Cracking The SAT Premium

solved   simultaneously  by  adding  or  subtracting     them.   The     key     is  to  get     one     variable    to

disappear.  Stack   the equations   and add them.

Therefore,  p   =   .

17. 
    

    6 Substitute x for y in the second equation to get (x − 2)^2 – 4 = –x. Expand the left side of the
    equation to get (x − 2)(x − 2) – 4 = –x or x^2 – 4x + 4 – 4 = –x. Simplify the equation to get
    x^2 – 4x = –x. Set the equation to 0 to get x^2 – 3x = 0. Factor an x out of the equation to get
    x(x − 3) = 0. Therefore, either x = 0 or x − 3 = 0, and x = 3. According to the question, the
    point of intersection is in quadrant I, where the x and y values are both positive. Therefore,
    x = 3 and y = 3. The sum of 3 + 3 = 6. The correct answer is 6.

    
    


18. 
    

    6 Try Plugging In different values of c to see which ones work. Make a table to keep track of
    all the numbers.

    
    

c   = (c    –   1)^2    =

1 (1    –   1)^2    =   0

2 (2    –   1)^2    =   1

3 (3    –   1)^2    =   4

4 (4    –   1)^2    =   9

5 (5    –   1)^2    =   16

6 (6    –   1)^2    =   25

7 (7    –   1)^2    =   36

The largest value   of  c   that    works   without hitting the boundaries  of  the inequality  is  6,  so  the

correct answer  is  6.

19. 8 When no picture is provided, it helps to draw one. First, rewrite each equation so that it is

in  the slope-intercept form    of  a   line,   which   is  y   =   mx  +   b,  where   m   is  the slope   and b   is  the

y-intercept of  the line.   The first   equation    becomes 2y  ≤   x   +   4,  or  y   ≤    x  +   2.  The second

equation    becomes y   ≥   2x  −   4.  The resulting   graph   looks   as  follows: