1001 Algebra Problems.PDF

525. b.



x

2

2

x^2

+

+

4 x

x

• 
  


• 
  

5

 3 

2

x

x

+

+

1

^3

– x+

2

 2 =

  – x+

2

 2 = 

x

x

+

+

5

 1 – x+

2

 2 =

– =

= =

526. c.



x

x

+

–3

^5 – x

x–

1

 3 – 

x

x

+

–3

^5 –

x–

1

 3

= (x– 3)

= – (x^2 – 4x– 5) – (x– 5)(x+ 1

527. c.

3 + x+

1

 (^3) 
x
x

• –2
  ^3 = [3(
  x
  x


• 
  


• 
  

  3
  3)+ 
  x+
  1
   3 ]
  x
  x

  
  


• 
  

  –2

  
  

^3 =

^3 xx++ 310 xx+–2^3 =

528. d.
     1 – ^2 x–  23 x–  61 x=

1 – ^2 x–  23 x+  61 x=

1 – (^26 (6x)– ^36 (3x)+  61 x) =

1 – 12 – 69 x+1= ^6 x 6 – x^4 = ^3 x 3 – x^2

Set 34 (Page 84)

529. a.First, clear the fractions from all terms in the
     equation by multiplying both sides by the least
     common denominator (LCD). Then, solve the
     resulting equation using factoring techniques:

^3 x= 2 + x

x^3 x= x(2 + x)

3 = 2x+ x^2

x^2 + 2x– 3= 0

(x+ 3)(x– 1) = 0

x+ 3 = 0 or x– 1 = 0

x= –3 or x= 1

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.

530. d.First, clear the fractions from all terms in the
     equation by multiplying both sides by the least
     common denominator. Then, solve the result-
     ing equation using factoring techniques:

^23 – ^3 x= ^12 

^23  6 x– ^3 x 6 x= ^12  6 x

4 x– 18 = 3x

x= 18

This value does not make any of the expres-

sions in the original equation, or any subse-

quent step of the solution, undefined, so we

conclude that it is indeed a solution of the

original equation.

^3 x+ 10

x+ 3

x+ 5 – x^2 + 3x

x– 3

x(x– 3)

x– 3

x^2 + 5x+ 8

(x+ 1)(x+ 2)

x^2 + 7x+ 10 – 2x– 2

(x+ 1)(x+ 2)

(x+ 5)(x+ 2) – 2(x+ 1)

(x+1)(x+2)

2(x+ 1)

(x+ 2)(x+ 1)

(x+ 5)(x+ 2)

(x+ 1)(x+ 2)

^2 x+ 3

x+ 1

(x= 5)(x– 1)

(2x+ 3)(x– 1)

ANSWERS & EXPLANATIONS–