1001 Algebra Problems.PDF

944. b.Identify the following determinants to be
     used in the application of Cramer’s rule for

the matrix equation :

So, from Cramer’s rule, we have:

Thus, the solution is x= – 5,y= 7.

Section 8—Common Algebra

Problems

Set 61 (Page 144)

945. b.The answer should be ^19 because (–3)–2=
     = ^19 .


946. c.There is no error. Any nonzero quantity
     raised to the zero power is 1.


947. a.The statement should be 0.00013 = 1.3
     10 –4because the decimal point must move to
     the left four places in order to yield 0.00013.


948. c.There is no error. The power doesn’t apply
     to the –1 in front of the 4. In order to square
     the entire –4, one must write (–4)^2.


949. a.This is incorrect because you cannot cancel
     members of a sum; you can cancel only factors
     that are common to the numerator and
     denominator.
     950. b.You must first get a common denominator
     before you add two fractions. The correct
     computation is:^34 +  2 a= ^34 + ^24 a= 3+ 4 ^2 a
     951. b.The placement of the quantities is incorrect.
     A correct statement would be “200% of 4 is 8.”


950. c.There is no error. In order to compute 0.50%
     of 10, you multiply 10 by 0.0050 to get 0.05.
     953. b.The sum  3 +  6 cannot be simplified
     further because the radicands are different.
     954. b.The first equality is incorrect; the radicals
     cannot be combined because their indices are
     different.
     955. a.The third equality is incorrect because the
     binomial was not squared correctly. The cor-
     rect denominator should be 2^2 + 2 3 + ( 3 )^2
     = 7 + 2 3 .
     956. a.The exponents should be multiplied, not
     added, so the correct answer should be x^10.
     957. c.There is no error.
     958. a.The first equality is wrong because you must
     multiply the numerator by the reciprocalof the
     denominator.
     959. b.The correct answer should be x^15 because
     = x^12 x^3 = x12+3= x^15.
     960. a.The correct answer should be e^8 xbecause
     (e^4 x)^2 = e^4 x^2 = e^8 x.

Set 62 (Page 146)

961. a.The inequality sign must be switched when
     multiplying both sides by a negative real num-
     ber. The correct solution set should be (–∞, 4).


962. a.There are two solutions of this equation,
     x= –1 and x= 3.


963. c.There is no error.


964. b. The value x= –7 cannot be the solution
     because it makes the terms in the original
     equation undefined—you cannot divide by
     zero. Therefore, this equation has no solution.

x^12

x–3

^1

(–3)(–3)

y== =DDy ––^5687

x===DDx^40 – 8 – 5

D ()( ) ()( )

0

4

14

y==– 20 0 –– 20 4 14 =– 56

Dx==–^142020 ( )() (14 0 ––20 2)()= 40

D==^0420 ()() ()() 00 –– 42 = 8

x

y

0

4

2

0

14

>>>HH= – 20 H

ANSWERS & EXPLANATIONS–