1001 Algebra Problems.PDF

531. 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:

(t– 7)(t– 1)t^2 – t 7 +t–^1  (^1)  = 2(t– 7)(t– 1)
t^2 – t 7 (t– 7)(t– 1) + (t–^1 1)(t– 7)(t– 1) =
2(t– 7)(t– 1) = 2t(t– 1) + (t– 7) = 2(t– 7)(t– 1)
2 t^2 – 2t+ t– 7 = 2t^2 – 16 + 14

• t– 7 = –16t + 14

15 t=21

t= ^2115 = ^75 

Since this value does not make any of the

expressions in the original equation, or any

subsequent step of the solution, undefined, it

is indeed a solution of the original equation.

532. a.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:

x(x+ 2)xx++^82 + x(x+ 2) = x(x+ 2)^2 x

x(x+ 8) + 12 = 2(x+ 2)

x^2 + 8 + 12 = 2x+ 4

x^2 + 6x+ 8 = 0

(x+ 4)(x+ 2) = 0

x= –4 or x= – 2

Note that x= –2 makes some of the terms in

the original equation undefined, so it cannot

be a solution of the equation. Thus, we con-

clude that the only solution of the equation is

x= –4.

533. 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:

xx–3+ ^2 x– x–^3  3

x–x 3 x(x– 3) + ^2 xx(x– 3) = x–^3  3 x(x– 3)

x^2 + 2(x– 3) = 3x

x^2 – x– 6 = 0

(x– 3)( x+ 2) = 0

x= 3, –2

Because x= 3 makes some of the terms in the

original equation undefined, it cannot be a

solution of the equation. Thus, we conclude

that the only solution of the equation is x= –2.

534. a.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:

x+^3  2 + 1 =

(2 – x)(2 + x) + 1(2 – x)(2 + x) =

(2 – x)(2 + x)

3(2 – x) + 4 – x^2 = 6

10 – 3x– x^2 = 6

x^2 + 3x– 4 = 0

(x+ 4)(x– 1) = 0

x= –4, 1

Neither of these values makes any of the

expressions in the original equation, or any

subsequent step of the solution, undefined, so

we conclude that both of them are solutions to

the original equation.

^6

(2 – x)(2 + x)

^3

x+ 2

^6

(2 – x)(2 + x)

^12

x^2 +2x

ANSWERS & EXPLANATIONS–