Expert: Ahmed Salami - 6/14/2004

Question
1) when administering a standard intelliegence test, we expect about 1/3 of the scores to be more than 12 units above 100 or more than 12 units below 100. Describe this situation by writing an absolute value inequality.

2) find an equation of the line that passes through the y-intercept and perpendicular to 2x-y+6=0

3)) Find the equation for the line that passes through the point (-5,2) and parallel to the line that passes throught the point (1,2) and (4,3)

3)Solve the rationalinequality
 x^2+x-2
_________
x^2-2x-3<0

4)5) Solve the inequality 8/3[(x-4)]<=2/9[(3x+2)]

2) The manager of a large apartment complex has found that the profit is given by P=-x^2+250x-15,000, where x is the number of apartments rented. For what values of x does the complex produce a profit?

5) solve the quadratic inequality 3x^2-5x>2

Answer
Hi Christysen,
Sorry about the delay.
I am answering your question in the order i feel appropriate.

(2)Before i start i would like you to note a few points. The equation of a straight line is always of the form
y = mx + c
where m is the slope and c is the y intercept
Parallel lines have the same slope.
for perpendicular lines with slopes m and n, mn = -1
For 2x - y + 6 = 0
y = 2x + 6 implying m = 2 and c = 6
Our new equation has a slope -1/2 and an equal intercept of 6
Therefore its equation is
y = (-1/2)x + 6

(3)The general equation of a line passing through two points (x1,y1) and (x2,y2) is
y-y1/x-x1 = y2-y1/x2-x1 = m
For the line passing through (1,2) and (4,3), we have
y-2/x-1 = 3-2/4-1
y-2/x-1 = 1/3  by cross multiplying
3y-6 =x-1
3y = x+5
y = x/3 + 5/3  implying m = 1/3
The new equation has an equal slope but passes through (-5,2)
Therefore
y-2/x+5 = 1/3
3y - 6 = x + 5
3y = x + 11
y = x/3 + 11/3

(6)For the complex to produce a profit,
p = -x^2 + 250x - 15000 should be greater than zero
-x^2 + 250x - 15000 > 0
multiplying both sides by -1 and changing the inequality sign
(NB:multiplying an inequality by a negative number requires  changing the inequality sign )
x^2 - 250x + 15000 < 0
(x-150)(x-100) < 0
Note that the roots of the equation (x-150)(x-100) = 0 are 100 and 150
The usual way is to take numbers in the range x<100, 100<x<150, x>150 and then test the product (x-150)(x-100) for each value. Our required range is the one that satisfies our problem, for this one (x-150)(x-100) < 0
However, i would give you a walkover method for two factors
For (x-a)(x-b) < 0, a<x<b
and for (x-a)(x-b) > 0, x<a and x>b
Therefore for our problem,
100 < x < 150

(7)3x^2 - 5x > 2
3x^2 - 5x - 2 > 0
The roots of this equation are 2 and -1/3
Therefore,
(x-2)(x+1/3) > 0
and by using our method, the solutions are
x < -1/3 and x > 2

(4)x^2 + x - 2 / x^2 -2x - 3   < 0
In fractional inequalities like this one, we must ensure that we multiply by only a positive quantity in order to retain our sign. The trick is therefore to multiply by the square of all denominators we have.
Therefore,
(x^2 + x - 2 / x^2 -2x - 3). ( x^2 -2x - 3) ^2 < 0
(x^2 + x - 2)(x^2 -2x - 3 ) < 0
Factorising results in
(x+2)(x+1)(x-1)(x-3) < 0
and we take numbers in the range x<-2, -2<x<-1, -1<x<1, 1<x<3 and x>3
and test accordingly to find our suitable range(s)

(5)To solve this, we multiply both sides by (x-4)^2.(3x+2)^2 to obtain
(8/3)(x-4).(3x+2)^2 <= (2/9)(x-4)^2.(3x+2)
multiplying by 9 and expanding gives
24(x-4)(9x^2+12x+4) <= 2(x^2-8x+16)(3x+2)
24(9x^3+12x^2+4x-36x^2-48x-16) <= 2(3x^3+2x^2-24x^2-16x+48x+32)
All you simply have to do now is rearrange, factorize and find solution by testing as usual.

(1)As for this one, i am not sure you have asked it well, but i am sure you would be able to do it now.
But as a note,
If |ax|< b
either ax < b or ax > -b

I hope you learn from all these.
Regards.