 

        -13x > -16

 

            x < 16/13

 

b). x² < 9

 

     x² - 9 < 0

 

     (x+3)(x-3) < 0

 

     (x+3)(x-3) = 0 when x = -3 or 3

 

     If x<-3, then x+3 and x-3 are both negative, so (x+3)(x-3) > 0

     If x>3, then x+3 and x-3 are both positive, so (x+3)(x-3) > 0

     If -3 < x < 3, then x+3 > 0 and x-3 < 0, so (x+3)(x-3) < 0

 

     Solution:  -3 < x < 3

 

c).  (x-3)/(x+5) < 0

 

      x-3 = 0 when x = 3   x+5=0 when x = -5

 

      If x < -5, the numerator is negative and the denominator is also            negative, so (x-3)/(x+5) > 0.

 

      If x > 3, the numerator is positive and so is the denominator.               Thus, (x-3)/(x+5) > 0.

 

      if -5 < x < 3, the numerator is negative and the denominator is           positive, so (x-3)/(x+5) < 0.

 

      Solution:  -5 < x < 3