Test B Problems and Solutions

I should have mentioned that #2 and #5 were the 4 point problems. Sorry.

1. What is the intersection of y = 3x+4 and y=x-2
   Since they are both equal to y, that means that:
   3x+4=x-2
   2x+4 = -2
   2x = -6
    x = -3 and plugging -3 into y=x-2 we get y=-3-2 or y=-5 so the point of intersection is (-3,-5)
Let's check our answer by plugging those x and y values into our original two equations.
 (-5)=3(-3) + 4 and (-5)=(-3) - 2
-5 = -9 +4 and -5 = -5 Well, those are both true so we have found the point that satisfies both equations.

2. Find the two points where y=|x-4| and y=2/3 x - 2 intersect. Leave the coordinates as impro
per fractions or integers.
Either y = x-4 or y= -x + 4  Now let's plug in 2/3 x - 2 for y in each of the two equations.
2/3 x - 2 = x-4  or 2/3 x - 2 = -x +4
-2 = 1/3 x -4    or   5/3 x - 2 = 4
2 = 1/3 x       or      5/3 x= 6
6 = x             or      x= 18/5     Now let's y=2/3x -2 to figure out the y value.
y = 2/3(6) - 2  and y=2/3(18/5) - 2
y = 4 - 2          and y = 36/15 - 2
y = 2              and y = 12/5 -10/5
                      and y = 2/5
The two points are (6,2) and (18/5,2/5)

3.Write as a trinomial in descending order: (2x+7) (9x-4)
 18x² - 8x + 63x -28 = 18x² + 55x -28 

4. (x² - 4)/(x  + 2)
There is no x term in the numerator so the coefficient is zero. So the question is what two numbers add to zero and multiply to negative four? Ans. 2 and -2. 
Hence simplify to ((x+2)(x-2))/(x+2) = x-2 =c

5. What is the length of the diagonal of a square of side length 2√15 m, in simplified radical form. Use the Pythagorean Theorem: (2√15)² + (2√15)² = c²
Hence 4x15 + 4x15 = c² or 60 + 60 = c² 
Hence 120 = c²  or c = √120 = √2x2x2x3x5= 2√30