1001 Algebra Problems.PDF

Since neither of these values makes any of the

expressions in the original equation, or any

subsequent step of the solution, undefined, we

conclude that both of them are solutions to

the original equation.

543. c.First, clear the fractions from all terms in
     the equation by multiplying both sides by the
     least common denominator. Then, solve the
     resulting equation using factoring techniques:

tt+–1^1 =

tt+–1^1 = (t–1)^4 (t+1)

(t– 1)(t+ 1)[tt+–1^1 ] = (t– 1)(t+ 1)

(t+1)^4 (t–1)

(t+ 1)(t+ 1) = 4

t^2 + 2t+ 1 = 4

t^2 + 2t– 3 = 0

(t+ 3)(t– 1) = 0

t= –3 or t= 1

Note that t= 1 makes some of the terms in the

original equation undefined, so it cannot be a

solution of the equation. Thus, we conclude

that the only solution of the equation is t = –3.

544. a.
     v=

v[1 + vc^1 v 2 ^2 ] = v 1 + v 2

v+ vvc^1 v 2 ^2 = v 1 + v 2

vvc^1 v 2 ^2 – v 1 = v 2 – v

v 1 (vcv 2 ^2 – 1) = v 2 – v

v 1 = = = (v 2 – v)vv 2 c–

2

c 2 =

Set 35(Page 86)

545. b.First, determine the x-values that make the
     expression on the left side equal to zero or
     undefined. Then, we assess the sign of the
     expression on the left side on each subinterval
     formed using these values. To this end, observe
     that these values are x= –3, –2, and 1. Now, we
     form a number line, choose a real number in
     each of the subintervals, and record the sign of
     the expression above each:

Since the inequality includes “equals,” we include

those values from the number line that make

the numerator equal to zero. The solution set

is [–2, 1].

546. c.First, we must make certain that the numer-
     ator and denominator are both completely fac-
     tored and that all common terms are canceled:
     x 2 x–^22 +x^9 –3= (x–x 32 )+(x^9 +1)

Next, determine the x-values that make this

expression equal to zero or undefined. Then,

we assess the sign of the expression on the left

side on each subinterval formed using these

values. To this end, observe that these values

are x= –1 and 3. Now, we form a number line,

choose a real number in each subinterval, and

record the sign of the expression above each:

Since the inequality does not include “equals,”

we do not include those values from the num-

ber line that make the numerator equal to zero.

Therefore, the solution set is (–∞,– 1)∪(3,∞).

1 3

+ –

+

3 21

+ +

––

c^2 (v^2 – v)

vv 2 – c^2

v 2 – v



vv^2 c– 2 c^2

v^2 – v

vcv 2 ^2 – 1

v^1 + v^2

1 +vc^1 v 2 ^2

^4

t^2 – 1

ANSWERS & EXPLANATIONS–