Solutions for Arithmetic
Review Problems

Volume 2, Three Skills for Algebra

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2.1 Basic Stuff

Perform the indicated calculations by hand. Then check your calculations with the aid of a calculator.

• 456+76+312 = 844
• 176·86 = 15136
• 4892-2396 = 2496. Check: 2496+2396 = 4892
• 1416¸813 = 1.742 to 3 decimal places after the decimal point.
• 2396-4892 = -(4892-2396) = -(2496) = -2496

2.2 More Basic Stuff

Simplify if possible. Remember that operations inside parentheses ( ) or brackets [ ] are to be done first.

Here 3² = 3 ×3. The other square root of 9 is -3. This question is perhaps ambiguous - oops, unless you we follow the convention that the phrase square root" here means the principal square root. We will follow this convention below.

in accordance with the convention that the square root of a positive number is always taken to be its principal square root. The answer -3 is not acceptable according to this principal square root convention. The number -3 is the other square root.

Therefore 

The numbers appearing in the calculation of M are identical to those appearing in the calculation of 

The number 

The factors of the number O and the number 

U = p- 3.1416 ¹ 0 as p is not exactly 3.1416        A better approximation to p is 3.141592654 but the latter is still not exact. The decimal expansion of p is infinite and non-repeating as the number p is not rational - why is a intellectual debt left to a higher mathematics course, if any. Here not rational means p is not a number of the form [(p)/(q)] where both p and q are whole numbers.

2.3 Calculator Button Exercises

1. A = sin(90^°) = 1
2. B = sin(180^°) = 0
3. C = sin(0^°) = 0
4. D = sin(270^°) = -1
5. E = sin(-90^°) = 0
6. F = sin(-720^°) = 0
7. G = cos(90^°) = 0
8. H = cos(180^°) = -1
9. I = cos(360^°) = 1
10. J = cos(0^°) = 1
11. K = cos(-90^°) = 0
12. L = cos(-720^°) = 1

1. a = sin([(p)/2] radians) = 1
2. b = sin(p  radians) = 0
3. c = sin(0  radians) = 0
4. d = sin([3/2]p radians) = -1
5. e = sin(-[1/2]p radians) = -1
6. f = sin(-4p radians) = 0
7. g = cos([1/2]p radians) = 0
8. f = cos(p radians) = -1
9. h = cos(2p radians) = 1
10. i = cos(1.5p  radians) = 0
11. j = cos(-[1/2]p radians) = 0
12. k = cos(-4p radians) = 1

2.4 More Calculator Button Work

1. A = exp( 2 ln(5)) = 25
2. B = e^{2 ln(5)} = 5² = 25
3. C = 10 ^{2 log(5)} = 25
4. D = 10^log(25) = 25
5. E = ln(exp(6.2)) = 6.2
6. F = ln( e^6.2) = 6.2
7. G = the sixth root of (16)¹² = (16)² = 256
8. H = [(16)¹²]^[1/6] = (16)² = 256
9. I = 1+3+3²+3³+3⁴+3⁵+3⁷ = 2551
10. J = [(-1+3⁷)/([-1+3])] = 1094
11. K = [([1-3⁷])/([1-3])] = 1094
12. M = 1+(1.06)+(1.06)²+(1.06)³ = 4.374616
13. N = [([-1+(1.06)⁴])/([-1+1.06])] = 4.374616
14. P = [([(1.06)⁴-1])/([1.06-1])] = 4.374616
15. Q = [([-1+(1.06)⁴])/[0.06]] = 4.374616
16. R = [1+(1.02)¹+(1.02)²+(1.02)³+(1.02)⁴]×(1.02)^(-4) = 4.480773
17. S = [([(1.02)⁽⁵⁾-1])/([1.02-1])] ×(1.02)^(-4) = 4.4416
18. T = 1+(1.02)^(-1)+(1.02)^(-2)+(1.02)^(-3)+(1.02)^(-4) = 4.4416
19. U = (1.02)^(-4)+(1.02)^(-3)+(1.02)^(-2)+(1.02)^(-1)+1 = 4.4416

2.5 More Arithmetic Examples

The expression

2.6 A Summation Shortcut

For the first task, S = 1+2³+3³+4³ = 1+8+27+64 = 100. For the second task

1. the sum of the cubes of the integers 1 to 5 equals

   S(5) = [ 1 2 5(6)]2 = 152 = 225

    

2. the sum of the cubes of the integers 1 to 15 equals

   S(15) = [( 1 2 )15(16)]2 = [15(8)]2 = [120]2 = 14400

3. the sum of the cubes of the integers 1 to 30 equals

   S(30) = [( 1 2 )30(31)]2 = [15(31)]2 = [465]2 = 216225

In the last calculation, use of the formula

2.7 Algebraic Exercises

Some may be harder than the previous ones. If the algebraic exercises are not understandable, try them later.

1.  

2. The rational expression (ratio of two polynomials) [(1+x+x²+x³)/(x-1)] can not be simplified further.
3. Factor x²+5x+6 Answer: Pairs of factors of 6 are

   6 and 1

   -6 and -1

   2 and 3

   -2 and -3.

   Observe the sum of 2 and 3 is five. Recall

   (x+a)(x+b) = x²+(a+b)x+ab

   But from this pattern,

   (x+2)(x+3) = x²+(2+3)x+(2)(3) = x²+5x+6

   So x²+5x+6 equals the product (x+2)(x+3). Factorization is done.
   
4. The product (x-1)(2x+4)(3-x) = 0 when and only when at least one of the factors is zero.
   
   • The first factor x-1 = 0 when and only when x = 1.
   • The second factor 2x+4 = 0 when and only when 2x = -4 or x = -2.
   • The third factor 3-x = 0 when and only when x = 3.

   Thus (x-1)(2x+4)(3-x) = 0 forces x to have the value 1, -2 or 3. To see each possible value is a solution, observe that x = 1, -2 or 3 implies one of the factors is zero and thus their product (x-1)(2x+4)(3-x) = 0.
   

5. x³-x = x (x² -1) = x(x-1)(x+1) factored.
6. Now

   (x+1)(x+3x²)-[(x+1)x+(x+1)3x²)]  =  [(x+1)(x+3x²)-(x+1)[x+3x²)] = 0

   simplified.
   
7. 13²-5²-(13+5)(13-5) = 0 simplified.
8. Here

   7 -

   ______ Ö(32+42)

   = 7 -

   _____ Ö(9+16)

   = 7 -

   __ Ö25

   = 7-5 = 2

   simplified.
   
9. Here 
   

   æ ç è

   3 7

   ö ÷ ø

   13

   ×

   æ ç è

   (4x2) (32 ×73)

   ö ÷ ø

   5

    

   =

   313 713

   ×

   45x10 310 ×715

   =

   313 ×45 ×x10 713 ×310×715

    

   =

   313-10 ×45 ×x10 713+15

   =

   33 45 x10 728

   
   Note parenthesis and grouping in products do not affect the result of their computation. So they can be put in and pulled out of products at convenience. This can be justified via deductive chains of reasons depending on two properties of multiplication - the associative law and the commutative law.

    

10. Here

    (9x²+3)(4+4x+4x²)][(x-1)(2x+2)-2x²+2)] 

    = [(9x²+3)(4+4x+4x²)][(x²-1)2-2x²+2)]
    = [(9x²+3)(4+4x+4x²)][0]
    = 0

    regardless of what value is used for x. This suggests that when the value of the original expression is requested, no computation is needed. The value will be zero. Of course, some computation would be needed if the simplification had not been done.
    
11. Here

    f(x) =   _____ Ö25-x2

    gives

    f(4) =   _____ Ö25-42 =   _____ Ö25 _ = Ö9 = 3

12. We are given x+y = p and y-x = 1. Observe by substitution, that is, replacement, that

    p+1 = (x+y)+(y-x) = 2y

    Therefore 2y = p+1 and so y = [(p+1)/2]. Now y-x = 1 yields y = 1+x and hence y-1 = x. Therefore,

    x = y - 1 = p+1 2 - 1 = p+1 -2 2 = p-1 2

    Thus in conclusion 

    (x,y) = ( p-1 2 , p+1 2 )

13. Here

    A = [2×32×y3z(-3)t3)(-2)] ×[33x4y(-5)]2 = [2(-2) ×3(-4)×y(-6)z6t(-6)] ×[36x8y(-10)] = 2(-2)3(-4+6) y(-6)z6t(-6)]x8y(-10) = z6 32 x8 22 y16 t6

    expressed with positive powers only (at the cost of introducing some division).

     

14. The equation (x-10)(x-3)=0 has two solutions namely x = 10 and x = 3 but only one satisfies the requirement x > 4, namely x = 10. So x = 10 is the answer. The other solution x = 3 is extra (or extraneous) due to this requirement or request that x > 4. If the latter requirement is replaced by the requirement x > 12, the equation (x-10)(x-3) = will have no solutions satisfying this new requirement. With each of the other requirements x > 2 and x > -99, we have two acceptable solutions, namely +3 and +10, which satisfy both of these requirements.
15. The equation 4 = [1/(x+1)] holds when and only when
    4(x+1) = 1, or x+1 = [1/4]. The latter holds when and only when x = ([1/4])-1 or x = -[3/4]. This solution can be verified or checked as follows: [1/(((-[3/4]+1))] = [1/(([1/4]))] = 1 ¸[1/4] = 1 ×[4/1] = 4
16. From z = 2x+3, t = 3², w-t = 10 and x = 4t+1 and y = y², different substitutions and replacements lead to different ways to find and compute the number z.

     

    • First Way: t = 3² gives x = 4t+1 = 4(3²)+1 and thus z = 2x+3 = 2(4(3²)+1)+3 Thus z = 2(4(9)+1)+3 = 2(36+1)+3 = 2(37)+3 = 74+3 = 77
    • Second Way: z = 2x+3 = 2(4t+1)+3 = 2(4(3²)+1)+3 = 2(4·9+1) + 3 = 
      (37)+3 = 74+3 = 77 as before.
      Third Way: z = 2x+3 = 2(4t+1)+3 = 8t+2+3 = 8t+5 = 8×9+5 = 72+5 = 77

    Each way is shown here to show you that a problem can be solved via slightly different reasoning paths.

Arithmetic - Ages 10+
1. Deciml Place Value - fun
2. Decimals for Tutors
3. Prime Factors - quickly
4. Fractions + Ratios
5. Arith with units - science