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168 The Nuts and Bolts of Proof, Third Edition
y z=z(af — ce)/{ad - be). If we substitute this representation of y into
the formula for x, we find that x = (de — bf)/{ad — be). The two
fractions are well defined because ad — be^O. Therefore, we have
found a solution of the given system. (Check what would happen
for other combinations of nonzero coefficients, in addition to
ad-be^O.)
Uniqueness: The solution just found is unique because the values of
X and y are uniquely determined by the two equations found above.
Note: The uniqueness of the solution can be established in another
way, which is considerably longer. Assume that {xi,yi) and ix2,y2)
are two solutions. Therefore,
axi + byi — e ^ 1 ^^'^ ~^ ^^^ ~ ^
exi + dyi =f 1 CX2 + dy2 =f'
Then
ax\ + by\ — ax2 + by2
cXi + dy\ — CX2 + dy2
This is equivalent to:
a(xi - X2) = b(yi - yi)
c(xi - X2) = d(yi - y2)'
Therefore, if a 7^ 0, xi - ^2 = b(yi - y2)/a. Substituting this expression
into the second equation yields (yi — y2)(cid - be) = 0. Because
ad-be^Q, this equality implies y\— y2 = 0; that is, j^i = y2. From
this conclusion, we obtain that x\ = X2, as well. Thus, the two
solutions coincide.
33. This statement will be proved by induction.
a. The smallest number we can use is k = 6. In this case, we obtain
2^ = 64>(6 + 1)^ = 49. Thus, the inequality is true in this case.
b We will now assume that 2" > (n + 1)^.
c. We need to prove that 2"+^ > [{n + 1) + 1]^. Using algebra rules and
the inductive hypothesis, we obtain 2"+^ = 2 x 2">2(w+ 1)^ =
2n^ + 4n + 2. We need to see how the expression we just obtained
compares with the right-hand side of the inequality we want to