1001 Algebra Problems.PDF

(Marvins-Underground-K-12) #1

  1. b.The phrase “p percent” can be represented
    symbolically as  1 p 00 . As such, since we are
    decreasing qby this quantity, the resulting
    quantity is represented by q– 1 p 00 .

  2. d.The cost of the three meals is (a+ b+ c)
    and a 15% tip is represented by. 0.15(a+ b+ c).
    This latter value is added to the cost of the three
    meals to obtain the total cost of the dinner,
    namely (a+ b+ c) + 0.15(a+ b+ c) = 1.15(a+
    b+ c). Now, splitting this cost evenly between
    the two brothers amounts to dividing this
    quantity by 2; this is represented by choices b
    andc.

  3. A 75% increase in enrollment Eis represented
    symbolically as 0.75E, which is equivalent to
    ^34 E. Adding this to the original enrollment E
    results in the sum E+ ^34 E, which is the new
    enrollment.

  4. a.The total cost of her orders, before the dis-
    count is applied, is represented by the sum W


+ X+ Y+ Z. A 15% discount on this amount
is represented symbolically as 0.15(W+ X+
Y+ Z). So, her total cost is (W+ X+ Y+ Z).
(W+ X+ Y+ Z) – 0.15(W+ X+ Y+ Z), which
is equivalent to 0.85 (W+ X+ Y+ Z).

Section 2—Linear Equations
and Inequalities

Set 10 (Page 18)



  1. a.
    z– 7 = –9
    z– 7 + 7 = –9 + 7
    z= –2
    146. e.
     8 k= 8
     8 k8 = 8 8
    k = 64
    147. a.
    –7k– 11 = 10
    –7k– 11 + 11 = 10 + 11
    –7k= 21
    –7k–^17 = 21–^17 
    k= –3
    148. c.
    9 a+ 5 = –22
    9 a+ 5 – 5 = –22 – 5
    9 a= –27
    9 a^19 = –27^19 
    a= –3
    149. d.
    p 6 + 13 = p– 2
    p 6 + 13 – 13 = p– 2 – 13
    p 6 = p– 15
    p 6 – p= p– p– 15

    • ^56 p= –15
      –5p 6 –^65 = –15–^65 
      p= 18



    1. a.
      2.5p+ 6 = 18.5
      2.5p+ 6 – 6 = 18.5 – 6
      2.5p= 12.5
      p= ^122 .. 55 =5

    2. a.
       130 x= ^1255 
       13  0 x(^130 ) = ^1255 (^130 )




x= ^1755 ^0 = 2

ANSWERS & EXPLANATIONS–
Free download pdf