Answers to Algebra Review

1.  First, regroup the terms according to the base (3 versus 5):
    12(3^(n-1)) - 2(3^n) - 15(5^(n-1)) + 5^n

    Now, we need to raise each exponent to n+1, instead of n and n-1.
    Remember that when you raise an exponent, you need to compensate
    by multiplying by a small coefficient.  For example, to re-write
    3^n as 3^(n+1), we need to multiply by 1/3:  3^n = (1/3) 3^(n+1).
    Similarly, to re-write 3^(n-1) as 3^(n+1), we need to multiply by 1/9:
    3^(n-1) = (1/9) 3^(n+1).  This technique applies to each term in
    the expression:
    (12/9) 3^(n+1) - (2/3) 3^(n+1) - (15/25) 5^(n+1) + (1/5) 5^(n+1)

    And we can conbine like terms and simplify.
    (12/9 - 2/3) 3^(n+1) + (-3/5 + 1/5) 5^(n+1)

    = (2/3) 3^(n+1) + (-2/5) 5^(n+1)

    And we see that p = 2/3 and q = -2/5.

2a.  Factor the equation:  (x+6)(x-2) = 0 --> x = -6 or 2

2b.  Use the quadratic formula:      3 +/- sqrt(21)
                                 x = --------------
                                            2
2c.  For a cubic equation, let's use synthetic substitution to identify
     factors.  +1 doesn't work as a root, but -1 does:

     -1 /  1   0   -7   -6
     ----     -1    1    6
          -----------------
           1  -1   -6    0      The "0" means a remainder of 0, so we see that
                                (x+1) is a divisor, and the quotient is
                                (x^2 - x - 6), which we can easily factor.
     Therefore, x^3 - 7x -6 = (x+1)(x-3)(x+2)

     (x+1)(x-3)(x+2) = 0 --> x = -1, -2, 3

3a.  To cancel the b's let's multiply the first equation by 4 and the second
     equation by -3:

     (4*)  5a - 3b = 7   -->   20a - 12b = 28
     (-3*) 7a - 4b = 11  -->  -21a + 12b = -33   And now we add:
                              -----------------
                               -a        = -5
                               So, a = 5.
     Substituting a = 5 into one of the original equations, we obtain:
     5(5) - 3b = 7
     25 - 3b = 7
     -3b = -18
     b = 6.
     So, our solution is a = 5 and b = 6.  We can check by substituting
     into the other equation:  7(5) - 4(6) = 11 --> 35 - 24 = 11.  It works!

3b.  First, let's cancel the c's by adding the first and third equations:

     3a + 5b - 3c = -4
     2a + 2b + 3c = 5
     ------------------
     5a + 7b      = 1

     Now, we need another equation involving only a and b.  This means
     we need to cancel c's in a different pair of equations.  Let's try
     the 2nd and 3rd.  To make the coefficients cancel, we need to
     multiply the 2nd equation by 3 and the 3rd equation by 2:

     (3*)  2a - 5b - 2c = 14  -->  6a - 15b - 6c = 42
     (2*)  2a + 2b + 3c = 5   -->  4a + 4b  + 6c = 10  And now we add:
                                   ------------------
                                  10a - 11b      = 52

     We now have a system of 2 equations in 2 variables:

     5a + 7b = 1
     10a - 11b = 52

     Let's multiply the first equation by -2, and add:

     (-2*)  5a + 7b  = 1   --> -10a - 14b = -2
           10a - 11b = 52  -->  10a - 11b = 52  And now we add:
                               ---------------
                                    -25b = 50
                                       b = -2
     Substitute b = -2 into an equation that has only one more variable:
     5a + 7(-2) = 1
     5a - 14 = 1
     5a = 15
     a = 3.

     We now have a = 3 and b = -2.  Substitute into an equation containing
     all three variables, such as the first original equation.
     3(3) + 5(-2) - 3c = -4
     9 - 10 - 3c = -4
     -1 - 3c = -4
     -3c = -3
     c = 1

     Therefore, our solution is (a, b, c) = (3, -2, 1).  We can check by
     substituting into the other equations:

     2(3) -5(-2) -2(1) = 14  -->  6 + 10 - 2 = 14 --> 16 - 2 = 14.  Yes!
     2(3) + 2(-2) + 3(1) = 5  -->  6 - 4 + 3 = 5  -->  5 = 5.  Yes!

5a.  The first term (a_1) of this geometric sequence is 10/16.  What is the 
     common ratio?  We could divide a_2/a_1 to give us r = (10/8) / (10/16) = 2.

     The formula for a_n is a_0 r^n, where a_0 is the mythical "zeroth"
     element.  This is 10/32.

     a_n = (10/32) 2^n

     Throughout this problem, you could of course reduce the fractions.
     I chose not to simply so that it would be easier to spot the
     common ratio.

5b.  The formula for the first n terms of a geometric series is
                  r^n - 1
     (first term) -------
                  r - 1

     To see why, note that the fraction above works out to
     r^(n-1) + r^(n-2) + r^(n-3) + ... + 1.
     And if you multiply each term of this series by a_1, the first term,
     you obtain all the terms in the series!

     a_1 (r^n - 1) / (r - 1) = (10/32) (2^n - 1) / (2-1),
     So our sum is simply (10/32)(2^n - 1).

6.   Use the binomial theorem:  C(10,3) = (10 * 9 * 8) / (3 * 2 * 1) =
     720 / 6 = 120.

7.   Set the parts equal.  The coefficients of x^2 must match, and the
     coefficients of x must match.  (The constant terms must also match,
     and we see they are zero on both sides).

     Coefficients of x^2:  7 = 2a + 5b
     Coefficients of x:    18 = 3a

     So, this question is a system in disguise.  The second equation
     gives us a = 6.  Substituting into the first equation, we have:
     2(6) + 5b = 7
     12 + 5b = 7
     b = -1.
     So, a = 6 and b = -1.

8.   In this problem, all the logs are to the base 10.
     Note that log 10 = 1, so that 
     log 5 = log (10/2) = log 10 - log 2 = 1 - 0.3 = 0.7

     The number of digits is approximated by the log.
     So, the question is asking for log(5^n) = n log 5 = 0.7 n.