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FOUR BASIC ARITHMETIC PROCESSES. ADDITION is the processes of combining numbers of the same kind. Measures like dollars and miles cannot be added because they are two different things.

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## FOUR BASIC ARITHMETIC PROCESSES

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**FOUR BASICARITHMETIC PROCESSES**• ADDITIONis the processes of combining numbers of the same kind. Measures like dollars and miles cannot be added because they are two different things. • SUBTRACTIONis the process of finding the difference between two like numbers by taking one (the subtrahend) from the other (the minuend). The answer is the remainder. • MULTIPLICATIONis the process of adding one number (the multiplicand) to itself as many times as there are units in another number (the multiplier). The answer is called the product. The multiplicand and multiplier are also called the factors of the product.**Arithmetic processes (cont…)**• DIVISIONis the process of finding how many times one number (the divisor) is contained in another number (the dividend). The result is the quotient. • Examples 7 – 3 = 4 5 * 3 = 15 Quotient Multiplier Divisor Minuend Remainder Dividend Multiplicand Subtrahend Product Remainder**Common Fraction**• A FRACTION is a portion of the whole; it signifies division. The top number is called the numerator (same as the dividend), and the bottom number is called the denominator (same as the divisor). The line between the numbers is the division sign. • A PROPER FRACTION has a numerator that is smaller than the denominator, as in ¾ or 2/7. • An IMPROPER FRACTION has a numerator that is equal to or greater than the denominator, as in 6/6 or 9/4. Numerator (or dividend) For convenience, this is often written as Denominator (or divisor)**Common Fraction (cont…)**• An improper fraction can be converted to a whole number (6/6 =1) or a mixed number ( ) by dividing the denominator into the numerator. A mixed number contains both a whole number and a fraction. • A complex fraction has one or more fractions in either the numerator or the denominator, or both. A complex fraction may be converted to a simple fraction by dividing the numerator by the denominator (remember the line in a fraction is a division sign). To divide by a fraction invert the denominator fraction (divisor) and multiply:**Common Fraction (Cont…)**NOTE:The answer to any example or problem should not be left as an improper fraction. It should be converted to a proper fraction or a mixed number, then reduce to lowest terms. 7 1**ADDITION OF FRACTIONS**• To add fractions with the same denominators, add the numerators. • To add fractions with different denominators, select a common denominator into which all the denominators will divide evenly. The smallest number into which all denominators divide evenly is called the least (or lowest) common denominator (L.C.D) or least common multiple (L.C.M).**ADDITION OF FRACTIONS (cont…)**• To find the L.C.M, spread out all the denominators in a short-division box. Divide the denominators by a prime number (a number divisible only by itself and 1) or a multiple of a prime number that goes into at least two of the denominators. When a denominator cannot be divided by a divisor (prime number), it is brought down into the quotient. Repeat the process until no two parts of the quotient are divisible by the same prime number. Then multiply all divisors and all parts fo the quotient together. The result is the L.C.M. • To find the L.C.M. of L.C.M**PROJECT 1 FRACTIONS**Change to the improper fractions. 1. 2. 3. 4. 6. 5. 7. 8.**PROJECT 1 (Cont…)**1. 2. Add the fractions and reduce the answers to the lowest terms. 3. 4. 5. 6. 7. 8.**SUBTRACTION OF FRACTIONS**• To subtract fractions that have the same denominator, simply subtract the numerators. • To subtract fractions with different denominators, select the least common denominator, and proceed with the operation. 6 25**MULTIPLICATION OF FRACTIONS**• There are three steps in multiplying fractions: (a) Multiply the numerators. (b) Multiply the denominators. (c) Reduce to the lowest terms. • “Cancelling” saves time and work. In the above example: • When multiplying by a whole number, think of the whole number as a fraction with a denominator of 1. 5 21 1 3 4 1**DIVISION OF FRACTIONS**• To divide fractions, invert the divisor, as explained earlier, and multiply. An inverted fraction is a fraction with the numerator and denominator interchanged. NOTE: When addition, subtraction, multiplication and division involves a mixed number, convert all mixed numbers to improper fractions. Then proceed with the operations. 1 2**PROJECT 2 FRACTIONS**• Perform the indicated operations. Reduce answers to the lowest terms where necessary. 1. 2. 3. 4. 5. 6. 7. 8. 11. 9. 10. 12. 13.**Project 3 Fractions**Mon. Tues. Wed. Thurs. Fri. Sat. Total. Ali • Compute the total hours worked by each employee and on each day, and verify the totals. Saeed Aina Sofia Jawad Iqrar Azhar Jamila Sana Total.**THE REAL NUMBER SYSTEM**• A set is a collection of elements listed within braces. The set {a, b, c, d, e} consists of five elements, namely a, b, c, d, and e. a set that contains no element is called an empty set (or null set). The symbol { } or the symbol Ø is used to represent the empty set. • There are many different sets of real numbers, two important sets are the natural numbers and the whole numbers. Natural numbers: {1, 2, 3, 4, 5, …. } Whole numbers: {0, 1, 2, 3, 4, 5, … } • To understand the sets of numbers we introduce the real number line. -4 -3 -2 -1 0 1 2 3 4 • The real number line continues indefinitely in both directions.**SETS OF REAL NUMBERS (cont…)**• Another important set of numbers is the integers. Integers: {… , -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, … } Negative integers Positive integers • The set of rational numbers consists of all numbers that can be expressed as the quotient of two integers, with denominator not zero. • Some numbers are not rational. Numbers such as square root of 2, written as , are not rational numbers. Any number that can be represented on the real number line that is not rational is called irrational number.**REAL NUMBERS**• Any numbers that can be represented on the real number line is real number. Real numbers Rational numbers Irrational numbers Integers Non-Integers Negative 0 Positive numbers numbers**PROJECT 4 NUMBER SYSTEM**• List the elements of the set that are: • Natural numbers b) Whole numbers • Integers d) Rational numbers • Irrational numbers f) Real numbers • List the numbers that are • Positive integers b) Whole numbers • Negative Integers d) Integers • Rational numbers f) Irrational numbers • Real numbers**ALGEBRA**• Some Basic Definitions • Variable A characteristic which changes from one individual to the other, e.g. the height of a student in your class, the temperature of different cities in Afghanistan, etc. Variable is denoted by the lower case letters, e.g. x, y, z, etc.**Constant**A characteristic which does not change, e.g. the dimension of your class room, the height of the chair you are setting on, the number of papers in BBA 1st semester exam, etc. A constant is denoted by the alphabets like a, b, c, d, etc.**Expression**It is the combination of operands and operators, e.g. x + y, 4 - 17, etc. Here x, y, 4 and 17 are called operands, while the symbols +, - are called operators. The other operators are *and ÷. The expression x + y is called algebraic expression and 4 - 17 is called the arithmetical expression.**Equation**An expression which involves the sign of equality is called equation. Examples 2x - 5y = 12 x2 + 3x - 5 = 0 x2 + y2 = r2(Equation of circle centered at the origin having radius r) 1/x + 7 = x/3**Operations on expressions**• Addition Add x + 2y and 3x – 7y x + 2y 3x - 7y ____________ 4x – 5y**Continue….**• Subtraction Subtract 9x – 2.5y from 4x + 5y 4x + 5y +9x – 2.5y -5x + 7.5y In addition and subtraction we combine the similar terms of the expressions. _ + _________________**Continue….**Example: Add 7x + 3y + z, -3x + y, and 5x – 4z 7x + 3y + z -3x + y + 0z 5x + 0y – 4z _____________ 9x + 4y – 3z**Distributive property**• For any real numbers a, b, c a (b + c) = ab + ac Examples • 2 (3 + 7) = 2 * 3 + 2 * 7 = 20 • 4 (2x - 4y) = 4 * 2x – 4 * 4y = 8x - 16y • -7 (3p - 5q) = -7(3p)-7(-5q)=-21p+35q • x (2.3y + 1.2z) = 2.3xy + 1.2xz • 3 (x + y + z) = 3x + 3y + 3z**Group Activity**Simplify • 2x2 – 4x +8x2 – 3 (x + 2) – x2 - 2 Sol: 2x2 -4x + 8x2 - 3x – 6 - x2 - 2 =(2x2 + 8x2 – x2) + (-4x - 3x) +(-6 - 2) = 9x2 - 7x - 8 • x2 + 2y - y2 + 3x + 5x2 + 6y2 + 5y • 2 [ 3 + 4 (x - 5) ] - [ 2 - (x - 3) ]**Linear Equation**• An equation of the form ax + b = c (1) where a, b and c, a≠0 are constants (real numbers) is called a linear equation. Examples • x + 4 = 7 (a = 1, b = 4, c = 7) • 2x – 3 = 5.4 (a = 2, b = -3, c = 5.4) • -(x + 3) - (x - 6) = 3x - 4.5 Though this equation is not of the form (1), but it is still a linear equation, since it can be rewritten as: 5x - 4.5 = 3 (a = 5, b = -4.5, c = 3)**Solution of a linear equation**By the solution of a linear equation we mean to find that value of the unknown ‘x’ which satisfies the equation. Ex • x = 3, is the solution of x – 2 = 1 • x = -6, is the solution of 2x + 1 = x - 5 • x = 11/2 is the solution of 3 (x + 2) = 5 (x-1) • y = 9 is not a solution of 2y – 3 = 5**How to solve a linear equation**We can solve a linear equation by • Substitution method • Graphical method • Algebraic method (explanation by examples) The substitution method In this method we make a guess for the solution of the linear equation, we then put the guessed value in the equation and check whether the guess is correct or not?**Continue….**Examples Let us try to solve the equation ‘4x+2=4’, An accurate guess for the solution is “x = ½”, we substitute this value in the given equation 4 * ½ + 2 = 4 2 + 2 = 4 4 = 4 ⁄ ⁄**Continue….**The substitution method is generally a time consuming method, i.e. sometimes it takes very long time to find an accurate guess. The graphical method In this method we sketch the graph of the linear equation, and then find the point where the graph cuts the x-axis, such a point is called the ‘x-intercept ’. This method is also not a very good method.**The algebraic method**In this method we use the algebraic operations to find the solution. The method is described in the following examples. Examples(Exercise Set 2.3, p#103) Solve the following equations and check your solution. • 2x=6 Sol: Given the equation 2x=6 Dividing both sides by 2 —x= 2 3 ⁄ 6 ⁄ — 2 ⁄ 2 ⁄**Continue….**Thus the solution is x=3. Q16. x/3 = -2 Sol: Multiplying both sides by 3, we get x = -6, which is the desired solution. Q19. -32x = -96 Sol: Dividing both sides by -32, we get x=3 as the required solution. Q24. -x = 9 Sol: Multiplying both sides by -1, we get the solution x = -9.**Continue….**Q25. -2 = -y Sol: Multiplying both sides by -1, we get y = 2 as solution. Q33. 13x = 10 Sol: Dividing both sides by 13, we get x = 10/13 the required solution. Q52. -2x = 3/5 Sol: Dividing both sides by -2, we get x = -3/10, which is the required equation.**Continue….**Q66. When solving the equation 3x = 5. Would you divide both sides of the equation by 3 or by 5? Explain. Q67. When solving the equation -2x = 5. Would you add 2 to both sides of the equation or divide both sides of the equation by -2? Explain. Q69. Consider the equation 4x = 3/5. Would it be easier to solve this equation by dividing both sides of the equation by 4 or by multiplying both sides of the equation by ¼, the reciprocal of 4? Explain your answer. Find the solution to the problem.**Exercise Set 2.4 (9-65), p# 110**Solve each equation. 11. -2x-5=7 Sol: Given -2x-5=7 Adding 5 to both sides -2x-5+5=7+5 -2x=12 Dividing both sides by -2 -2x/2=12/-2 => x=-6 The desired solution.**Continue….**• -4.2 = 2x + 1.6 Sol: We have -4.2 = 2x + 1.6 Subtracting 1.6 from both sides -4.2 - 1.6 = 2x + 1.6 – 1.6 -5.8 = 2x Now dividing both sides by 2 -5.8/2 = 2x/2 => -2.9 = x (Using calculator) Or x = -2.9 as desired.**Continue….**Q33. x + 0.05x = 21 Sol: 1.05x = 21 Dividing both sides by 1.05 (the coefficient of x) x = 21/1.05 => x = 20 as required. Q36. -2.3 = -1.4 + 0.6x Sol: Adding 1.4 to both sides 0.6x = -2.3 + 1.4 0.6x = -0.9 Dividing through out by 0.6 x = -0.9/0.6 => x = -1.5**Continue….**Q38. 32.76 = 2.45x – 8.75x Sol: Given 32.76 = 2.45x – 8.75x => 32.76 = -6.30x Dividing both sides by -6.30 x = -32.76/6.30 x = -5.2 the answer.**Continue….**Q46. -2(x+4) + 5 = 1 Sol: Using distributive property -2x – 8 + 5 = 1 or -2x – 3 = 1 Adding 3 to both sides -2x -3 + 3 = 1 + 3 i.e. -2x = 4 Now dividing both sides by -2 x = -4/2 or x = -2 as desired.**Continue….**Q58. 0.1(2.4x + 5) = 1.7 Sol: given the equation 0.1(2.4x + 5) = 1.7 Using distributive property 0.24x + 0.5 = 1.7 Subtracting 0.5 from both sides 0.24x = 1.7 – 0.5 • 0.24x = 1.2 Finally dividing both sides by 1.2 x = 1.2/0.24 x = 5 the required answer.**Continue….**Q65. 5.76 – 4.24x – 1.9x = 27.864 Sol: Given that 5.76 – 4.24x – 1.9x = 27.864 • 5.76 – 6.14x = 27.864 Now subtracting 5.76 from both sides 5.76 - 5.76 – 6.14x = 27.864 – 5.76 -6.14x = 22.104 Finally dividing both sides by -6.14 -6.14x/-6.14 = -22.104/6.14 x = 3.454 Which is the desired result.**Challenging problems**Solve each equation. • 3(x-2) – (x+5) – 2(3-2x) = 18 • -6 = -(x-5) – 3(5+2x) – 4(2x-4) • 4[3 – 2(x+4)] – (x + 3) = 13**Solving linear equations with the variable on both sides of**the equation Hints: • Use the distributive property to remove the parentheses. • Combine like terms on the same side of the equal sign. • Rewrite the equation with all terms containing the variable on one side of the equation and all the terms not containing the variables on the other side of the equation. • Isolate the variable using the multiplication property, this gives the solution. • Check your answer, by putting the value of ‘x’ in the original equation.**Exercise Set 2.5(9-54), p#118/119**Solve each equation. 9. 4x = 3x + 5 Sol: Given the equation 4x = 3x + 5 Combining the terms involving variables on one side if the equation 4x – 3x = 5 x = 5 Which is the required answer.**Continue….**• 15 – 3x = 4x – 2x Sol: Given that 15 – 3x = 4x – 2x Shifting 3x to the right side of the equation. 15 = 4x – 2x + 3x • 15 = 5x Isolating ‘x’ using the multiplication property 15/5 = x or x = 3 as required.**Continue….**• x – 25 = 12x + 9 + 3x Sol: We have x – 25 = 12x + 9 + 3x Shifting ‘x’ to the right & 9 to left of the equation -25 – 9 = 12x + 3x – x -34 = 14x Using multiplicative property -34/14 = x or x = 2.428 as required.**Continue….**• 4r = 10 – 2(r-4) Sol: Given that 4r = 10 – 2(r-4) Using distributive property 4r = 10 – 2r + 8 Shifting 2r to the left of the equation 4r + 2r = 10 + 8 6r = 18 Dividing both sides by 6 r = 18/6 or r = 3 as required.

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